What volume of concentrated nitric acid (15.0M) is requiredfor
1 answer:
Answer:
0.133 mL
Explanation:
Given data
- Initial concentration (C₁): 15.0 M
- Initial volume (V₁): to be determined
- Final concentration (C₂): 0.001 M
- Final volume (V₂): 2.00 L
We can find the volume of the concentrated solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.001 M × 2.00 L / 15.0 M
V₁ = 1.33 × 10⁻⁴ L = 0.133 mL
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Answer: 0.18 M
Explanation:
Initial molarity, M1 = 0.8 M
Initial olume, V1 = 225 ml
Final volume, V2 = 1000 ml
Final Molarity, M2 = M1V1/V2
= 0.8 x 225/1000
= 0.18 M
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Anything that has mass and volume (takes up space) is called matter.
Answer:
8.96 L
Explanation:
At STP, 1 mole = 22.4 L
0.400 mole * (22.4 L. /1 mole of gas) = 8.96 L