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elena-s [515]
3 years ago
8

A chemist uses hot hydrogen gas to convert chromium(III) oxide to

Chemistry
1 answer:
Darya [45]3 years ago
7 0

Answer: 3.024 g grams of hydrogen are needed to  convert 76 grams of chromium(III) oxide, Cr_{2}O_{3}

Explanation:

The reaction equation for given reaction is as follows.

Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O

Here, 1 mole of Cr_{2}O_{3} reacts with 3 moles of H_{2}.

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide (Cr_{2}O_{3}) is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of Cr_{2}O_{3} is calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol

Now, moles of H_{2}.given by 0.5 mol of Cr_{2}O_{3} is calculated as follows.

0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}

As molar mass of H_{2} is 2.016 g/mol. Therefore, mass of H_{2} is calculated as follows.

No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g

Thus, we can conclude that 3.024 g grams of hydrogen are needed to  convert 76 grams of chromium(III) oxide, Cr_{2}O_{3}.

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A chemist determines by measurements that 0.030 moles of nitrogen gas participate in a chemical reaction. calculate the mass of
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Remember that: 
number of moles = mass/molar mass

First, we get the molar mass of the nitrogen gas molecule:
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Second we calculate the mass of the precipitate:
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3 0
2 years ago
How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
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Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

Now put all the given values in the above expression, we get:

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

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3 years ago
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Answer:

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2 years ago
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