Explanation:
I won't give you the answers. It won't help you in the long run. Instead, let's look at it logically:
You have two of something and you're gonna turn it into something else. That thing doesn't just disappear, it has to stay there. You can't act as if there's only one of them in the end result, that doesn't make sense.
Balancing chemical equations is about finding the ratios at which things occur. In the first example,
N2 + F2 -> NF3
You have two N on the left, one on the right. Better double the one on the right so you have equal amounts.
N2 + F2 -> 2NF3
Now, since there are two Ns on each side, there must be 6 Fs on the right (2×3). You only have 2 on the left side, so the next question is: how do I get from 2 to 6?
Simple! Multiply by 3:
N2 + 3F2 -> 2NF3
Let's check left side vs right side to see if we're right:
Left:
2 N
6 F (3×2)
Right:
2 N
6 F (2×3)
Left side = Right side, so that's the right answer.
Don't let questions like these stress you out. Just balance out the numbers so that there's the same amount of stuff on each side.
A helium balloon initially at room temperature is dunked into a bucket of liquid nitrogen (T = 77 K). Which of the following will occur? A) Volume increases B) Volume decreases C) The number of moles increases D) The number of moles decreases E) Temperature increases 2126970 Chem101 lu.co Introductory Che... Question 2 of 25 If two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have: A) Higher pressure B) Lower presure C) A greater universal gas constant D) A smaller universal gas constant Q Introductory Che... Question 3 of 25 Which one of the following changes would cause the volume of a gas to double, assuming moles were held constant? A) Doubling pressure while keeping temperature constant. B) Doubling absolute temperature while keeping pressure constant. C) Doubling absolute temperature while cutting the pressure in half. D) Doubling the absolute temperature while doubling the pressure.
Answer:
Question 1
B) Volume decreases
Question 2
A) Higher pressure
Question 3
A and B
Explanation:
Question 1
B) Volume decreases
Charles' Law: The Temperature-Volume Charles Law. It states that at constant pressure, the volume of a given amount of gas is directly proportional to temperature in Kelvin. As the temperature goes down, the volume also goes down, and vice-versa.
ogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the volume and amount (moles) of
Question 2
A) Higher pressure
Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the amount (moles) and the volume of the gas are directly proportional at constant temperature and pressure
Increasing the number of moles increases the volume to maintain the same volume of the container with less moles the gas has to be compressed, increasing its pressure
Boyle's law is a gas law, that states that at constant temperature, the pressure and volume of a gas are inversely proportional. increasing the volume , decreases the pressure and vice versa.
Question 3
A and B
A) Doubling pressure while keeping temperature constant. B) Doubling absolute temperature while keeping pressure constant.
The volume of a given mass of gas decreases as pressure increases but increases as absolute temperature increases.
In the first question, a helium balloon initially at room temperature is dunked into a bucket of liquid nitrogen, the volume of the immediately decreases because volume and absolute temperature are directly proportional according to Charles law.
In the second question, two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have higher pressure because the pressure of a gas is directly proportional to the number of molecules of a gas present.
In the third question, the change that would cause the volume of a gas to double, assuming moles were held constant is doubling absolute temperature while keeping pressure constant. This follows from Charles law where the volume of a given mass of gas is directly proportional to its volume at constant pressure.
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A hydrogen atom in an excited state emits a photon of frequency ν = 3.084 x 1015 s-1. If the electron returns to the ground state, in which level was it before the photon was emitted?
The Rydberg formula can calculate the wavelengths of light emitted by hydrogen atoms. However, with some modifications, it can also be used to calculate the wavelengths of light emitted by atoms of other elements. Hence, the initial energy level is n = 3
Rydberg equation:
1/λ = f/c = R(1/m² - 1/n²)
where c is the speed of light, f is frequency, λ is the wavelength, R is Rydberg constant, and n and m are the quantum numbers of the energy levels.
The initial energy level has quantum number n.
The final energy level is a ground state with the quantum number m = 1.
2.924 x 1015/2.998 x 108 = 1.097 x 107 x (1/12 - 1/n²)
(1/12 - 1/n²) = 0.8891
1/n² = 0.1109
=> n²
= 9
=> n = 3
Thus, the initial energy level is n = 3
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The electron in the hydrogen atom was in the 9th energy level (n = 9) before emitting the photon.
