A compound contains 38.7% K, 13.9% N, and 47.4% O by mass. What is the empirical formula of the compound?
1 answer:
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Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
- V₂ = 295 mL
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
<em>The volume of the glucose solution to be prepared</em> = 500
<em>Molarity of the glucose solution to be prepared</em> = 1 M
i. Molar mass of glucose () = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii.<em> mole = molarity x volume</em>. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole