22.37 g of chromium(II) nitrate, must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt.
According to the definition, the molar concentration of a substance in a solution is the ratio of the number of the moles to the volume of the solution:
c=n/V.
The number of the moles is related to the mass with the molar mass:
n=m/M;
m=n·M.
Thus, given the volume of the solution of chromium(II) nitrate, its concentration and molar mass is 238.011 g/mol we can calculate the mass of chromium(II) nitrate needed for the preparation :
∴ Cr(NO₃)₃ = cVM
= 0.188 M × 0.5 L × 238.011 g/mol
= 22.37 g
Therefore, 22.37 g of chromium(II) nitrate, must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt.
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From the equation of te reaction, we know that the mass percent of NaOH in the mixture is 1.4%.
<h3>What is neutralization?</h3>
Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.
In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O
Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles
Since the reaction is 1:1, 0.0075 moles of NaOH reacted.
Mass of NaOH = 0.0075 moles of NaOH * 40 g/mol = 0.3 g
Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%
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Answer:
c. all matter is composed of small particles,which combine to make larger objects
Answer:
517.3 K
Explanation:
Initial volume of gas V1= 114 L
Initial temp. T1= 273 K
Final volume V2= 216 L
Final temp. T2= ?
From Charles law, Volume is directly proportional to temperature provided pressure is kept constant
V1/T1 = V2/T2
T2 = V2T1/V1
T2 = (216×273)/114
T2 = 517.3 K
