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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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Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9
EleoNora [17]

Answer : The current passing between the electrodes is, 1.056\times 10^{-2}A

Explanation :

First we have to calculate the charge of sodium ion.

q=ne

where,

q = charge of sodium ion

n = number of sodium ion = 2.68\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

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

q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C

Now we have to calculate the charge of chlorine ion.

q'=ne

where,

q' = charge of chlorine ion

n = number of chlorine ion = 3.92\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

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

q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C

Now we have to calculate the current passing between the electrodes.

I=\frac{q}{t}+\frac{q'}{t}

I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}

I=1.056\times 10^{-2}A

Thus, the current passing between the electrodes is, 1.056\times 10^{-2}A

4 0
3 years ago
1.00 mole of an ideal gas at STP is cooled to -41°C while the
qwelly [4]

Answer:

V₂ = 18.13 L

Explanation:

Given data:

Mole of gas = 1 mol

Initial temperature = 273 K

Initial pressure = 1 atm

Final volume = ?

Final temperature = -41°C (-41+273 = 232 K)

Final pressure = 805 mmHg (805/760 = 1.05 atm)

Solution:

First of all we will calculate the initial volume of gas.

PV = nRT

V = nRT/P

V = 1 mol × 0.0821 mol.L/atm.K × 273 K / 1 atm

V = 22.4 L/atm / 1 atm

V = 22.4 L     ( initial volume)

Now we will determine the final volume by using equation,

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Now we will put the values.

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 22.4 L ×  232 K / 273 K × 1.05 atm

V₂ = 5196.8 atm .L. K / 286.65 atm.K

V₂ = 18.13 L

5 0
3 years ago
118g of o2 gas is held at STP what is the volume of gas
shutvik [7]

Answer:

At STP one mole of any gas occupies a volume of 22.4 L: this is the molar volume.

Explanation:

4 0
4 years ago
Helpppp me please please
kipiarov [429]
The answer is letter d    
3 0
3 years ago
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How many megagrams are in 90.532 g?
Fed [463]
9.0532e-5 should be correct! hope this helps
7 0
3 years ago
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