Answer:
1. By Pressure factor: if we double the pressure volume become half of its original
2. 2.14 L
3. 2.15 L
Explanation:
part 1
Data Given:
volume of container change
temperature of remain constant
The pressure doubles
Solution:
This problem can be explained by Boyle's Law that at constant temperature pressure and volume has an inverse relation with each other.
So the volume change due to change in Pressure.
P1V1 = P2V2
if we consider conditions at STP, as follows
initial volume V1 = 22.42 L
and
initial pressure P1 = 1 atm
if the pressure doubles then
final pressure P2 = 2 atm
Put values in Boyle's law equation
(1 atm) (22.42L) = (2 atm) (V2)
Rearrange the above equation to find V2
V2 = (1 atm) (22.42L) / 2 atm
V2 = 11.12 L
So it is clear from calculation if we double the pressure volume become half of its original. So its the pressure due to volume become effected and decrease by its increase.
_____________
Part 2
Data Given:
Initial temperature T1= 250 K
final Temperature T2= 350 K
initial volume V1 = ?
final volume V2 = 3.0 L
Solution:
This problem will be solved by Charles' Law equation at constant pressure
V1 / T1 = V2 / T2 . . . . . . . . (1)
put values in above equation
V1 / 250 K = 3.0 L / 350 K
Rearrange the above equation to calculate V1
V1 = (3.0 L / 350 K) x 250 K
V1 = (0.0086 L . K) x 250 K
V1 = 2.14 L
So the initial volume = 2.14 L
_________________
part 3
Data Given:
Initial temperature T1= 20 ºC
Convert Temperature from ºC to Kelvin
T = ºC + 273
T = 20 + 273 = 293 K
final Temperature T2= -5.00 ºC
Convert Temperature from ºC to Kelvin
T = ºC + 273
T = - 5.00 + 273 = 268 K
initial volume V1 = 2.35 L
final volume V2 = ?
Solution:
This problem will be solved by Charles' Law equation at constant pressure
V1 / T1 = V2 / T2 . . . . . . . . (1)
put values in above equation
2.35 L / 293 K = V2 / 268 K
Rearrange the above equation to calculate V1
V2 = (2.35 L / 293 K) x 268 K
V2 = (0.008 L . K) x 268 K
V2 = 2.15 L
So the volume at -5.00ºC = 2.15 L
Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculate the standard enthalpy of formation of acetone.
The question is incomplete, here is the complete question:
Using this information together with the standard enthalpies of formation of [tex]O_2(g)[/tex], [tex]CO_2(g)[/tex], and [tex]H_2O(l)[/tex] from Appendix C. Calculate the standard enthalpy of formation of acetone.
Complete combustion of 1 mol of acetone [tex](C_3H_6O)[/tex] liberates 1790 kJ:
[tex]C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l);\Delta H^o=-1790kJ[/tex]
Answer: The enthalpy of the formation of [tex]CO_2(g)[/tex] is coming out to be -247.9 kJ/mol
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-1790kJ[/tex]
Putting values in above equation, we get:
[tex]-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol[/tex]
Hence, the enthalpy of the formation of [tex]C_3H_6O(g)[/tex] is coming out to be -247.9 kJ/mol.
Below are 4 Lewis Structures for CO2. Only ONE of them is correct. Identify the correct one, and explain why it is correct. For each of the three incorrect ones, explain why they are incorrect.
Answer:
B.
Explanation:
Carbon = 4
Oxygen = 6x 2
Total e- = 16 - 4 ( the number of initial single bonds)
Final number of available e- = 12
The total number of available electrons is 12 and is distributed to oxygen atoms having 6 e- each. To make C octet each oxygen atoms donates its pair of electrons to form another bond. So we have a double bond on each side making C octet and two O atoms.
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.
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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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) .
How many moles of \ce{AgCl}AgClA, g, C, l will be produced from 60.0 \text{ g}60.0 g60, point, 0, start text, space, g, end text of \cu{AgNO3}AgNO3 , assuming \ce{NaCl}NaCl, a, C, l is available in excess
Answer:
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
Explanation:
How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.
