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lisov135 [29]
3 years ago
15

A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de

creased to 3.8 m3. the temperature is decreased to 200 K. What is the final pressure of the nitrogen gas?
Chemistry
1 answer:
zhannawk [14.2K]3 years ago
7 0

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

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Explanation:

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In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
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Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

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From the given reaction, we conclude that

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3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
3 years ago
PLEASE HELP ME ASAP!
uranmaximum [27]

your answer is b hope this helps


7 0
3 years ago
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Explanation:

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