Hey there!
The best way to balance chemical equations is to first start by balancing polyatomic ions such as OH and SO₄.
Next, balance other elements, but save elements that are by themselves for last, such as H₂ or Fe. Once you balance everything else you can do the ones by themselves, it's much easier.
Hope this helps!
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is 
Explanation:
The chemical equation for this decomposition of ammonia is
↔ 
The initial concentration of ammonia is mathematically represented a
![[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%20%20%5Cfrac%7Bn_1%7D%7BV_1%7D%20%20%3D%20%5Cfrac%7B29%7D%7B75%7D)
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
The initial concentration of nitrogen gas is mathematically represented a
![[N_2] = \frac{n_2}{V_2}](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%20%5Cfrac%7Bn_2%7D%7BV_2%7D)
![[N_2] = 0.173 \ M](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%200.173%20%20%5C%20%20M)
So looking at the equation
Initially (Before reaction)


During reaction(this is gotten from the reaction equation )
(this implies that it losses two moles of concentration )
(this implies that it gains 1 moles)
(this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium


Now since
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
Now the equilibrium constant is
![K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=K_c%20%20%3D%20%20%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
substituting values


Answer:

Explanation:
When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.
So:
24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O
The next step is to divide each mass by their molar mass to convert your grams to moles.
24.42/40.08 = 0.6092 mol
17.07/14.01 = 1.218 mol
58.85/15.99 = 3.680 mol
Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.
0.6092 mol/0.6092 mol = 1
1.218 mol/0.6092 mol = 2
3.680 mol/0.6092 mol = 6
So the empirical formula is 
Answer:
The answer to your question is Ferns
Explanation:
Ferns are called nonflowering plants and produce spores instead of seeds.
Metallic properties head to the left.