Explanation:When a hydrogen atom transitions from an excited state to the ground state, it emits a photon with a certain frequency. In this case, the frequency is given as ν = 3.084 x 10^15 s^-1. We can use the equation ν = E/h, where E is the energy of the photon and h is Planck's constant, to find the energy of the emitted photon. Once we have the energy, we can determine the energy level the electron was in before the photon was emitted.
Using the energy-frequency relationship and Planck's constant, we have:
E = hν = (6.626 x 10^-34 J s)(3.084 x 10^15 s^-1) ≈ 2.041 x 10^-18 J
From this energy, we need to find the corresponding energy level. The energy levels of hydrogen are given by the formula: E = -13.6 eV/n^2, where E is the energy of the level and n is the principal quantum number. Rearranging the formula, we have:
n^2 = -13.6 eV/E ≈ -13.6 eV/(2.041 x 10^-18 J) ≈ -6.662 x 10^17
Taking the square root of both sides, we find:
n ≈ -8.159 x 10^8
Since n must be a positive integer, we can conclude that the electron was in the 9th level (n = 9) before emitting the photon.
Fe₃O₄ reacts with CO according to the equation
Fe₃O₄ + CO --> Fe + CO₂ (unbalanced)
If 478.9 g Fe₃O₄ is reacted with excess CO, what mass of CO₂ will be produced assuming 100% yield?
1. 91.02 g CO₂
2. 182.05 g CO₂
3. 364.1 g CO₂
4. none of these
5. 273.1 g CO₂
Answer:
Option 3.
364.1 g of CO₂ are produced by the reaction
Explanation:
Let's verify the balance equation for this reaction:
Fe₃O₄ + 4 CO → 3 Fe + 4CO₂
Let's convert the mass of Fe₃O₄ to moles (mass / molar mass)
478.9 g / 231.55 g/mol = 2.07 moles
Ratio is 1:4, so let's make a rule of three to determine the moles of CO₂ produced and then its mass.
1 mol of Fe₃O₄ is needed to produce 4 moles of CO₂
Then, 2.07 moles of Fe₃O₄ will produce (2.07 .4) /1 = 8.27 moles of CO₂.
Molar mass . moles = mass
44 g/mol . 8.27 mol = 364 g of CO₂
If your hot coffee sits on your desktop and loses 50 kJ of energy duringcooling, what is the quantitative change in enthalpy of the coffee?
Answer : The change in enthalpy of the coffee is negative.
Explanation :
Endothermic reaction : It is defined as the chemical reaction in which the energy is absorbed from the surrounding.
In the endothermic reaction, the energy of reactant are less than the energy of product.
Exothermic reaction : It is defined as the chemical reaction in which the energy is released into the surrounding.
In the exothermic reaction, the energy of reactant are more than the energy of product.
Enthalpy change : It is the difference between the energy of product and the reactant. It is represented as [tex]\Delta H[/tex].
When the system gains energy in the form of heat then the change in enthalpy is positive.When the system loses energy in the form of heat then the change in enthalpy is negative.As per question, if hot coffee sits on your desktop and loses 50 kJ of energy during cooling, the change in enthalpy of the coffee is negative.
Hence, change in enthalpy of the coffee is negative.
A polymer is a large molecule that forms when smaller molecules known as monomers bond covalently in a repeating pattern. There are many biological polymers such as nucleic acids, proteins, and starches. What are the monomer units that make up starches?
Answer:
glucose
Explanation:
Starch -
Starch is generated by the plants , and is a granular , white , organic compound .
The general formula of the starch is (C₆H₁₀O₅)ₙ , the n shows it to be a polymer , and it is composed of small monomers of glucose , that are linked at the alpha 1 , 4 linkages.
Starch is is colorless , and tasteless powder.
The most simplest form of starch is the linear polymer amylose , and the branched one amylopectin.
Propane (C₃H₈) burns in oxygen to produce carbon dioxide and water. What is the correct form of the chemical equation that describes this reaction?
