Answer:
The answer is 375.54 g of AgBr
Explanation:
Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)
Mass = 2M x 1L x 187.77 g/mol
Mass = 375.54g
HCl = H⁺ + Cl⁻
c(HCl)=9.8*10⁻⁵ mol/l
pH=-lg[H⁺]
[H⁺]=c(HCl)
pH=-lg{c(HCl)}
pH=-lg{9.8*10⁻⁵}=4.009
pH=4.009
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 transverse and longitudinal wave combine to form a Water Wave.
