Answer:
V₂ = 15.00 atm
Explanation:
Given data:
Initial pressure = 5.00 atm
Initial volume = 3.00 L
Final pressure = 760 mmHg ( 760/760 = 1 atm)
Final volume = ?
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 5.00 atm × 3.00 L / 1 atm
V₂ = 15.00 atm
The percentage of yield was 777.78%
<u>Explanation:</u>
We have the equation,
Be
[s] + 2
HCl
[aq] → BeCl
2(aq] +
H
2(g] ↑ Be
(s] +
2
HCl
[aq] → BeCl
2(aq] +
H
2(g]
↑
To find the percent yield we have the formula
Percentage of Yield= what you actually get/ what you should theoretically get x 100
=3.5 g/0.45 g 100
= 777.78 %
The percentage of yield was 777.78%
Answer:
alright that sounds like a recipe i gota try
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