Answer:
The answer to your question is below
Explanation:
Propane is a hydrocarbon that has only single bonds in its structure so it is an alkane.
Propane formula = C₃H₈
Oxygen formula = O₂
Carbon dioxide formula = CO₂
Water formula = H₂O
Reaction:
C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂O
Reactants Elements Products
3 C 3
8 H 8
10 O 10
What type of chemical bond joins a functional group to the carbon skeleton of a large molecule?
Answer:
Covalent bond
Explanation:
Functional groups can be defined as a group of atoms or ions responsible for the properties particular to a certain group of organic compounds. What we are saying in essence is that it is the functional groups that decides the behavior of the organic compound in question.
Covalent linkages are the mechanism through which these functional groups are linked to the carbon skeleton of the compound to which they belong. Covalent bonding is a type of chemical bonding which in there is sharing of electrons between atoms
Covalent bond
Aliphatic hydrocarbons are divided into three main groups according to the types of bonds they contain: alkanes, alkenes, and alkynes. Alkanes have only single bonds, alkenes contain a carbon-carbon double bond, and alkynes contain a carbon-carbon triple bond.A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. Carbon skeletons are the backbones of organic molecules. They are composed of carbon-carbon atoms that form chains to make an organic compound.Learn more:
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A 31.4−g sample of ethylene glycol, a car radiator coolant, loses 607 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5°C? (c of ethylene glycol = 2.42 J/g·K
The initial temperature of the ethylene glycol is equal to 40.5°C.
What is the specific heat capacity?The specific heat capacity is defined as the amount of heat required to raise the temperature of one unit of material by one degree Celsius. The specific heat capacity of the material depends upon the nature of the material.
The mathematical expression is used to calculate the specific heat is equal to:
[tex]Q = mC \triangle T[/tex]
Given, the mass of the sample of ethylene glycol, m = 31.4 g
The final temperature of the sample, T₂ = 32.5°C = 305.5 K
The specific heat capacity of ethylene glycol, C = 2.42 J/g·K
The heat lost from the sample, Q = - 607 J
The initial temperature of the sample:
- 607 = 31.4 × 2.42 × (305.5 - T₁)
305.5 - T₁ = - 7.988
T₁ = 313.49 K
T₁ = 40.5°C
Therefore, the initial temperature of the ethylene glycol is 40.5°C.
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The initial temperature of ethylene glycol can be found by rearranging the equation q = mcΔT and dividing the heat lost by the product of the mass and specific heat capacity, considering the known final temperature and that the ethylene glycol is cooling down.
Explanation:To calculate the initial temperature of ethylene glycol when it loses heat, we can use the equation that relates heat loss to temperature change, q = mcΔT, where q is the heat lost, m is the mass, c is the specific heat capacity of the substance, and ΔT is the change in temperature. Given that ethylene glycol loses 607 J of heat (q), has a mass of 31.4g (m), and a specific heat capacity of 2.42 J/g·°C (c), and the final temperature is 32.5°C, we can rearrange the equation to solve for the initial temperature, T_initial.
The heat lost is negative because the ethylene glycol is cooling down, so we have: -607 J = 31.4g × 2.42 J/g·K × (32.5°C - T_initial). Solving for T_initial we find that the initial temperature of the ethylene glycol is higher than the final temperature of 32.5°C by an amount resultant from dividing the heat lost by the product of the mass and the specific heat capacity.
How many moles of potassium chloride are in 28 grams of KCl?
A.) .265 mole KCl
B.) .856 mole KCl
C.) .376 mole of KCl (think it's this but idk)
D.) 1.2 mole KCl
Answer:
0.3758moles
Explanation:
moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles
Answer:
We have 0.376 moles in 28.0 grams of KCl
Option C is correct.
Explanation:
Step 1: Data given
Mass KCl = 28.00 grams
Molar mass KCl = 74.55 g/mol
Step 2: Calculate moles KCl
Moles KCl = mass KCl / moalr mass KCl
Moles KCl = 28.0 grams / 74.55 g/mol
Moles KCl = 0.376 moles KCl
We have 0.376 moles in 28.0 grams of KCl
Option C is correct.
