Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry ) 
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
Answer:
Explanation:
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In this case, considering the partial Dalton's law of partial pressures, we can notice that the total pressure equals the pressure of steam and the pressure of hydrogen, which can be determined as shown below:
Thus, by using the ideal gas law, we can compute the moles of hydrogen as shown below:
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Answer:
U= 238g/mol
U2O5= 556g/mol
Explanation:
Since U= 238
O=16
U3O5= 2(238)+3(16)=556g/mol
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