Explanation:
1)  + 7 H_2(g)](https://tex.z-dn.net/?f=%202%20Al%28s%29%20%2B%202%20NaOH%28aq%29%20%2B%206%20H_2O%28l%29%20%5Clongleftrightarrow%202%20Na%5BAl%28OH%29_4%5D%28aq%29%20%2B%207%20H_2%28g%29)
![Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNa%5BAl%28OH%29_4%5D%5D%5E2%2A%5BH_2%5D%5E7%7D%7B%5BNaOH%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

2) 
![Kc=\frac{[H_2SO_4]}{[SO_3]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2SO_4%5D%7D%7B%5BSO_3%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

3)
![Kc=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
Answer:
The answer to your question is 41.6 g of AgCl
Explanation:
Data
mass of NH₄Cl = 15.5 g
mass of AgNO₃ = excess
mass of AgCl = 35.5 g
theoretical yield = ?
Process
1.- Write the balanced chemical reaction.
NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃
2.- Calculate the molar mass of NH₄Cl and AgCl
NH₄Cl = 14 + 4 + 35.5 = 53.5 g
AgCl = 108 + 35.5 = 143.5 g
3.- Calculate the theoretical yield
53.5 g of NH₄Cl -------------------- 143.5 g of AgCl
15.5 g of NH₄Cl ------------------- x
x = (15.5 x 143.5) / 53.5
x = 2224.25 / 53.5
x = 41.6 g of AgCl
Answer:
19.8% of Nitrogen
Explanation:
In the Al(NO₃)₃ there are:
1 atom of Al
3 atoms of N
And 9 atoms of O
The molar mass of Al(NO₃)₃ is:
1 Al * (26.98g/mol) = 26.98g/mol
3 N * (14g/mol) = 42g/mol
9 O * (16g/mol) = 144g/mol
26.98 + 42 + 144 = 212.98g/mol
We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:
2.57g * (42g/mol / 212.98g/mol) =
0.51g N
Percent composition of nitrogen is:
0.51g N / 2.57g * 100
= 19.8% of Nitrogen
NaBrO3 is the chemical formula for Sodium Bromate.