How did Dalton describe the relationship between atoms and elements? An element is made up of one kind of atom. Atoms are made up of combinations of elements. Elements are made up of atoms arranged in whole-number ratios. Different kinds of atoms chemically combine to form elements.
Dalton described the relationship between atoms and elements in terms of the composition of elements and the combination of atoms to form compounds.
Explanation:Dalton described the relationship between atoms and elements in the following way:
An element is always made up of one kind of atom. Each element is composed of atoms that have the same number of protons.Atoms are made up of combinations of elements. Atoms combine with other atoms of different elements to form compounds.Elements are made up of atoms arranged in whole-number ratios. The atoms of different elements combine in fixed ratios to form compounds.Different kinds of atoms always chemically combine to form elements. Atoms of different elements can combine in various ways to form new substances.Learn more about Dalton's atomic theory here:https://brainly.com/question/13157325
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Dalton described elements as being made up of a single, unique type of atom. Atoms of different elements differ in properties. Atoms combine in fixed, whole-number ratios to form compounds.
John Dalton, the English chemist, proposed a theory that is fundamental to understanding chemical elements and atoms. According to Dalton's atomic theory:
Elements are made of extremely small particles called atoms.Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and other properties.Atoms cannot be subdivided, created, or destroyed by chemical means.Atoms of different elements combine in fixed, small, whole-number ratios to form compounds.Hence, Dalton described the relationship between atoms and elements by stating that each element is composed of one kind of atom unique to that element, which can combine to form more complex structures.
Identify the true statements about colloids.
a.) Emulsions are a type of colloid
b.) The particals of a colloid are larger than the particles of a solution
c.) The particles of a colloid will settle over time
d.) Many colloids scatter light (tyndal effect)
Answer : The true statements are:
(a) Emulsions are a type of colloid
(b) The particles of a colloid are larger than the particles of a solution
(d) Many colloids scatter light (tyndal effect)
Explanation :
Colloid : It is defined as the solution in which the one substance is insoluble in another solution that means the insoluble substance rotating in the solution.
The particles of a colloid are larger than the particles of a solution.
Colloid do not separate on standing.
Cannot be separated by filtration.
Scatter light (Tyndall effect).
For example :
Milk is considered as a colloid because various substances (fats, proteins etc..) are present in milk which are suspended in a solution.
Suspension : It is a heterogeneous mixture in which some of the particles are settle down in the mixture on standing or over time.
The particles in a suspension are far larger than those of a solution.
Emulsion : It is a mixture of two or more liquids that are normally immiscible.
Emulsions are a type of colloid.
True or False? The fatal 1930 incident in Muese Valley, Belgium, was associated with leakage of toxic chemicals from a former dump site.
Answer:
FALSE
Explanation:
The incident of Muese Valley occured in 1930 due to air pollution.
Muese Valley lies along the river Muese which is situated Huy and Liege, Belgium. This region was crowded with industries including steel manufacturers, glass manufacturers, explosives plants, zinc smelter, etc.
The increase number of industries and population lead to the sources of pollution. Also increase in burning of domestic coal increased pollution surrounding the area.
Air pollution became so severe at this region that people have severe respiratory problems. Residents suffered from vomiting, retrosternal pain, coughing fits and several experienced nausea. There were fog and smog all over and many people died.
Hence the answer is FALSE.
The solubility of barium fluoride in water is 1.32 grams per liter. If a barium fluoride solution had a concentration of 1.32 grams per liter, it would be said to be:________.
A barium fluoride solution with a concentration of 1.32 grams per liter, equivalent to its solubility, would be considered saturated. This means that no more barium fluoride can be dissolved in the water at the same temperature.
Explanation:The solubility of barium fluoride in water is 1.32 grams per liter. Therefore, if a barium fluoride solution had a concentration of 1.32 grams per liter, it would be referred to as being saturated. A saturated solution is a solution in which no more solute can be dissolved in the solvent at a given temperature. In this case, the maximum amount of barium fluoride that can be dissolved in water at the given temperature has been reached.
