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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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The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p
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Answer:

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Explanation:

This is a direct application of the equation for ideal gases.

  • PV=nRT

Where:

  • P = pressure = 1.25 atm
  • V = volume = 25.2 liter
  • R = Universal constant of gases = 0.08206 atm-liter/K-mol
  • T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
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Solving for n:

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Substituting:

n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol

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