The answer is ethyl methyl ketone.
Answer:
http://hyperphysics.phy-astr.gsu.edu/hbase/Biology/imgbio/treecycle.p ng
Explanation:
I guess it is the second one but you missed the state symbols.
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%
C=2.41 g/L
m(NaCl)=0.291 g
c=m(NaCl)/v
v=m(NaCl)/c
v=0.291/2.41= 0.1207 L = 120.7 mL