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Which of the following is NOT possible? a. compressing 10 liters of oxygen gas into a 1-liter volume b. compressing 2 liters of water into a 1-liter volume c. filling a balloon using helium gas from a pressurized tank d. allowing 5 liters of compressed air to expand to a volume of 100 liters
Answer:b
Explanation:
Answer: B. Compressing 2 liters of water into a 1 liter volume.
Explanation: Liquids have little compressibility as compared to gases. Their molecules are quite near each other compared to gases which far apart. When liquids are compressed up to their capacity it will begin to resist. But gases have higher compressibility rate than liquids.
Which extraction procedure will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride?
A) Add 0.1M NaOH (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the ether layer by evaporating the solvent.
B) Add 0.1 M HCl (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from ether layer by evaporating the solvent.
C) Add 0.1M NaOH (aq) to quench unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl (aq).
D) Add 0.1M HCl (aq) to quench the unreacted anhydride. Then add diethyl ether and separate the layers. The amide can be obtained from the aq. layer by neutralizing the NaOH (aq).
Option C, which involves adding 0.1M NaOH to quench unreacted anhydride, then adding diethyl ether and separating the layers, is the extraction procedure that will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride.
Explanation:The extraction procedure that will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride is option C. First, you would add 0.1M NaOH (aq) to quench unreacted anhydride. Then, you would add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl (aq).
Barium-142 is used as a GI radiocontrast agent. After 1.25 hours, 9.25 μg remains in the patient. Determine the original dose given the half-life of Ba-142 is 10.6 minutes.
Answer:
The answer to your question is the original dose of Ba-142 was 1184 μg
Explanation:
Data
Total time = 1.25 hours
The Final amount of Ba = 9.25 μg
The Half-life of Ba = 10.6 minutes
Process
1.- Convert total time to minutes
1 h ----------------- 60 min
1.25 h ------------ x
x = 75 min
2.- Draw a table of this process
Final amount of Ba Time
9.25 75 min
18.5 64.4 min
37 53.8 min
74 43.2 min
148 32.6 min
296 22 min
592 11.4 min
1184 0.8 min
Find the total number of atoms in a sample of cocaine hydrochloride, C17H22CINO4 of mass 23.0 mg
Answer:
The answer to your question is 4.07 x 10²² atoms
Explanation:
Process
1.- Get the molecular weight of Cocaine hydrochloride
C = 12 x 17 = 204 g
H = 22 x 1 = 22 g
Cl = 36 x 1 = 36 g
N = 14 x 1 = 14 g
O = 16 x 4 = 64 g
Molecular mass = 340 g
2.- Calculate the moles of the mass given
340 g ------------------- 1 mol
23 g ------------------- x
x = (23 x 1) / 340
x = 0.068 moles of Cocaine
3.- Calculate the atoms
1 mol -------------------- 6 .023 x 10 ²³ atoms
0.068 moles ------------- x
x = (0.068 x 6.023 x 10²³) / 1
x = 4.07 x 10²² atoms
When the equation, ___O2 + ___C 2H 6 → ___CO2 + ___H2O is balanced, the coefficient of O2 is: When the equation, ___O2 + ___C 2H 6 ___CO2 + ___H2O is balanced, the coefficient of O2 is:
Final answer:
To balance the equation C₂H₆ + O₂ → CO₂ + H₂O, we end up with a balanced chemical equation of 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, where the coefficient of O₂ is 7.
Explanation:
The question concerns the balancing of a chemical equation involving ethane (C₂H₆) and oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). Balancing chemical equations requires ensuring that the number of atoms of each element is the same on both sides of the equation. Starting with the unbalanced equation C₂H₆ + O₂ → CO₂ + H₂O, we first balance the carbons (C) and hydrogens (H) by adjusting the coefficients of the products. For example, the equation can be balanced by placing a coefficient of 3 before H₂O and 2 before CO₂, resulting in C₂H₆ + O₂ → 2CO₂ + 3H₂O. However, this leads to seven oxygen atoms on the product side. We need an even number of oxygen atoms to balance them with the O₂ reactant, so we use a fractional coefficient of 3.5 (which is 7/2) in front of O₂.
To remove the fractional coefficient, we can multiply all coefficients by 2, resulting in the balanced chemical equation: . Thus, the coefficient of O₂ when the equation is balanced is 7.
