Answer:
no of neutron = atomic mass - atomic number
Explanation:
here
atomic mass = 64
atomic number = 30
no of neutron = <span>64−30</span>
no of neutron = 34
Q1)
molarity is defined as the number of moles of solute in 1 L of solution.
the NaCl solution volume is 1.00 L
number of moles NaCl = NaCl mass present / molar mass of NaCl
NaCl moles = 112 g / 58.5 g/mol = 1.91 mol
the number of moles of NaCl in 1.00 L of solution is - 1.91 mol
therefore molarity of NaCl is 1.91 M
Q2)
molality is defined as the number of moles of solute in 1 kg of solvent.
density is mass per volume.
density of the solution is 1.08 g/mL.
therefore mass of the solution is = density x volume
mass = 1.08 g/mL x 1000 mL = 1080 g
since we have to find the moles in 1 kg of solvent
mass of solvent = 1080 g - 112 g = 968 g
number of moles of NaCl in 968 g of solvent - 1.91 mol
therefore number of NaCl moles in 1000 g - (1.91 mol / 968 g) x 1000 g/kg = 1.97 mol/kg
molality of NaCl solution is 1.97 mol/kg
Q3)
mass percentage is the percentage of mass of solute by total mass of the solution
mass percentage of solution = mass of solute / total mass of the solution
mass of solute = 112 g
total mass of solution = 1080 g
mass % of NaCl = 112 g / 1080 g x 100%
therefore mass % of NaCl = 10.4 %
answer is 10.4 %
Answer:
0.221M
Explanation:
From the question ,
The Molarity of AgNO₂ = 0.310 M
Hence , the concentration of Ag⁺ = 0.301 M
The volume of AgNO₂ = 250 mL
and,
The Molarity of Sodium chromate = 0.160 M
The volume of Sodium chromate = 100 mL.
As the solution is mixed the final volume becomes ,
250mL +100mL = 350mL
Now, using the formula , to find the final molarity of the mixture ,
M₁V₁ ( initial ) = M₂V₂ ( final )
substituting the values , in the above equation ,
0.310M * 250ml = M₂ * 350ml
M₂ = 0.221M
Hence , the concentration of the silver in the final solution = 0.221M
Answer:
156 g of AlCl₃ will be produced from 3.5 moles of HCl.
Explanation:
Given data:
Number of moles of HCl = 3.5 mol
Grams of AlCl₃ produced = ?
Solution:
Balanced Chemical equation:
3HCl + Al(OH)₃ → AlCl₃ + 3H₂O
Now we will compare the moles of AlCl₃ with HCl from balanced chemical equation.
HCl : AlCl₃
3 : 1
3.5 : 1/3×3.5 = 1.17 mol
Mass of AlCl₃ produced:
Mass = number of moles ×molar mass
Mass = 1.17 mol × 133.341 g/mol
Mass = 156 g
Thus 156 g of AlCl₃ will be produced from 3.5 moles of HCl.
