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1 year ago

A sample of carbon dioxide occupies 200 L at 22°C and 1.0 atm. At what temperature will

Chemistry

1 answer:

xxMikexx [17]1 year ago

3 0

Answer:

1328 K or 1055 °C

Explanation:

To find the new temperature, you need to use the Combined Gas Law:

P₁V₁ / T₁ = P₂V₂ / T₂

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the final pressure, volume, and temperature. After converting the temperature (T₁) from Celsius to Kelvin, you can plug the values into the equation and simplify.

P₁ = 1.0 atm P₂ = 0.90 atm

V₁ = 200 L V₂ = 1000 L

T₁ = 22 °C + 273.15 = 295.15 K T₂ = ? K

P₁V₁ / T₁ = P₂V₂ / T₂

(1.0 atm)(200 L) / 295.15 K = (0.90 atm)(1000 L) / T₂

0.6776 = (0.90 atm)(1000 L) / T₂

(0.6776) x T₂ = (0.90 atm)(1000 L)

(0.6776) x T₂ = 900

T₂ = 1328 K

T₂ = 1055 °C

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What is the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm

Virty [35]

The volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.

<h3>How to calculate volume?</h3>

The volume of a given mass of gas can be calculated using the following formula:

PV = nRT

Where;

P = pressure
V = volume
R = gas law constant
T = temperature
n = number of moles

According to this question, 0.98 moles of oxygen gas at 275 k contains a pressure of 2.0 atm. The volume is calculated as follows:

2 × V = 0.98 × 0.0821 × 275

2V = 22.13

V = 11.06L

Therefore, the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.

Learn more about volume at: brainly.com/question/12357202

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2 years ago

What type of heat is transferred from hot water to metal and plastic forks?

patriot [66]

Answer:

Kinetic energy is an energy that can be applied to another object, often considered the energy of motion.

Explanation:

Metals are generally good conductors of heat.

7 0

3 years ago

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Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

3 years ago

Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2

Mkey [24]

Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

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3 years ago

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How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?

kramer

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

$\\ \Large\sf\longmapsto 250\times 15\%$

$\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}$

$\\ \Large\sf\longmapsto 37.5g$

<u>37.5g of NaCl present in 250g of solution.</u>

8 0

3 years ago

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