For hot vacuum filtration, the filter paper should be completely dry when pouring the hot solution into the Buchner funnel to filter.
A) True
B) False
Answer:
False.
Explanation:
The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).
The ratio of sizes between the ionic radii of anions and cations in a cell has no influence on the manner of packing for that cell. a. True b. False
Answer:
The answer is B. False
Explanation:
The ratio of sizes between the ionic radii of cations and anions in a cell influences the manner of packing for that cell thereby predicting the possible cation/anion coordination number in any compound and establishing the structure of ionic solids.
Many classic experiments have given us indirect evidence of the nature of the atom. Which of the experiments listed below did not give the results described?
A. The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.
B. The Rutherford experiment was useful in determining the nuclear charge on the atom.
C. Millikan's oil-drop experiment showed that the charge on any particle was a simple multiple of the charge on the electron.
D. The electric discharge tube proved that electrons have a negative charge.
E. All of the above experiments gave the results described.
Answer:
Option A, The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.
Explanation:
Thomson's plum pudding model:
Plum pudding model was proposed by J.J Thomson. In Thomson's model, atoms are proposed as sea of positively charge in which electrons are distributed through out.
Result of Rutherford experiment:
As per Rutherford's experiment:
Most of the space inside the atom is empty.
Positively charge of the atom are concentrated in the centre of the atom known as nucleus.
Electrons are present outside the nucleus and revolve around it.
As it is clear that, result of Rutherford experiment did not supported the Thomson model.
The Rutherford experiment did not prove the Thomson 'plum-pudding' model of the atom to be essentially correct. Instead, it disproved this model by showing that atoms have a small, dense, positively charged nucleus.
Explanation:The statement 'The Rutherford experiment proved the Thomson 'plum-pudding' model of the atom to be essentially correct.' (option A) is not accurate. In fact, the Rutherford experiment disproved the Thomson 'plum-pudding' model of the atom. Rutherford's experiment revealed that atoms have a small, dense, positively charged nucleus, contradicting Thomson's model which proposed that positive charge was spread evenly throughout the atom. All other listed experiments (options B, C, and D) did indeed provide the results described.
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A 1.50 g sample of a compound containing only C, H and O was burned in an attempt to determine its simplest formula. 3.00 g of CO2 and 1.23 g of H2O were collected. Find the simplest formula for the compound.
Explanation:
The mass of carbon and hydrogen is calculated from the mass of their oxides ([tex]CO_{2}[/tex] and [tex]H_{2}O[/tex]) as follows.
Mass of C = [tex]3.00 g CO_{2} \times \frac{12.01 g C}{44.0 g CO_{2}}[/tex]
= 0.818 g C
Mass of H = [tex]1.23 g CO_{2} \times \frac{2.016 g H}{18.02 g H_{2}O}[/tex]
= 0.137 g H
So, the mass of C + mass of H is as follows.
0.818 g + 0.137 g = 0.955 g
This mass is actually less than the mass of sample. And, the missing mass must be caused by O. Hence, the mass of O will be calculated as follows.
Mass of O = 1.50 g - 0.955 g
= 0.545 g
Now, we convert masses to moles and find their moles and ratio as follows.
Element Mass/g Moles Ratio [tex]\times 2[/tex] Integers
C 0.818 0.068 1 2 2
H 0.137 0.068 1 2 2
O 0.545 0.0340 0.5 1 1
Thus, we can conclude that simplest formula for the given compound is [tex]C_{2}H_{2}O[/tex].
Answer:
C2H4O (ut quest)
Which of the following precautions is important when using a Bunsen burner or Meker burner?a. Set up your work space so that wires and cables cannot accidentally make contact with the flame or hot glassware and meltb. Set up your work space with the burner in a secure location away from the edge of the benchc. Set up the work space so that flammable materials are away from the burnerd. Never leave an open flame unattendede. Remember that any glassware heated by the burner will look the same when hot as cold, and it will take a while to coolf. Always tie back hair
Answer:
Never leave an open flame unattended
Explanation: if an open flame is left unattended it can cause a fire outbreak so we have to watch it at all times to prevent the fire outbreak
Calculate the number of moles of potassium permanganate (KMnO4) corresponding to 230.8 g of the substance.
