Answer:
A. How the concentration of the reactants affects the rate of a reaction
Explanation:
Let's consider a generic reaction.
A + B ⇒ Products
The generic rate law is:
rate = k × [A]ᵃ × [B]ᵇ
where,
- rate: rate of the reaction
- [A] and [B]: molar concentrations of the reactants
As we can see, the rate law shows how the concentration of the reactants affects the rate of a reaction.
Answer:
Give them each one so all of you is 8
Explanation:
I hope it helps:)
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>
m 9CH₃COOH: 60u×9 = <u>540u</u>
<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g
</u></span>
The alcohol concentration of the mixed solution is 20%
Simplification :
Based on the given condition, formulate :
35% ×0.40 + 0.6 ×10% ÷{ 0.4+0.6}
Calculate the product :
Calculate the sum or difference : 
Any fraction with denominator 1 is equal to numerator : 0.2
Multiply a number to both numerator, denominator : 0.2 ×
Calculate the product or quotient : 
A fraction with denominator equals to 100 to a percentage 20%.
How do you find the concentration of a mixed solution?
In general when your are mixing two different concentrations together first calculate number of moles for each solution (n=CV ,V-in liter) then add them together it will be total moles,then concentration of mixture will be = total moles / total volume(liter).
Learn more about concentration of alcohol :
brainly.com/question/13220698
#SPJ4
Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
