Answer:
13.4 milimoles.
Explanation:
The following data were obtained from the question:
Volume = 1.88 L
Molarity = 0.00713 M
Millimoles of NaCN =?
Next, we shall determine the number of mole NaCN in the solution. This can be obtained as follow:
Molarity = mole /Volume
Volume = 1.88 L
Molarity = 0.00713 M
Mole of NaCN =?
Molarity = mole /Volume
0.00713 = moles of NaCN /1.88
Cross multiply
Mole of NaCN = 0.00713 × 1.88
Mole of NaCN = 0.0134 mole
Finally, we shall convert 0.0134 mole to Millimoles. This can be obtained as follow:
1 mole = 1000 millimoles
Therefore,
0.0134 mole = 0.0134 × 1000
0.0134 mole = 13.4 milimoles
Therefore, the millimoles of the solute, NaCN in the solution is 13.4 milimoles
Answer:
1. 0.74mol
2. 0.42mol
3. 2.125mol
4. 0.301mol
5. 4.52 × 10^23 particles
Explanation:
Number of moles (n) in a substance can be found using the formula:
mole (n) = mass/molar mass
Using this formula, the following moles are calculated:
1. Molar of Na = 23g/mol
mole = 17/23
mole = 0.74mol
2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)
= 46 + 32 + 64
= 142g/mol
Mole = 60/142
mole = 0.42mol
3. Molar mass of CO2 = 12 + 16(2)
= 12 + 32
= 44g/mol
mole = 93.5/44
mole = 2.125mol
4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)
= 23 + 14 + 48
= 85g/mol
mole = 25.6/85
mole = 0.301mol
5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;
0.75mol × 6.02 × 10^23
= 4.515 × 10^23
= 4.52 × 10^23 particles
The balanced chemical reaction is:
Zn + 2AgNO3 = Zn(NO3)2 + 2Ag
To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:
5.65 g Zn ( 1 mol / 65.38 g) = 0.09 mol Zn
21.6 g AgNO3 (1 mol / 169.87 g) = 0.13 mol AgNO3
The limiting reactant would be silver nitrate since it is consumed completely in the reaction. The excess reactant would be zinc.
Excess Zinc = 0.09 mol Zn - (0.13 / 2) mol Zn = 0.025 mol Zn left
Answer:
72.5Kg
Explanation:
that is the procedure above