A graduated cylinder( approximate as a regular cylinder) has a radius of 1. 045 cm and a high of 30.48 cm. what is the volume of
1 answer:
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a)
A: Copper
B: CuO
C:
D: $\mathrm{CuCO_3}$
E: $\mathrm{CO_2}$
F: $\mathrm{Cu(NO_3)_2}$
b)
$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$
c)
$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$
Answer:
Mass of SO₄⁻² = 0.123 g.
Mass percentage of SO₄⁻² = 41.2%
Mass of Na₂SO₄ = 0.0773 g
Mass of K₂SO₄ = 0.1277 g
Explanation:
Here we have
We place Na₂SO₄ = X and
K₂SO₄ = Y
Therefore
X +Y = 0.205 .........(1)
Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have
Amount of BaSO₄ from Na₂SO₄ is therefore;
Amount of BaSO₄ from K₂SO₄ is;
Molar mass of
BaSO₄ = 233.38 g/mol
Na₂SO₄ = 142.04 g/mol
K₂SO₄ = 174.259 g/mol
=
= 1.643·X
=
= 1.339·Y
Therefore, we have
1.643·X + 1.339·Y = 0.298 g.....(2)
Solving equations (1) and (2) gives
The mass of SO₄⁻² in the sample is given by
Mass of sample = 0.298
Molar mass of BaSO₄ = 233.38 g/mol
Mass of Ba = 137.327 g/mol
∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g
Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412
Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.
Percentage mass of SO₄⁻² = 41.2%
Solving equations (1) and (2) gives
X = 0.0773 g and Y = 0.1277 g.