Step 1: Data given
Mass of AgNO3 = 60.0 grams
Molar mass AgNO3 = 169.87 g/mol
NaCl is in excess, so AgNO3 is the limiting reactant
Step 2: The balanced equation
AgNO3 + NaCl → AgCl + NaNO3
Step 3: Calculate moles AgNO3
Moles AgNO3 = mass AgNO3 / molar mass AgNO3
Moles AgNO3 = 60.0 grams / 169.87 g/mol
Moles AgNO3 = 0.353 moles
Step 4: Calculate moles AgCl
For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
Final answer:
To find the moles of AgCl produced from a reaction, calculate using stoichiometry and molar mass.
Explanation:
To determine the moles of AgCl produced:
Step 1: Data given
Mass of AgNO3 = 60.0 grams
Molar mass AgNO3 = 169.87 g/mol
NaCl is in excess, so AgNO3 is the limiting reactant
Step 2: The balanced equation
AgNO3 + NaCl → AgCl + NaNO3
Step 3: Calculate moles AgNO3
Moles AgNO3 = mass AgNO3 / molar mass AgNO3
Moles AgNO3 = 60.0 grams / 169.87 g/mol
Moles AgNO3 = 0.353 moles
Step 4: Calculate moles AgCl
For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
Will the pH at the equivalence point of the 50 mL titration of 0.10 M HCl be the same, more, or less than the pH at the equivalence point of the 50 mL titration of 0.10 M acetic acid?
It will be less
Explanation:When equivalence point is achieved the acid or base is completely neutralized, but its salt (conjugate acid or base) can alter the pH of the solution. In the comparison of two different acidic or basic species, their conjugate is evaluated. If the base dissociation coefficient (Kb) of one conjugate base is greater than other, then the pH change due to it will be more basic.
The neutralization of HCl can be given as
HCl + OH⁻ -------------- > H₂O + Cl⁻
Here Cl⁻ is the remaining ion at the equivalence point, and it is the conjugate base of HCl. It has a Kb value of 1.0 X 10⁻²⁰ (that is why it is not considered basic).
The neutralization of acetic acid is given as
CH₃COOH + OH⁻ -------------- > H₂O + CH₃COO⁻
Here CH₃COO⁻ is the remaining ion at the equivalence point, and it is the conjugate base of acetic acid. Its Kb value is 5.6 X 10⁻¹⁰, which is higher than the Kb value of Cl⁻. As the amount of HCl and acetic acid is the same, so the solution containing chloride ions will have a lower pH than the solution containing acetate ions.
The pH at the equivalence point of the 50 mL titration of 0.10 M HCl will be less than the pH at the equivalence point of the 50 mL titration of 0.10 M acetic acid.
Hydrochloric acid (HCl) is a strong acid, which means it dissociates completely in water to form [tex]H_3O+[/tex] ions and Cl- ions. At the equivalence point of the titration of HCl with NaOH, all the [tex]H_3O+[/tex] ions from HCl have been neutralized by OH- ions from NaOH, resulting in a solution of its conjugate base, Cl-.
Therefore, the pH at the equivalence point is determined by the autoionization of water, which gives a pH of 7.0 at 25°C.
On the other hand, acetic acid [tex](CH_3COOH)[/tex] is a weak acid. It does not dissociate completely in water, and at the equivalence point of its titration with NaOH, the solution contains the conjugate base of acetic acid, the acetate ion [tex](CH_3COO-)[/tex].
The acetate ion is a moderately strong base, and it can react with water to form OH- ions, which increases the pH of the solution above 7.0. Therefore, the pH at the equivalence point of the titration of acetic acid is basic, typically around 8.7 for a 0.10 M acetic acid solution.
Carbon, pictured here, is considered versatile when it comes to bonding with other atoms and making important compounds. Which feature of carbon most contributes to this versatility?
Answer:
Catenation
Explanation:
Catenation is one the most important bonding characteristics of carbon. It is the ability of carbon atoms to join with themselves to form chains of different chain length.
This is the principal reason why there are a lot of organic compounds. The chain forming capability of carbon, otherwise known as catenation is responsible for this
The versatility of carbon in bonding with other atoms and forming important compounds is primarily attributed to its tetravalent nature, stable covalent bonds with other carbon atoms, and the formation of double and triple bonds.
Explanation:The most significant feature of carbon that contributes to its versatility in bonding with other atoms and forming important compounds is its tetravalent nature. Carbon has four valence electrons in its outermost energy level, allowing it to form up to four covalent bonds with other atoms. This ability to form multiple bonds makes carbon the building block of a wide variety of organic molecules.
