At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 55 ∘C ?

Answers

Answer 1

Answer:

Taking into account the Charles's law, the same amount of gas at the same pressure and 55 ∘C would occupy a volume of 26.91 L.

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure.

This law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$(V)/(T)=k$

Studying an initial state 1 and a final state 2, it is satisfied:

$(V1)/(T1)=(V2)/(T2)$

In this case, you know:

V1= 22.4 L
T1= 0 C= 273 K
V2= ?
T2= 55 C= 328 K

Replacing:

$(22.4 L)/(273 K)=(V2)/(328 K)$

Solving:

$V2=328 Kx(22.4 L)/(273 K)$

V2= 26.91 L

Finally, the same amount of gas at the same pressure and 55 ∘C would occupy a volume of 26.91 L.

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Answer 2

Answer:

Answer:26.9L

Explanation: this is Charles' law which states that the volume of a gas is directly proportional to the absolute temperature at contant pressure. The expression is V1/T1 = V2/T2

Making V2 the subject of the formula we have

V2 = V1 xT2/T1

= 22.4 x 328/273

= 26.9L

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The decomposition of a substance x is a second order reaction. it takes 0.5 h for a solution that is initially 1.00 m to decrease to 0.50 m. how much time will it take for the concentration of a solution of x to decrease from 1.00 m to 0.25 m?

A student obtained the following data for the rearrangement of cyclopropane to propene at 500 °C. (CH2)3(g)CH3CH=CH2(g) [(CH2)3], M 0.128 6.40×10-2 3.20×10-2 1.60×10-2 time, min 0 14.4 28.8 43.2 (1) What is the half-life for the reaction starting at t=0 min? min What is the half-life for the reaction starting at t=14.4 min? min Does the half-life increase, decrease or remain constant as the reaction proceeds? _________ (2) Is the reaction zero, first, or second order? _______ (3) Based on these data, what is the rate constant for the reaction? min-1

Answers

Explanation:

CH2)3(g)CH3CH=CH2(g) [(CH2)3], M time, min

0.128 0

6.40×10-2 14.4

3.20×10-2 28.8

1.60×10-2 43.2

(1) What is the half-life for the reaction starting at t=0 min? min

Half life is the amount of time required for a substance to decay by half of it's initial concentration.

Starting form 0, the initial concentration = 0.128

After 14.4 mins, the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=0min

What is the half-life for the reaction starting at t=14.4 min?

Starting form 14.4min, the initial concentration = 6.40×10-2

After 14.4 mins (28.8 - 14.4), the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=14.4min

Does the half-life increase, decrease or remain constant as the reaction proceeds?

The half life is a constant factor, hence it remains constant as the reaction proceeds.

(2) Is the reaction zero, first, or second order?

Because the half life is independent of the concentration, it is a first order reaction.

In a zero order reaction, the half life Decreases as the reaction progresses; as concentration decreases.

In a first order reaction, the half life Increases with decreasing concentration.

(3) Based on these data, what is the rate constant for the reaction? min-1

The realtionship between the half life and rate onstant is;

k = 0.693 / half life

k = 0.693 / 14.4

k = 0.048125 min-1

What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?

Answers

Answer:

K.E. = 5.4362 × 10⁻¹⁹ J

Explanation:

The expression for Bohr velocity is:

$v=(Ze^2)/(2 \epsilon_0* n* h)$

Applying values for hydrogen atom,

Z = 1

Mass of the electron ( $m_e$ ) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

$\epsilon_0$ = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

$v=\frac {2.185* 10^6}{n}\ m/s$

Given, n = 2

So,

$v=\frac {2.185* 10^6}{2}\ m/s$

$v=1.0925* 10^6\ m/s$

Kinetic energy is:

$K.E.=\frac {1}{2}* mv^2$

So,

$K.E.=\frac {1}{2}* 9.1093* 10^(-31)* ({1.0925* 10^6})^2$

K.E. = 5.4362 × 10⁻¹⁹ J

Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Answers

The number of oxygen atoms in 19.3 g of sodium sulfate (Na₂SO₄) is 3.27×10²³ atoms

We'll begin by calculating the number of mole in 19.3 g of sodium sulfate (Na₂SO₄).

