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:
Mass of NaBr produced = 23.67 g
Explanation:
Given data:
Mass of AgBr = 42.7 g
Mass of NaBr produced = ?
Solution:
Chemical equation:
2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂
Number of moles of AgBr:
Number of moles = mass/molar mass
Number of moles = 42.7 g/ 187.7 g/mol
Number of moles = 0.23 mol
now we will compare the moles of AgBr with NaBr.
AgBr : NaBr
1 : 1
0.23 : 0.23
Mass of NaBr:
Mass = number of moles × molar mass
Mass = 0.23 mol × 102.89 g/mol
Mass = 23.67 g
Answer:
D. 1:1
Explanation:
For every 1 mole of chlorine (Cl₂), there is one mole of calcium chloride (CaCl₂).
So, the mole ratio of chlorine to calcium chloride is 1:1.
Hope this helps. :)
Since the given formula is . According to cross method formula, magnesium has +2 charge so, is multiplied by 2.
Thus, 1 molecule of magnesium phosphate will contain 2 atoms of phosphorus.
Therefore, three molecules of magnesium phosphate contains following number of atoms.
Mg = 9
P = 6
O = 24
Hence, we can conclude that there are 6 atoms of phosphorus in three molecules of magnesium phosphate, .
