<u>Answer and Explanation:</u>
Mercury combines with sulfur as follows -
Hg + S = HgS
Hg = 200,59
S = 32,066 Therefore 1.58 g of Hg will react with -
1.58 multiply with 32,066 divide by 200,96 of sulfur.
= 0.25211 g S
This will form 1.58 + 0.25211 g HgS = 1.83211 g HgS
The amount of S remaining = 1.10 - 0.25211 = 0.84789 g
Answer:
F=ma where F = force in N, m = mass in g and a = acceleration in 
Answer:
Explanation:
The formation of ammonia that occurred by the reaction of nitrogen and ammonia is expressed as:
⇄ 
where;
The reactants are:
Hydrogen and nitrogen
The product is ammonia.
For the reaction, the equilibrium constant can be expressed as:
![K = \dfrac{[product]}{[reactants]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cdfrac%7B%5Bproduct%5D%7D%7B%5Breactants%5D%7D)
![K = \dfrac{[NH_3]^2}{[N_2]^3[H_2]^3}](https://tex.z-dn.net/?f=K%20%3D%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5E3%5BH_2%5D%5E3%7D)
From the equilibrium constant conditions, the formation of ammonia and its decomposition due to its reversible reaction back to hydrogen and nitrogen are equal. It implies that the rate of the forward reaction is also equal to that of the backward reaction.
Thus, during when equilibrium is obtained;
Hydrogen, Nitrogen, and Ammonia are present.
Answer:
For the production of 77.4 L water 90.3 L oxygen is required.
Explanation:
Given data:
Volume of oxygen required = ?
Volume of water produced = 77.4 L
Solution:
Chemical reaction equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
1 mole = 22.414 L
There are 6 moles of water = 6×22.414 = 134.5 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
Now we will compare the litters of water and oxygen:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
Answer:
The answer to your question is: P2 = 846 kPa
Explanation:
Data
V1 = 2.5 l
P1 = 300 kPa
T1 = 273 °K
P2 = ?
V2 = 1 l
T2 = 308°K
Formula




P2 = 846 kPa