Mass of Na₂SO₄ = 19.3 g

Molar mass of Na₂SO₄ = (23×2) + 32 +(16×4)

= 46 + 32 + 64

= 142 g/mol

Mole of Na₂SO₄ =?

Mole = mass / molar mass

Mole of Na₂SO₄ = 19.3 / 142

Mole of Na₂SO₄ = 0.136 mole

Recall:

1 mole of Na₂SO₄ contains 4 moles of O.

Therefore,

0.136 mole of Na₂SO₄ will contain = 0.136 × 4 = 0.544 mole of O

Finally, we shall determine the number of atoms in 0.544 mole of O.

From Avogadro's hypothesis,

1 mole of O = 6.02×10²³ atoms

Therefore,

0.544 mole of O = 0.544 × 6.02×10²³

0.544 mole of O = 3.27×10²³ atoms

Thus, 19.3 g of sodium sulfate (Na₂SO₄) contains 3.27×10²³ atoms of oxygen.

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Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05 $(g)/(mol)$ .

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · $(1 mol Na^2SO^4)/(142.05g Na^2SO^4)$ · $(4 mol O)/(1 mol Na^2SO^4)$ · $(6.022x10^2^3)/(1 mol O)$

After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.

Determine total H for bonds broken and formed, the overall change in H, and the final answer with units. Is it ENDOthermic or EXOthermic? Reaction:
2H2(g) +O2(g) -> 2H2O^(g)
Given:
H-H 436
O+O 499
O-H 463

Answers

E(Bonds broken) = 1371 kJ/mol reaction
E(Bonds formed) = 1852 kJ/mol reaction

ΔH = -481 kJ/mol.
The reaction is exothermic.

Explanation

2 H-H + O=O → 2 H-O-H

There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb

E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
ΔH(Breaking bonds) = +1371kJ/mol

Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release

E(Bonds formed) = 2 × 2 × 463 = 1852kJ/mol reaction.
ΔH(Forming bonds) = - 1852 kJ/mol

Heat of the reaction:

$\Delta H_{\text{rxn}} = \Delta H(\text{Breaking bonds}) + \Delta H(\text{Forming bonds})\n\phantom{ \Delta H_{\text{rxn}}} = +1371 + (-1852) \n\phantom{ \Delta H_{\text{rxn}}} = -481 \; \text{kJ} / \text{mol}$

$\Delta H_{\text{rxn}}$ is negative. As a result, the reaction is exothermic.

Station 9: Iodine ______________ g I2 Perform the calculations in the space between each lettered problem. Enter the answer with correct significant figures and units on the line provided.

Answers

Answer:

a) Moles of iodine molecules = 0.0748 moles of Iodine molecules

b) Moles of iodine atoms = 0.150 moles of iodine atoms

c) Number of iodine atoms = 9.03 * 10²² atoms

d) Number of iodine molecules = 4.50 * 10²² molecules

Note: The complete question is found in the attachment below.

Explanation:

a. Number of moles of iodine molecules in 19.0 g of I₂

Molar mass of iodine molecule = 2 * 127 g/mol = 254g/mol

Number of moles = mass / molar mass

Number of moles = 19.0 g / 254 g/mol

Moles of iodine molecules = 0.0748 moles of Iodine molecules

b) Number of moles of iodine atoms

I mole of iodine molecules contains 2 moles of iodine atoms

Therefore, 0.0748 moles of iodine molecules will contain 2 * 0.0748 moles of iodine atoms

Moles of iodine atoms = 0.150 moles of iodine atoms

c) Number of iodine atoms = number of moles of iodine atoms * 6.02 * 10²³

Number of iodine atoms = 0.150 * 6.02 * 10²³

Number of iodine atoms = 9.03 * 10²² atoms

d) Number of iodine molecules = number of moles of iodine molecules * 6.02 * 10²³

Number of iodine molecules = 0.0748 * 6.02 * 10²³

Number of iodine molecules = 4.50 * 10²² molecules

From which type of clouds do hail pellets form?

Answers

Answer:

Hail pallets forms inside of cumulonimbus clouds

Explanation:

Answer: cumulonimbus hope it helps pls stay safe

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