Answer:
The answer to your question is 1.46 moles of KMnO₄
Explanation:
Data
number of moles = ?
mass = 230.8 g
molecular mass of KMnO₄ = 39 + 55 + (16 x 4) = 158 g
Process
1.- Use proportions and cross multiplication to answer this problem.
158 g of KMnO₄ ---------------- 1 mol
230.8 g of KMnO₄ ----------- x
x = (230.8 x 1) / 158
2.- Simplifying
x = 230.8 / 158
3.- Result
x = 1.46 moles
Methane gas is produced from the reaction of solid carbon and hydrogen gas: C(s)+2H2(g)→CH4(g) . How many liters of hydrogen gas at standard temperature and pressure (STP) are required to produce 40 liters of methane?
Answer:
80 liters
Explanation:
At STP, 1 mole of ideal gas has a volume of 22.4 liters.
Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.
40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂
it is difficult to evaluate the toxicity of substances what factors can vary how harmful a chemical is dose of exposure age and genetic makeup will affect response to a toxin.
Answer:
The explanation is given below
Explanation:
Toxicity is a measure of how harmful a substance is on a living organism (causing illness, injury or even death). There are several factors that must be considered when evaluating toxicity:
- Dose of exposure: the amount of substance to which the organism has been exposed to;
- Frequency of exposure: how many times (and for how long) the individual has been exposed to the substance;
- Age and health condition of the individual exposed;
- Genetic makeup: the genetic backgroun of an organism will determine its response and degree of sensitivity to a given substance.
Moreover, there are five main characteristics of substances that determine its toxicity: Solubility (hidrophilic or lipophilic substances), persistence (for how long the substance remains the same and cause the same damage), bioaccumulation (when the substance is progresivelly incorporated in tissues), biomagnification (when the amount of substance rise along trophic levels) and other chemical interactions (different interactions with other chemicals can increase the degree of damage) .
NEED HELP A 25.0 mL solution of nitric acid (HNO3) with an unknown concentration is titrated with 12.5 mL of a 1.0x10-4 M solution of lithium hydroxide (LiOH). Calculate the molar concentration of the HNO3.
3.80x10-10 M HNO3
5.00x10-5 M HNO3
2.80x10-3 M HNO3
2.30x10-5 M HNO3
2.50x10-5 M HNO3
To calculate the molar concentration of HNO3 in the solution, use the balanced chemical equation to determine the mole ratio and then calculate the moles of HNO3 used. Finally, divide the moles of HNO3 by the volume of the solution to get its molar concentration.
Explanation:To calculate the molar concentration of HNO3, we first need to use the balanced chemical equation to determine the mole ratio between HNO3 and LiOH. The balanced equation is 2HNO3 + 2LiOH → 2LiNO3 + H2O. From the equation, we can see that 2 moles of HNO3 react with 2 moles of LiOH. Since we know the molar concentration of LiOH and the volume used, we can calculate the moles of LiOH used: (12.5 mL)(1.0x10^-4 M) = 0.00125 moles of LiOH. Since the mole ratio is 1:2, the moles of HNO3 used would be half of that, which is 0.000625 moles of HNO3. Now, we can use the volume of the HNO3 solution to calculate its molar concentration:
Molar concentration of HNO3 = (0.000625 moles) / (25.0 mL) = 0.025 M HNO3
Final answer:
The molar concentration of HNO3 is 5.00x10-5 M.
Explanation:
To calculate the molar concentration of HNO3, we first need to determine the number of moles of LiOH used in the titration. The molarity of the LiOH solution is given as 1.0x10-4 M, and the volume used is 12.5 mL (0.0125 L).
moles of LiOH = Molarity x Volume = (1.0x10-4 M) x (0.0125 L) = 1.25x10-6 mol
Since the balanced chemical equation shows a 1:1 ratio between LiOH and HNO3, the number of moles of HNO3 is also 1.25x10-6 mol.
To find the molar concentration of HNO3, we use the volume of the HNO3 solution, which is 25.0 mL (0.025 L).
