What are the products in the equation for cellular respiration? A. Oxygen and lactic acid b. Carbon dioxide and water c. Glucose
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Answer:
The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.
Explanation:
Suppose there are 100 moles in solution:
Moles of sulfuric acid = 6% of 100 moles = 6 moles
Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g
Moles of water = 100%- 6% = 94%= 94 moles
Mass of water = 94 mol × 18 g/mol = 1692 g
Specific gravity of the solution ,S.G= 1.07
Density of solution = D
= density of water = 1 g/mL
Mass of the solution = 588 g + 1692 g = 2280 g
Volume of the solution = V
Volume =
1 mL = 0.001 L
n = number of moles of compound
V = volume of the solution in L
here we have ,n = 6 moles of sulfuric acid
V = 2.13084 L
So, the molarity of the solution is :