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3 years ago

How many moles of neon (Ne) gas have a volume of 4.5L and a pressure of 2.1 atm at 302 K

Chemistry

1 answer:

Montano1993 [528]3 years ago

7 0

Answer:

0.3811 mol.

Explanation:

To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

∵ P = 2.1 atm, V = 4.5 L, T = 302 K, R = 0.0821 L.atm/mol.K.

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the latest weather report includes the following statement: the temperature is 78°f, barometric pressure is 29 and the relative

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3 years ago

At what temperature does 0.019135 moles of Ne in a 878.3 mL container exert a pressure of 0.946 atm?

Reika [66]

<h3>Answer:</h3>

Temperature is 529.164 K

<h3>Explanation:</h3>

We are given

Number of moles of Ne (n) = 0.019135 moles

Volume (V) = 878.3 mL

Pressure (P) = 0.946 atm

We are required to calculate the temperature;

We can do this using the ideal gas law equation which is;

PV = nRT, where P is the pressure, n is the number of moles, V is the volume, R is the ideal gas constant (0.082057 Latm/mol/K) and T is the temperature.

From the equation;

$T=\frac{PV}{nR}$

$T = \frac{(0.946)(0.8783)}{(0.082057)(0.019135)}$

$T=529.164 K$

Therefore, the temperature will be 529.164 K.

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3 years ago

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At 1.00 atmosphere pressure, a certain mass of a gas has a temperature of 100oC. What will be the temperature at 1.13 atmosphere

Inessa [10]

Answer: Final temperature of the gas will be 330 K.

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$ (At constant volume and number of moles)

${P_1\times T_1}={P_2\times T_2}$

where,

$P_1$ = initial pressure of gas = 1.00 atm

$P_2$ = final pressure of gas = 1.13 atm

$T_1$ = initial temperature of gas = $100^0C=(100+273)K=373K$ K

$T_2$ = final temperature of gas = ?

${1.00\times 373}={1.13\times T_2}$

$T_2=330K$

Therefore, the final temperature of the gas will be 330 K.

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3 years ago

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid

bogdanovich [222]

Explanation:

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H₂O). They will react according to the following equation.

HBr + NaOH ---> NaBr + H₂O

0.81 g of HBr are mixed with 0.568 g of NaOH. We have to find the mass of NaBr that can be produced. To do that we have to find which of the reactants is limiting the reaction. First, we will convert their grams into moles using their molar masses.

molar mass of HBr = 80.91 g/mol

molar mass of NaOH = 40.00 g/mol

mass of HBr = 0.81 g

mass of NaOH = 0.568 g

moles of HBr = 0.81 g * 1 mol/(80.91 g)

moles of HBr = 0.0100 moles

moles of NaOH = 0.568 g * 1 mol/(40.00 g)

moles of NaOH = 0.0142 moles

HBr + NaOH ---> NaBr + H₂O

Now if we take a quick look at the coefficients of the reaction we will see that 1 mol of HBr will react with 1 mol of NaOH since both coefficients are 1. Then their molar ratio is 1 : 1. That also means that 0.0100 moles of HBr will only react with 0.0100 moles of NaOH, and we have mixed 0.0142 moles of it. So, NaOH is in excess and HBr is the limiting reagent.

1 mol of HBr : 1 mol of NaOH molar ratio

moles of NaOH = 0.0100 moles of HBr * 1 mol of NaOH/(1 mol of HBr)

moles of NaOH = 0.0100 moles < 0.0142 moles ----> NaOH is in excess

And now that we know that HBr is the limiting reagent we can find the number of moles of NaBr that will be produced by 0.0100 moles of HBr. And finally convert those moles into grams using the molar mass.

1 mol of HBr : 1 mol of NaBr molar ratio

moles of NaBr = 0.0100 moles of HBr * 1 mol of NaBr/(1 mol of HBr)

moles of NaBr = 0.0100 moles

molar mass of NaBr = 102.89 g/mol

mass of NaBr = 0.0100 moles * 102.89 g/mol

mass of NaBr = 1.0289 g

mass of NaBr = 1.0 g

Answer: the maximum mass of sodium bromide that could be produced is 1.0 g.

7 0

1 year ago