Another important feature of carbon is its ability to form stable covalent bonds with other carbon atoms, resulting in the formation of long chains, branched structures, and rings. This property gives rise to the diversity and complexity of organic compounds found in nature.
Furthermore, carbon is capable of forming double and triple bonds with other atoms, such as oxygen and nitrogen. These multiple bonds contribute to the unique properties and reactivity of specific carbon compounds, such as aldehydes, ketones, and aromatic compounds.
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The balanced equation below shows the products that are formed when pentane (C₅H₁₂) is combusted. [tex]C_5H_{12} + 8O_2 \rightarrow 10CO_2 + 6H_2O[/tex]What is the mole ratio of oxygen to pentane? a. 1:6 b. 6:8 c. 8:1 d. 10:8
Answer:
Option c → 8:1
Explanation:
This is the reaction:
C₅H₁₂ + 8O₂ → 10CO₂ + 6H₂O
1 mol of pentane needs 8 moles of oxygen to be combusted and this combustion produces 10 mol of carbon dioxide and 6 moles of water.
To determine the ratio, look the stoichiometry.
For every 8 moles of oxygen, I need 1 mole of pentane gas.
All samples of a given compound, regardless of their source or how they were prepared, have the same proportions of their constituent elements. Which law does this refer to?
Answer:
Law of constant compositions
Explanation:
The law of constant composition is the law that guides this principle. It is one of the laws of chemical combinations. The others being law of multiple proportion, law of reciprocal proportion and law of conservation of matter.
What the law is trying to say regardless of where or when a particular substance is obtained, it contains exactly the same proportions and constituent of elements. This means its identity remains constant regardless
The statement refers to the Law of Definite Proportions or Proust's Law. This law states that a chemical compound always contains its component elements in fixed ratios and is not dependent on its source or method of preparation.
Explanation:The statement that all samples of a given compound, regardless of their source or how they were prepared, have the same proportions of their constituent elements, refers to the Law of Definite Proportions, also known as Proust's Law. Essentially, this law states that a chemical compound always contains the same elements in the same proportions by mass. For example, water (H2O) is always made up of two parts hydrogen to sixteen parts oxygen by mass, regardless of the source of the water.
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What is the concentration in parts per million? Assume the density of the solution is the same as that for pure water (1.00 g/mL). ppm
The question is incomplete, here is the complete question:
What is the concentration, in parts per million, of a solution prepared by dissolving 0.00040 mol HCl in 2.2 L [tex]H_2O[/tex] ? Assume that the volume of the solution does not change when the HCl is added. Assume the density of the solution is the same as that for pure water (1.00 g/mL)
Answer: The concentration of HCl in the solution is 6.64 ppm
Explanation:
To calculate the number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of HCl = 0.00040 moles
Molar mass of HCl = 36.5 g/mol
Putting values in above equation, we get:
[tex]0.00040mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.00040mol\times 36.5g/mol)=0.0146g[/tex]
To calculate the mass of water, we use the equation:[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of water = 1.00 g/mL
Volume of water = 2.2 L = 2200 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]1.00g/mL=\frac{\text{Mass of water}}{2200mL}\\\\\text{Mass of water}=(1.00g/mL\times 2200mL)=2200g[/tex]
ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.
To calculate the ppm of HCl in the solution, we use the equation:
[tex]\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6[/tex]
Both the masses are in grams.
We are given:
Mass of HCl = 0.0146 g
Mass of solution = 2200 g
Putting values in above equation, we get:
[tex]\text{ppm of HCl in solution}=\frac{0.0146g}{2200g}\times 10^6\\\\\text{ppm of HCl in solution}=6.64[/tex]
Hence, the concentration of HCl in the solution is 6.64 ppm
Answer:
6.6
Explanation:
I NEED HELP. PLEASE HELP ME!!!!
How many rubidium nitrate molecules (RbNO3) are needed to balance the equation shown below?
