Answer:
The answer to your question is given after the questions so I just explain how to get it.
Explanation:
a)
Get the molecular weight of Phosphoric acid
H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)
= 3 + 31 + 64
= 98 g
98 g ----------------- 1 mol
0.045 g --------------- x
x = (0.045 x 1) / 98
x = 0.045 / 98
x = 0.00046 moles or 4.6 x 10 ⁻⁴
b)
Molarity = 
Molarity = 
Molarity = 0.0013 or 1.31 x 10⁻³
c)
Formula C₁V₁ = C₂V₂
V₁ = C₂V₂ / C₁
Substitution
V₁ = (0.0013)(1) / 0.01
Simplification and result
V₁ = 0.0013 / 0.1
V₁ = 0.13 l = 130 ml
Answer:
Kc = 2.34 mol*L
Explanation:
The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.
A + B ⇄ C + D
Kc = [C] * [D] / [A] * [B]
According to the reaction
Kc = [SO2]^2 * [O2]^2 / [SO3]^2
Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3
0.450 --> 0 + 0 (Beginning of the reaction)
0.260 --> 0.260 + 0.130 (During the reaction)
0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)
Kc = [0.260]^2 + [0.130]^2 / [0.190]^2
Kc = 2.34 mol*L
Answer:
67.5%
Explanation:
Step 1: Write the balanced equation for the electrolysis of water
2 H₂O ⇒ 2 H₂ + O₂
Step 2: Calculate the theoretical yield of O₂ from 17.0 g of H₂O
According to the balanced equation, the mass ratio of H₂O to O₂ is 36.04:32.00.
17.0 g H₂O × 32.00 g O₂/36.04 g H₂O = 15.1 g O₂
Step 3: Calculate the percent yield of O₂
Given the experimental yield of O₂ is 10.2 g, we can calculate its percent yield using the following expression.
%yield = (exp yield / theoret yield) × 100%
%yield = (10.2 g / 15.1 g) × 100% = 67.5%
I think O and Cl will form covalent bonds since they are both non-metals and don't have a large enough difference in electronegativities to create an ionic bond.
I hope this helps. Let me know if anything is unclear.
Answer:
Three things that can generate electrical energy is <u>coal, natural gas, and petroleum.</u> (They generate from fossil fuels).
