N= m/v n=0.077/.200 = 0.385
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate
It will benefit the flower because it will always get Polonated but it can be bad for the bees because if the flower dies so will the bees
Answer:
3.5 moles Fe
Explanation:
From the equation, Reaction of 2 moles of Fe₂O₃ with 1 mole of C produces 1 mole of Fe. When excess Fe₂O₃ is used, the only liming factor is C.
The ratio of amount of C used to the amount of Fe produced is 1:1
Therefore, if 3.5 moles of C are used, 3.5 moles of Fe are also produced.