10 Rb + _____ —> 6 Rb2O + N2
A. 2 RbNO3
B. 6 RbNO3
C. 10 RbNO3
D. 12 RbNO3
Answer:
We have to add 2RbNO3 (option A)
10 Rb +2RbNO3 → 6 Rb2O + N2
Explanation:
Step 1: The equation:
10 Rb + _____ —> 6 Rb2O + N2
Step 2: Balancing the equation
On the right side we have 6x2 = 12 Rb atoms.
On the left side we have 10x Rb
This means we need to add 2x Rb on the left side.
On the right side we have 2x N, On the left side 0x N.
This means we need to add 2x N on the left side.
On the right side we jave 6x O, on the left side we have 0x O.
This means we need to add 6x O on the left side.
We add this by adding RbNO3
This means we have to add 2x RbNO3 (option A)
10 Rb +2RbNO3 → 6 Rb2O + N2
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 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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The flame in a torch used to cut metal is produced by burning acetylene (C2H2) in pure oxygen. Assuming the combustion of 1 mole of acetylene releases 1251 kJ of heat, what mass of acetylene is needed to cut through a piece of steel if the process requires 20.7 × 104 kJ of heat?
Answer:
We need 4.31 kg of acetylene
Explanation:
Step 1: Data given
Combustion of 1 mol acetylene releases 1251 kJ of heat
Molar mass of acetylene = 26.04 g/mol
Step 2: Calculate moles of acetylene
1251kJ /mol * x moles = 20.7 * 10^4 kJ
x moles = 20.7 * 10^4 kJ / 1251 kJ/mol
x moles = 165.47 moles
Step 3: Calculate mass of acetylene
Mass acetylene = moles acetylene * molar mass acetylene
Mass acetylene = 165.47 moles * 26.04 g/mol
Mass acetylene = 4308.8 grams = 4.31 kg of acetylene
We need 4.31 kg of acetylene
The mass of acetylene needed is 4290 g of acetylene.
We can see that 1 mole of acetylene produces 1251 kJ of heat, so we can obtain the number of moles of acetylene required from stoichiometry as follows;
1 mole of acetylene produces 1251 kJ of heat
x moles of acetylene produces 20.7 × 10^4 kJ of heat
x = 1 mole × 20.7 × 10^4 kJ / 1251 kJ
x = 165 moles
Now;
Molar mass of acetylene = 26 g/mol
Mass of acetylene = 165 moles × 26 g/mol
= 4290 g of acetylene
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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.
If 4.05 g of KNO₃ reacts with sufficient sulfur (S₈) and carbon (C), how much P-V work will the gases do against an external pressure of 1.00 atm given the densities of nitrogen and carbon dioxide are 1.165 g/L and 1.830 g/L, respectively, at 20°C?
Explanation:
Chemical reaction equation for the given reaction is as follows.
[tex]2KNO_{3}(s) + \frac{1}{8}S_{8}(s) + 3C(s) \rightarrow K_{2}S(s) + N_{2}(g) 3CO_{2}(g)[/tex]
Therefore, we will calculate the number of moles of [tex]KNO_{3}[/tex] as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{4.05 g}{101.1 g/mol}[/tex]
= 0.04 mol
Number of moles of nitrogen gas formed is calculated as follows.
No. of moles = [tex]\frac{1}{2} \times \text{no. of moles of KNO_{3}}[/tex]
= [tex]\frac{1}{2} \times 0.04 mol[/tex]
= 0.02 mol
Mass of [tex]N_{2}[/tex] gas formed will be calculated as follows.
No. of moles × Molar mass of [tex]N_{2}[/tex]
= [tex]0.02 mol \times 28.0 g/mol[/tex]
= 0.56 g
Now, the number of moles of [tex]CO_{2}[/tex] formed is as follows.
= [tex]\frac{3}{2} \times 0.04[/tex]
= 0.06 mol
Hence, mass of [tex]CO_{2}(g)[/tex] formed will be as follows.
[tex]0.04 mol \times 44 g/mol[/tex]
= 1.76 g
Volume of [tex]N_{2}(g)[/tex] is calculated as follows.
Volume = [tex]\frac{mass}{density}[/tex]
= [tex]\frac{0.56 g}{1.165 g/L}[/tex]
= 0.48 L
And, volume of [tex]CO_{2}[/tex] is calculated as follows.
Volume = [tex]\frac{mass}{density}[/tex]
= [tex]\frac{1.76 g}{1.830 g/L}[/tex]
= 0.96 L
Let us assume that the volume of solids are negligible. Therefore, total volume will be as follows.
