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3 years ago

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a 15.0 L tank of gas is contained at a high pressure of 8.20*10^4. The tank is opened and the gas expands into an empty chamber

with a volume of 6.00*10^4 L. Calculate the new pressure of the gas.

1 answer:

Nostrana [21]3 years ago

8 0

Answer:

$P_2=20.5$

Explanation:

Assuming that temperature is constant

According to Boyle's Law, at constant temperature pressure is inversly proportional to the volume and mathematically it can be expressed as:

$P_1V_1 = P_2V_2$ ..........1

$P_1=8.2 \times 10^4$

$P_2=?$

$V_1=15L$

$V_2=6.00 \times 10^4L$

from the first equation after putting all the value

we get,

$P_2=20.5$

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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

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<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

Chlorine pentafluoride gas is collected at -17.0 °C in an evacuated flask with measured volume of 35.0 L. When all the gas has b

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Answer:

1. The mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of Chlorine pentafluoride, ClF5, is 0.3mole

Explanation:

1. To solve the mass of Chlorine pentafluoride, ClF5, first, let us calculate the molar mass of ClF5. This is illustrated below:

Molar Mass of ClF5 = 35.5 + (5 x 19) = 35.5 + 95 = 130.5g/mol

From the ideal gas equation:

PV = nRT (1)

Recall:

Number of mole(n) = mass (m) /Molar Mass(M)

n = m/M

Now substituting the value of n in equation 1, we have:

PV = nRT

PV = mRT/M

Now we can obtain the mass of Chlorine pentafluoride ClF5 as follow:

Data obtained from the question include:

T (temperature) = -17.0 °C = - 17 + 273 = 256K

V (volume) = 35L

P (pressure) = 0.180 atm

R (gas constant) = 0.082atm.L/Kmol

m (mass of Chlorine pentafluoride) =?

M (molar mass of Chlorine pentafluoride) = 130.5g/mol

PV = mRT/M

0.180 x 35 = m x 0.082 x 256/ 130.5

Cross multiply to express in linear form as shown below:

m x 0.082 x 256 = 0.180x35x130.5

Divide both side by 0.082 x 256

m = (0.180x35x130.5) /(0.082x256)

m = 39.16g

Therefore, the mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of ClF5 can be obtained as follow:

Mass of ClF5 = 39.16g

Molar Mass of ClF5 = 130.5g/mol

Mole of ClF5 =?

Number of mole = Mass /Molar Mass

Mole of ClF5 = 39.16/130.5g

Mole of ClF5 = 0.3mole

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insens350 [35]

Answer:

401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.

Explanation:

The expression for the standard change in free energy is:

$\Delta G=\Delta H-T\times \Delta S$

Where,

$\Delta G$ is the change in the Gibbs free energy.

T is the absolute temperature. (T in kelvins)

$\Delta H$ is the enthalpy change of the reaction.

$\Delta S$ is the change in entropy.

Given at:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25.0 + 273.15) K = 298.15 K

$\Delta H$ = 128.9 kJ/mol

$\Delta G$ = 33.1 kJ/mol

Applying in the above equation, we get as:-

$33.1=128.9-298.15\times \Delta S$

$-29815\Delta S=-9580$

$\Delta S=-9580$ = 0.32131 kJ/Kmol

So, For reaction to be spontaneous, $\Delta G$

Thus, For minimum temperature:-

$\Delta H-T\times \Delta S=0$

$128.9-T\times 0.32131=0$

$T=\frac{128.9}{0.32131}=401.17\ K$

<u>Hence, 401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.</u>

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Answer:

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