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PtichkaEL [24]
3 years ago
6

T he thermochemical equation for the reaction between hydrazine, N2H4, and dinitrogen tetroxide is given as: 2N2H4(2) N204 (1) 3

N2 (g) + 4H20 (g),AHO-1049 kJ What is the corresponding thermochemical equation for this reaction when 6 mol of nitrogen are formed?
Chemistry
1 answer:
SpyIntel [72]3 years ago
6 0

Answer:

4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ

Explanation:

2N2H4(2) + N204 (1) --> 3N2 (g) + 4H20 (g),AHO-1049 kJ

when 6 mol of nitrogen are formed?

The ratio of Nitrogen in the previous therrmochemical equation to when 6 mol f nitrogen are formed is 2; 6/3 = 2.

So to get the new thermoochemical equation, multiply all parameters in the given equation by 2.

We have;

4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ

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How many quarts of pure antifreeze must be added to 5 quarts of a 20 ntifreeze solution to obtain a 50 ntifreeze solution?
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Quarts of pure antifreeze must be added to 5 quarts of a 20 antifreeze solution to obtain a 50 antifreeze solution is 3.

Antifreeze solution is a solute that lowers the freezing point of the liquid in a solution.

For calculating the amount of quarts of pure antifreeze solution added can be given by:

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Where, x is the amount of antifreeze required for addition and pure antifreeze is 100% of the antifreeze taken.

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2 years ago
What is the cell that is outlined in black?
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A teapot with a surface area of 785 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic
lina2011 [118]

Answer:

Given info: The surface area of teapot plated with silver is 700 cm2700 cm2, the cell is powered by 12.0-V12.0-V, the resistance of the cell is 1.80 Ω1.80 Ω, thickness of silver layer is 0.133-mm0.133-mm and density of silver is 10.5×103 kg/m310.5×103 kg/m3.

Write the expression for the mass of the silver.

m=ρAdm=ρAd (1)

Here,

ρρ is the density of silver.

AA is the surface area.

dd is the thickness of the silver layer.

Substitute 10.5×103 kg/m310.5×103 kg/m3 for ρρ, 700 cm2700 cm2 for AA and 0.133-mm0.133-mm for dd in equation (1) to find mass of the silver.

m=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 gm=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 g

Thus, the mass of silver is 97.8 g97.8 g.

Write the expression for number of moles of silver.

n=mWan=mWa (2)

Here,

mm is the mass of silver.

WaWa is the atomic weight.

The atomic weight of silver is 107.8 g/mole107.8 g/mole

Substitute 97.8 g97.8 g for mm, and 107.8 g/mole107.8 g/mole for WaWa in equation (2) to find number of moles of silver.

n=97.8 g107.8 g/mole=0.907 moln=97.8 g107.8 g/mole=0.907 mol

Thus, the number of moles of silver is 0.907 mol0.907 mol.

7 0
4 years ago
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