[tex]\Delta V[/tex] = (0.48 L + 0.96 L)
= 1.44 L
Relation between work, pressure and volume is as follows.
w = -[tex]P_{ext} \times \Delta V[/tex]
= -[tex]1.00 atm \times 1.44 L[/tex]
= -1.44 atm L
As 1 tm L = 101.3 J. So, convert 1.44 atm L into joules as follows.
[tex]1.44 \times 101.3 J[/tex]
= 145.87 J
Thus, we can conclude that the given gases will do 145.87 J of work.
An incomplete chemical equation is given here. Which set of numbers, placed in the blanks as the coefficients, will balance the equation shown? __HCl + __Na2CO3 → __H2CO3 + __NaCl 1, 2, 1, 2 2, 1, 2, 1 2, 1, 1, 2 2, 2, 1, 1
Answer:
The answer to your question is the coefficients are 2, 1, 1, 2
Explanation:
Chemical Reaction
HCl + Na₂CO₃ ⇒ H₂CO₃ + NaCl
Reactants Elements Products
1 Cl 1
2 Na 1
1 C 1
1 H 2
3 O 3
This reaction is unbalanced
2 HCl + Na₂CO₃ ⇒ H₂CO₃ + 2 NaCl
Reactants Elements Products
1 Cl 1
2 Na 2
2 C 2
2 H 2
3 O 3
Now, the reaction is balanced.
Answer:
The coefficients are: 2,1,1,2 (Option 3)
Explanation:
Step 1: Unbalanced equation
HCl + Na2CO3 → H2CO3 + NaCl
Step 2 : Balancing the equation
On the right side we have 2x H (in H2CO3), on the left side we have 1x H (in HCl). To balance the amount of H, we have to multiply HCl, on the left side, by 2.
2 HCl + Na2CO3 → H2CO3 + NaCl
On the left side we have 2x Na (in Na2CO3), on the right side, we have 1x Na (in NaCl). To balance the amount of Na, we have to multiply NaCl, on the right side, by 2. Now the equation is balanced.
2 HCl + Na2CO3 → H2CO3 + 2NaCl
The coefficients are: 2,1,1,2 (Option 3)
Provide the most likely dominant bonding mechanism in the following solid compounds:a.CaO b. InAs c. Al2O3 d.Bronze
Answer:
CaO- ionic
InAs-covalent
Al2O3-ionic
Bronze- metallic
Explanation:
CaO and Al2O3 are mostly ionic even though the posses a little covalent character but ionic bonding is the main bonding scheme. Bronze is an alloy of two metals hence it contains a metallic bond. InAs has an electro negativity difference of 0.4 between the atoms so it is a polar covalent bond.
You are performing a titration by adding a Strong Base to a solution of Weak Acid. At the equivalence point, will your solution be acidic, basic, or neutral?
Answer:
Basic
Explanation:
The titration of a strong base with a weak acid would always result in a solution whose pH is greater than 7,i.e a basic solution. For this reason its advisable to use an indicator whose color change will occur in that pH range.
For other cases, the titration of a stong acid and strong will result in a neutral solution and the titration of a weak base versus a strong acid would result in an acidic solution.
Final answer:
At the equivalence point in a titration between a weak acid and a strong base, the solution will be basic due to the formation of a conjugate base from the weak acid which then increases the OH- concentration.
Explanation:
When performing a titration involving a weak acid and a strong base, the resulting solution at the equivalence point will typically be basic. This is due to the fact that the weak acid has a less complete dissociation in water compared to a strong base. At the equivalence point, a weak acid has been completely neutralized by the strong base, which means it has converted to its conjugate base by losing a proton. The conjugate base may then react with water to form OH- (hydroxide ions), making the solution basic.
Indicators are chosen according to the expected pH at the equivalence point. When titrating a weak acid with a strong base, an indicator that changes color in the basic pH range would be suitable, such as phenolphthalein. It is importantly noted that in a titration of a strong acid with a strong base, the equivalence point occurs at a neutral pH of 7.00; however, this is not the case for a titration involving a weak acid.
An example of an element is ____. a) chicken noodle soup b) powerade c) air inside a balloon d) lead pipe e) baking soda (NaHCO3)
Answer:
d) lead pipe
Explanation:
An element -
It is a substance , which have exactly same atoms , is referred to as an element.
