Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.
Answer:
Empirical formula is C₃H₃O
Explanation:
Given data:
Mass of ethyl butyrate = 1.95 mg
Mass of CO₂ = 4.42 mg
Mass of water = 1.81 mg
Empirical formula = ?
Solution:
Percentage of C = 4.42/1.95 × 12/44×100
= 2.27× 0.273 ×100
= 62
percentage of H = 1.81/1.95 × 2/18 × 100
= 0.93 × 0.11 ×100
= 10.23
percentage of oxygen = 100 - (72.23)
= 27.77
number of gram atoms of C = 62/12 =5.17
number of gram atoms of H = 10.23 / 2 =5.115
number of gram atoms of O = 27.77 / 16 =1.74
Atomic ratio:
C : H : O
5.17/1.74 : 5.115/1.74 : 1.74/1.74
3 : 3 : 1
Empirical formula is C₃H₃O
A. What life may have been like before the evoltuon of photosynthesis
D=m/v d=63.5g/11.6cm3=5.47g/cm3