Answer:
2 C4H10 + 5 O2 → 4 CH3CO2H + 2 H2O.
Explanation:
Light naphtha components are readily oxidized by oxygen or even air to give peroxides, which decompose to produce acetic acid according to the chemical equation, illustrated with butane .
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
Answer:
10/9
Explanation:
First, let's convert 1/3 and 7/9 so that the have the same denominator. To do this let's find the least common multiple of 3 and 9.
List the multiples of 3 and 9:
3: 3, 9
9: 9
They have a least common multiple of 9
We need to convert 1/3 so it has a denominator of 9:
1/3*3/3 (we can multiply it by 3/3 because any number over itself is 1) = 3/9
s-3/9=7/9
Add 3/9 to both sides to isolate s
s=10/9
a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
