Answer:
Hg(NO₃)₂(aq) + Na₂SO₄(aq) → 2NaNO₃(aq) + HgSO₄(s)
Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles
Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles
Limiting reagent is Na₂SO₄ as it controls product formation
Moles of HgSO₄ formed = 0.1172 moles
= 0.1172 x 296.65
= 34.757g
Explanation:
Clap your hands each part word you eat and you will get your answer.
The height of the described cylinder is 0.85 cm
Mass= 98g
Density= 8.9g/cm^3
Density= (mass/volume)
Substitute the values
8.94= 98/ volume
Volume= 
Volume= 10.97
Volume of a cylinder can be found with (V = π r 2 h.)
h= height of cylinder ?
radius= 2cm
Substitute the values
10.97= π ×
× h
h= 10.97/12.57
h= 0.85 cm
Therefore, height of the described cylinder is 0.85 cm
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