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3 years ago

In an exothermic reaction, an increase in temperature favors the formation of products. True or False?

1 answer:

6 0

For exothermic reaction, heat is one of the products. According to Le Chatelier's principle, the reaction would shift to production that would create equilibrium. The answer to this question is FALSE. Increasing the temperature of an exothermic reaction favors the reactants.

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Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Sladkaya [172]

Explanation:

The given balanced reaction is as follows.

$2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O$

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of $NH_{4}NO_{3}$ present will be as follows.

Moles of $NH_{4}NO_{3}$ = $\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}$

= $\frac{86000 g}{80.043 g/mol}$

= 1074.42 mol

Therefore, moles of $N_{2}$ , $O_{2}$ and $H_{2}O$ produced by 1074.42 mole of $NH_{4}NO_{3}$ will be as follows.

Moles of $O_{2}$ = $\frac{1}{2} \times 1074.42 mol$

= 537.21 mol

Moles of $N_{2}$ = $\frac{2}{2} \times 1074.42 mol$

= 1074.42 mol

Moles of $H_{2}O$ = $\frac{4}{2} \times 1074.42 mol$

= 2148.84 mol

Therefore, total number of moles will be as follows.

537.21 mol + 1074.42 mol + 2148.84 mol

= 3760.47 mol

According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.

PV = nRT

1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as $307^{o}C$ = 307 + 273 = 580 K)

V = 179066.06 L

Thus, we can conclude that total volume of the gas is 179066.06 L.

3 0

3 years ago

A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715

andrey2020 [161]

Answer:

$w_{Ar}=0.814$

Explanation:

Hello,

In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\ \\n=0.41mol$

Next, using the molar fraction of argon, we compute the moles of argon:

$n_{Ar}=0.41mol*0.715=0.29mol$

And the moles of methane:

$n_{CH_4}=0.41mol-0.29mol=0.12mol$

Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:

$w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol} }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}} \\\\w_{Ar}=0.814$

Regards.

7 0

3 years ago

Please answer this question it’s due in 5 minutes thanks if you do ! :))

Oksi-84 [34.3K]

Answer:

you are correct

Explanation:

3 0

3 years ago

The part that includes land and water

luda_lava [24]

Answer: this is all of it

Explanation: i know this because its all of it

6 0

3 years ago

Read 2 more answers

6. Give reasons.

MArishka [77]

Answeryou understand but id ont if your know the answer plz help me

Explanation:

3 0

3 years ago