Using the law of dilution :
Mi x Vi = Mf x Vf
2.00 x Vi = 0.15 x 100.0
2.00 x Vi = 15
Vi = 15 / 2.00
Vi = 7.5 mL
hope this helps!
99% of the filtrate's water that enter bowman's capsule is reabsorbed into the blood. This is because the water and the salts contained in the filtrate are needed for optimal functioning of the body system.
Answer:
75.15 mol.
Explanation:
- Firstly, we need to write the balanced equation of the reaction:
<em>Fe₂O₃ + 3CO → 2Fe + 3CO₂.</em>
It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.
∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.
- we need to calculate the no. of moles of (4000 g) of Fe₂O₃:
<em>no. of moles of Fe₂O₃ = mass/molar mass</em> = (4000 g)/(159.69 g/mol) = <em>25.05 mol.</em>
<u>Using cross multiplication:</u>
1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,
∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.
<em>∴ The no. of moles of CO needed</em> = (3.0 mol)(25.05 mol)/(1.0 mol) =<em> 75.15 mol.</em>
Word equation - calcium + oxygen -> calcium oxide
Chemical equation - Ca + O2 -> CaO2
Hope this helps!
Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