Element is the most simplest form of any substance and hence , can not be further broken down into simpler form by any chemical reaction.
All the elements known are all placed in a synchronized manner in the periodic table.
Hence, from the question,
Lead pipe is an example of an element , as it is composed only one lead elements , rest other like , baking soda , air , noodle soap , all are the that are mainly composed of the mixture of many atoms.
Final answer:
The correct answer is a lead pipe because lead is an element found on the periodic table, distinguishing it from mixtures and compounds such as chicken noodle soup, Powerade, air, and baking soda. (Option d)
Explanation:
The question "An example of an element is ____." seeks to identify a substance that cannot be broken down into chemically simpler components. Among the given choices, lead pipe is the correct answer. This is because lead, represented by the symbol Pb, is a basic chemical element found on the periodic table. In contrast, chicken noodle soup, Powerade, the air inside a balloon, and baking soda ([tex]NaHCO_3[/tex]) are not elements. Chicken noodle soup and Powerade are considered mixtures, the air inside a balloon is a mixture of different gases, and baking soda is a compound consisting of sodium, hydrogen, carbon, and oxygen.
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).
Can anyone help with 5 through 8? Please :(
Answer:
The answer to your question is below
Explanation:
5) Fe₂O₃(s) + 3H₂O ⇒ 2Fe(OH)₃ (ac) Synthesis reaction
6) 2C₄H₁₀(g) + 13O₂(g) ⇒ 8CO₂ (g) + 10H₂O Combustion reaction
7) 2NO₂ (g) ⇒ 2O₂ (g) + N₂ (g) Decomposition reaction
8) H₃P (g) + 2O₂ (g) ⇒ PO (g) + 3H₂O Single replacement reaction
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.
Nitrogen gas and hydrogen gas undergo synthesis to form ammonia gas. If 1.40 g of nitrogen are used in the reaction, how many grams of hydrogen will be needed?
Answer:
The answer to your question is 0.3 g of H₂
Explanation:
Data
N₂ (g) = 1.4 g
H₂ (g) = ?
Balanced Reaction
N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)
Process
1.- Calculate the atomic mass of nitrogen and hydrogen.
N₂ = 14 x 2 = 28 g
H₂ = 1 x 6 = 6 g
2.- Use proportions and cross multiplication to solve it
28 g of N₂ ------------------- 6 g of H₂
1.4 g of N₂ -------------------- x
x = (1.4 x 6) / 28
x = 0.3 g of Hydrogen
What is the conecntration of Fe3 and the concentration of No3- present in the solution that result when 30.0 ml of 1.75M Fe(No3)3 are mixed with 45.0 ml of 1.00M Hcl?
Answer:
[tex][Fe^{+3}]=0.700 M[/tex]
[tex][NO_{3}^{-}]=2.10 M[/tex]
Explanation:
Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:
[tex]C_{1}. V_{1}= C_{2}.V_{2}[/tex]
Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.
Solving for [Fe],
[tex][Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }[/tex]
[tex][Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }[/tex]
[tex][Fe^{+3}]=0.700 M[/tex]
For [NO₃⁻],
There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.
[tex][NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }[/tex]
[tex][NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }[/tex]
[tex][NO_{3}^{-}]=2.10 M[/tex]
What mass of Al2O3 (alumina, a finely divided white powder that is produced as billows of white smoke) is formed in the reaction of 2800 kg of aluminum?
Answer:
5290.7 kg of Al₂O₃ are produced in the reaction of 2800 kg of Al
Explanation:
The reaction to produce alumina is:
4 Al + 3O₂ → 2 Al₂O₃
So, let's convert the mass of Al to moles (mass / molar mass)
2800 kg = 2800000 g (Then, 2.8 ×10⁶ g)
2.8 ×10⁶ g / 26.98 g/mol = 103780.5 moles of Al
Ratio is 4:2, so with the moles I have, I will produce the half of moles of alumina.
