C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
Flower and sugar
flowers in sugar water? something along these lines
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer:
1.332 g.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):
<em>∴ (n) of CO₂ = (n) of C₂H₆</em>
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∵ n = mass/molar mass
<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>
mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.
mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.
<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>
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