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3 years ago

What pressure (in atm) will 0.44 moles of co2 exert in a 2.6 l container at 25Â°c?

1 answer:

6 0

We can use the ideal gas law equation to find the pressure
PV = nRTwhere
P - pressure
V - volume - 2.6 x 10⁻³ m³
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³ = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm

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Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

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Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

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3 years ago

"A crosslinked copolymer consists of 57 wt% ethylene (C2H4) repeat units and 43 wt% propylene (C3H6) repeat units. Determine the

ch4aika [34]

Explanation:

Percentage ethylene by weight = 57%

Percentage propylene by weight = 43%

Suppose in 100 grams of polymer:

Mass of ethylene = 57 g

Mass of propylene = 43 g

Moles of ethylene = $\frac{57 g}{28 g/mol}=2.036 mol$

Moles of propylene = $\frac{43g}{42g/mol}=1.024 mol$

1 mole = $N_A =6.022\times 10^{23}$ molecules/ atoms

Units of ethylene = $2.036 mol\times N_A$

Units of propylene = $1.024 mol\times N_A$

a) Fraction of ethylene units:

$=\frac{2.036 mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{509}{765}$

b ) Fraction of propylene units:

$=\frac{1.024mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{256}{765}$

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3 years ago