Molarity of HNO3 = moles of HNO3 / Volume = (1.25x10-6 mol) / (0.025 L) = 5.00x10-5 M HNO3
Combustion reactions are exothermic. The heat of reaction for the combustion of cyclopropane, C3H6, is 499.8 kcal/mol. What is the heat of combustion for cyclopropane in kcal/gram?
To find the heat of combustion for cyclopropane in kcal/gram, divide the heat of reaction (499.8 kcal/mol) by the molar mass of cyclopropane (42.08 g/mol).
Explanation:The heat of combustion for cyclopropane in kcal/gram can be calculated using its molar mass and the given heat of reaction. The molar mass of cyclopropane, C3H6, is 42.08 g/mol. With a heat of reaction of 499.8 kcal/mol, we can calculate the heat of combustion per gram by dividing the heat of reaction by the molar mass:
Heat of combustion (kcal/g) = Heat of reaction (kcal/mol) / Molar mass (g/mol)
Thus,
Heat of combustion (kcal/g) = 499.8 kcal/mol / 42.08 g/mol
Once you perform the division, you'll have the heat of combustion for cyclopropane in kcal/gram.
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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooling system holds 4.90 gal, what is the boiling point of the solution?
Answer:
108.25 ºC
Explanation:
The boiling point elevation for a given solute in water is given by the expression:
ΔTb = i Kbm
where ΔTb is the boiling point elevation
i is the van´t Hoff factor
Kb the boiling constant which for water is 0.512 ºC/molal
m is the molality of the solution
The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.
Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.
Our equation is then:
ΔTb = Kbm
So lets calculate the molality and then ΔTb:
m = moles ethylene glycol / Kg solvent
Converting gal to L
4.90 g x 3.785 L/gal = 18.55 L
in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.
We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³
9.27 L x 1000 cm³ / L = 9273.25 cm³
mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g
mol ethylene glycol = 9273.25 g/ M.W ethylene glycol
= 9273.25 g / 62.07 g/mol =149.4mol
molality solution = 149.4 mol / 9.27 Kg H₂O = 16.12 m
( density of water 1 kg/L )
Finally we can calculate ΔTb:
ΔTb = Kbm = 0.512 ºC/molal x 16.12 molal = 8.25 ºC
boiling point = 100 º C +8.25 ºC = 108.25 ºC
( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )
What would be the pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container found using the van der Waals equation? For Cl₂, a = 6.49 L²･atm/mol² and b = 0.0652 L/mol.
Answer:
57.478atm
Explanation:
T = 400k
n = 5mol
v = 2.00
a = [tex]6.49L^{2}[/tex]
b = 0.0652L/mol
R = 0.08206
Formula
P = [tex]\frac{RT}{(\frac{v}{n} )-b} - \frac{a}{(\frac{v}{n} )^{2} }[/tex]
[tex]P = \frac{0.08206 * 400}{(\frac{2.00}{5.00} )-0.0652} - \frac{6.49}{(\frac{2.00}{5.00} )^{2} }[/tex]
[tex]P = \frac{32.824}{0.3348} - \frac{6.49}{0.16}[/tex]
[tex]P = 98.041 - 40.563[/tex]
[tex]P = 57.478atm[/tex]
The van der Waals equation, which accounts for the actual volume of gas molecules and the attractions between them, can be used to find the pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container.
Explanation:The pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container can be found using the van der Waals equation. The van der Waals equation is a modification of the ideal gas law that takes into account the volume of the gas molecules themselves (represented by the parameter 'b') and the attractions between gas molecules (represented by 'a').
In this problem, the values of 'a' and 'b' for Cl₂ are given to be 6.49 L²･atm/mol² and 0.0652 L/mol respectively. The van der Waals equation is given by:
[P + a(n/V)²] (V/n - b) = RT
where P is the pressure we are trying to find, n is the number of moles of gas, V is the volume of the container, R is the gas constant (0.0821 L･atm/mol･K), and T is the temperature in kelvins. Substituting given values into this equation and simplifying it, we get the value of the pressure. Please note, this type of problem often requires some algebraic manipulation.
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