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3 years ago

Consider the balanced equation.

Chemistry

2 answers:

vaieri [72.5K]3 years ago

8 0

The balanced chemical reaction is:

N2 + 3H2 = 2NH3

We are given the amount of H2 being reacted. This will be our starting point.

26.3 g H2 (1 mol H2 / 2.02 g H2) 2 mol O2/3 mol H2) ( 17.04 g NH3 / 1mol NH3) = 147.90 g O2

Percent yield = actual yield / theoretical yield x 100

Percent yield = 79.0 g / 147.90 g x 100

Percent yield = 53.4%

Slav-nsk [51]3 years ago

3 0

The answer is D 53.4%
just took the test

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P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:

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Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

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3 years ago

4) What volume will the gas in the balloon at right occupy at 250k? balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

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