103780.5 mol / 2 = 51890.25 moles of Al₂O₃
Now we can convert these moles to mass ( molar mass . mol )
101.96 g/mol . 51890.25 mol = 5290729.8 g
5290729.8 g → 5290.7 kg
The mass of alumina Al₂O₃ formed from the reaction between Aluminum and oxygen is 5290.73 kg
The equation for the reaction that leads to the formation of alumina(Al2O3) can be expressed as:
[tex]\mathbf{2Al + \dfrac{3}{2}O_2 \to Al_2O_3}[/tex]
Given that:
the mass of aluminium Al = (2800 kg = 2800 × 1000) grams
= 2800000 grams
The molar mass of Aluminium Al = 26.98 g/mol
Number of moles of Al = 2800000 g/ 26.98 g/mol
Number of moles of Al = 103780.5782 moles
Since the ratio of the aluminium in the reactant and product is 2:1
There will be half moles of alumina;
∴
Moles of alumina will be:
[tex]\mathbf{\dfrac{1}{2} \times 103780.5782 \ moles}[/tex]
= 51890.2891 moles of Al₂O₃
Mass = number of moles × molar mass
The molar mass of Al₂O₃ = 101.96 g/mol
Mass of Al₂O₃= 51890.2891 g × 101.96 g/mol
Mass of Al₂O₃ = 5290733.877 grams
Mass of Al₂O₃ = (5290733.877/1000 ) kg
Mass of Al₂O₃ = 5290.73 kg
Learn more about the mass of a substance here:
https://brainly.com/question/5833529?referrer=searchResults
Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction: C(s)+2H2O(g)→2H2(g)+CO2(g) Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol.
The question is incomplete , complete question is:
Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:
[tex]C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?[/tex]
Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.
Answer:
The ΔH of the reaction is -626 kJ/mol.
Explanation:
[tex]C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?[/tex]
We are given with:
[tex]\Delta H_{H-O}=459 kJ/mol[/tex]
[tex]\Delta H_{H-H}=432 kJ/mol[/tex]
[tex]\Delta H_{C=O}=799 kJ/mol[/tex]
ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)
[tex]\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})[/tex]
[tex]=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol[/tex]
[tex]\Delta H=-626 kJ/mol[/tex]
The ΔH of the reaction is -626 kJ/mol.
The enthalpy for the formation of carbon dioxide has been -626 kJ/mol.
[tex]\Delta[/tex]H has been the energy required for the breaking of the bonds in the dissociation reaction, and the energy for the formation of bond.
Given, [tex]\Delta[/tex]H H-O bond = 459 kJ/mol
[tex]\Delta[/tex]H for H-H bond = 432 kJ/mol
[tex]\Delta[/tex]H for C=O bond = 799 kJ/mol
[tex]\Delta[/tex]H = Energy for breaking bond - energy for bond formation
[tex]\Delta[/tex]H = (4 times H-O bond) - (2 time H-H bond + 2 times C=O bond formation)
[tex]\Delta[/tex]H = (4 [tex]\times[/tex] 459 kJ/mol) - (2 [tex]\times[/tex] 432 kJ/mol + 2 [tex]\times[/tex] 799 kJ/mol)
[tex]\Delta[/tex]H = 1,836 - (1,598 + 864) kJ/mol
[tex]\Delta[/tex]H = 1,836 - 2,462 kJ/mol
[tex]\Delta[/tex]H = -626 kJ/mol
The enthalpy for the formation of carbon dioxide has been -626 kJ/mol.
For more information about the production of hydrogen, refer to the link:
https://brainly.com/question/19790865
Dry ice sublimes into carbon dioxide gas. If the proper conditions are maintained and the system is closed, the dry ice and the carbon dioxide gas will eventually.
Dry ice sublimes into carbon dioxide gas. If the proper conditions are maintained and the system is closed, the dry ice and the carbon dioxide gas will eventually.
1 )become the same phase
2) reach equilibrium
3)same properties and composition throughout
4) both become same phase and reach equilibrium
Answer:
Under the the proper conditions maintained over the closed system , the dry ice and the carbon dioxide gas will eventually reach equilibrium.
Explanation:
The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.
Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.
For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equals to rate of the backward reaction.
Under the the proper conditions maintained over the closed system , the dry ice and the carbon dioxide gas will eventually reach equilibrium.
[tex]CO_2(s)\rightleftharpoons CO_2(g)[/tex]
Amount of carbon dioxide changing from solid to gas will be equal to amount of carbon dioxide changing from gas to solid.
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.