NaOH(aq) + MgSO4(aq) →
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The question is incomplete, complete question is :
A chemist must dilute 73.9 mL of 400 mM aqueous sodium carbonate solution until the concentration falls to 125 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.
Answer:
The final volume of the solution will be 0.236 L.
Explanation:
Concentration of sodium carbonate solution before dilution =
Volume of sodium carbonate solution before dilution =
Concentration of sodium carbonate solution after dilution =
Volume of sodium carbonate solution after dilution =
Dilution equation is given by:
1 mL = 0.001 L
236 mL = 0.236 L
The final volume of the solution will be 0.236 L.
Answer: There are atoms of hydrogen are present in 40g of urea,
.
Explanation:
Given: Mass of urea = 40 g
Number of moles is the mass of substance divided by its molar mass.
First, moles of urea (molar mass = 60 g/mol) are calculated as follows.
According to the mole concept, 1 mole of every substance contains atoms.
So, the number of atoms present in 0.67 moles are as follows.
In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.
Thus, we can conclude that there are atoms of hydrogen are present in 40g of urea,
.