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Volgvan
3 years ago
12

Balance the following equation by inspection: Cu + HNO3 --> Cu(NO3)2 + H2O +NO

Chemistry
2 answers:
lorasvet [3.4K]3 years ago
8 0

Answer:

3Cu + 8HNO_{3} → 3Cu(NO_{3} )_{2} +4H_{2}O + 2NO

Explanation:

Considering the number of each atom in both reactants, it must be equal to the number of each atom in the product. Thus, the number of copper atom, hydrogen atom, nitrogen atom and oxygen atom must be equal on both sides of the equation.

pashok25 [27]3 years ago
7 0

Answer:

f is this

Explanation:

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Name two compounds in unpolluted air?
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The two compounds present in unpolluted air are carbon dioxide and water. You may think that carbn dioxide is a greenhouse gas, yes it is and at the same time not. There animals in the earth that produces carbon dioxide into the atmosphere, the carbon dioxide too, is also need for plan respiration. 

3 0
2 years ago
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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Answer:

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