Answer:
The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g
Explanation:
providing a solution to 5% weight-volume as found in commerce:
⇒ % 5 = (5g d-glucose/ 100 mL sln)×100
⇒ 0.05 = g C6H12O6/mL sln
⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)
⇒ g C6H12O6 = 1 E-4 g C6H12O6
Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
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Answer:
0.534
Explanation:
Mole fraction can be calculated using the formula:
Mole fraction = number of moles of solute ÷ number of moles of solvent and solute (solution).
In this question, solute is dimethyl ether while the solvent is methanol.
Mole (n) = mass (M) ÷ molar mass (MM)
Mole of solute (dimethyl ether) = 148.5 ÷ 46.07
= 3.22moles.
Mole of solvent (methanol) = 90 ÷ 32.04
= 2.81moles.
Total moles of solute and solvent = 3.22 + 2.81 = 6.03moles.
Mole fraction of dimethyl ether = number of moles of dimethyl ether ÷ number of moles of solution (dimethyl ether + methanol)
Mole fraction = 3.22/6.03
= 0.534
