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3 years ago

A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa

Chemistry

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

$V_{2} = 8.92 L$

Explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

$P_{1} V_{1} =nRT$

$320Kpa*0.003m^{3} =nRT$

$1000L = 1m^{3}$

$nRT= 0.96$

Now, as for the final condition:

$P_{2}V_{2}=nRT$

$P_{2} V_{2} =0.96$

clearing $V_{2}$

$V_{2} =\frac{0.96}{P_{2} }$

$V_{2} =0.00872m_{3}$

$V_{2} = 8.92 L$

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A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

3 0

3 years ago

What do you call the family of elements that takes up the column on the far right of the periodic table?

ivann1987 [24]

Noble gases.
group 18. elements that are all unreactive .
fixed naming!

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3 years ago

Which base has the lowest ionization constant (Kb)?

inysia [295]

Answer:D.Blood

Explanation:

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3 years ago

Oxygen gas is collected over water. The total pressure (the O2 pressure the water vapor pressure) is 736 torr. The temperature o

tigry1 [53]

Answer:

The value of the partial pressure of the oxygen $P_{O_{2} }$ = 690 torr

Explanation:

Total pressure of the mixture of gases = 736 torr

The partial pressure of water vapor = 46 torr

From the law of pressure we know that

Total pressure = The partial pressure of water vapor + The partial pressure of oxygen $O_{2}$

Put the values of pressures in above equation we get,

⇒ 736 = 46 + $P_{O_{2} }$

⇒ $P_{O_{2} }$ = 736 - 46

⇒ $P_{O_{2} }$ = 690 torr

This is the value of the partial pressure of the oxygen.

3 0

3 years ago

Enter your answer in the provided box. From the data below, calculate the total heat (in J) needed to convert 0.304 mol of gaseo

diamong [38]

Answer:

-35,281.5 J

Explanation:

To convert the gaseous ethanol to liquid ethanol, three steps will occur. First, it will lose heat and the temperature will decrease until its boiling point, so from 300.0°C to 78.5°C. Thus, more heat will be lost, but now, with the temperature constant, so the gas will be converted to liquid. And then, the liquid will lose heat to decrease the temperature from 78.5°C to 25.0°C.

The total heat loss is the sum of the heats of each step. Because the heat is being removed from the system, it's negative. The first and last step occurs with a change in temperature, and so the heat is calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat of the gas (first step) or liquid (last step), and ΔT the temperature variation (final - initial). The mass of ethanol is the molar mass 46.07 g/mol multiplied by the number of moles, so:

m = 46.07 * 0.304 = 14.00 g

The second step occurs without a change in temperature, and the heat is then:

Q = -n*ΔH°vap

Where n is the number of moles, ΔH°vap is the heat of vaporization, and the minus signal indicates that the heat is being lost. Then, the heat of each step is:

Q1 = 14.00*1.43*(78.5 - 300,0) = -4434.43 J

Q2 = -0.304*40.5 = -12.312 kJ = -12312 J

Q3 = 14.00*2.45*(25.0 - 78.5) = -1835.05 J

Q = Q1 + Q2 + Q3

Q = -35,281.5 J

5 0

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