Answer:
I would say the last one because mass is not created nor destroyed.
Explanation:
Answer:
I think the answer is boiling
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answer:
The expression of an equilibrium constant will given as:
![K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
K is the constant of a certain reaction when it is in equilibrium

The expression of an equilibrium constant will given as:
![K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)