The partial pressure of hydrogen is 0.31 atm
calculation
find the number of hydrogen moles the container, that is
25/100 x 6.4 =1.6 moles of hydrogen
find the partial pressure for hydrogen in 1.6 moles
that is 6.4 moles= 1.24 atm
1.6 moles= ?
by cross multiplication
1.6moles x1.24 atm/ 6.4 moles= 0.31 atm
Answer:
CU
Explanation:
Cu is the correct answer if I'm correct
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>
Answer:
The answer to your question is: letter c
Explanation:
Data
V1 = 612 ml n1 = 9.11 mol
V2 = 123 ml n2 = ?
Formula
n2 = 1.83 mol
Answer:
do it by yourself 6637373
