This is true there are many factors that influence the amount of a solute.
Answer:
32.8%
Explanation:
All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).
First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:
- 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄
There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.
We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:
- 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb
Finally we calculate the percentage composition of Pb:
- 0.083 g Pb / 0.254 g salt * 100% = 32.8%
A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O
Let the total mass of compound = 100g
The mass of Aluminum = 22.10 g
Moles of Al = mass / atomic mass = 22.10 /23= 0.96
The mass of P = 25.40 g
Moles of P = 25.40 / 31 = 0.819
The mass of oxygen = 52.50 g
Moles of oxygen = 52.50 / 16= 3.28
The mole ratio of the two elements will be
Al = 0.96/0.819 = 1.17
P = 0.819/0.819 = 1
O = 3.28 / 0.819 = 4
The whole number ratio will be 1 : 1 : 4
So formula will be AlPO4
$6.72/lb = $6.72/16 oz = $0.42/oz = 42 cents per ounce
<span>RE: "numerically to the hundredths place." </span>
<span>Since, in this case, you're dealing with money, "the hundredths place" simply means, "to the nearest cent."</span>
Answer:
I can't answer that Po sorry
