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Answer:
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Explanation:
As we know
1 liter = 1000 grams
2H2 + O2 --> 2H2O
Weight of H2 molecule = 2.016 g/mol
Weight of water = 18.01 gram /l
2 mole of oxygen react with 2 mole of H2
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Answer:
The endpoint volume is 50.52 ± 0.14 mL
Explanation:
In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:
V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)
V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL
V = 50.52 ± 0.14 mL
It is necessary to consider the sum of the errors too.
Answer:
The answer to your question is 3% H2SO4 solution
Explanation:
Data
Concentration 2 = C₂ = ?
Concentration 1 = C₁ = 15 %
Volume 1 = V₁ = 50 ml
Volume 2 = V₂ = 250 ml
Formula
C₁V₁ = C₂V₂
Solve for C₂
C₂ = C₁V₁ / V₂
Substitution
C₂ = (15)(50) / 250
Simplification and result
C₂ = 3 %
