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Yanka [14]
2 years ago
10

How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

Chemistry
1 answer:
zaharov [31]2 years ago
7 0

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

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Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

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2 years ago
When atoms lose or gain electrons in chemical reactions they form?
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8 0
2 years ago
Express the quantity 556.2 x 10^-12 in each units<br> A.) ms<br> B.) ns<br> C.) ps<br> D.) fs
zavuch27 [327]
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we should know the meaning of each abbreviation:
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ns means  means  nanosecond,   its value is 10^-9 s
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<span>Expressions of the quantity 556.2 x 10^-12 are</span>
556.2 x 10^-12 =556.2 ps
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6 0
2 years ago
Elena makes the table below to determine the number of atoms of each element in the chemical formula 3(NH4)2SO4
Hunter-Best [27]

Answer:

She should not have multiplied the nitrogen atom by subscript 2.

Explanation:

Chemical formula:

3(NH₄)₂SO₄

Elements present in given formula:

Nitrogen

Hydrogen

Sulfur

Oxygen

Total number of atoms of elements:

N = 3×1×2 = 6

H = 4×2×3 = 24

S = 1×3 = 3

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The number nitrogen atoms are six. Elena did mistake by counting the number of nitrogen. She should didn't multiplied the nitrogen atom by subscript 2.

5 0
2 years ago
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