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natima [27]
3 years ago
12

(?) NH4NO3 ---> (?) N2O + (?) H2O

Chemistry
1 answer:
miskamm [114]3 years ago
6 0

Answer:

NH4NO3 -> N2O + 2H2O

Explanation:

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2. When heated above 500 ºC, potassium nitrate decomposes according to the equation below. 4KNO3 2K2O + 2N2 + 5O2A.If oxygen is
AysviL [449]

Answer:

(a) The rate of formation of K2O is 0.12 M/s.

The rate of formation of N2 is also 0.12 M/s

(b) The rate of decomposition of KNO3 is 0.24 M/s

Explanation:

(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.

Rate of formation of O2 is 0.3 M/s

Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s

Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.

Rate of formation of N2 = (2×0.3/5) = 0.12 M/s

(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.

Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s

3 0
2 years ago
Revise the following statement to make it true:
tino4ka555 [31]

Answer:

All the individuals of a species living within a specific area are collectively called a population.

Explanation:

6 0
2 years ago
How many joules would it take to raise 2 grams of AL 1 degree?
denis23 [38]

Explanation:

Air has a heat capacity of about 700 Joules per kg per °K and a density of just 1.2 kg/m3, so its initial energy would be 700 x 1 x 1.2 x 293 = 246,120 Joules

4 0
2 years ago
How many grams of Cl are in 345 g of CaCl2
swat32
Hope this helps you.

3 0
3 years ago
A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o
Andre45 [30]

Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

Regards.

7 0
3 years ago
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