molar concentration of AgNO₃ solution = 0.118 mole/L
Explanation:
Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.
number of moles = mass / molecular weight
number of moles of AgNO₃ = 10 / 170 = 0.0588
molar concentration = number of moles / volume (L)
molar concentration of AgNO₃ solution = 0.0588 / 0.5
molar concentration of AgNO₃ solution = 0.118 mole/L
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We can calculate for temperature by assuming the equation
for ideal gas law:
P V = n R T
Where,
P = pressure = 1.80 atm
V = volume = 18.2 L
n = number of moles = 1.20 moles
R = gas constant = 0.08205746 L atm / mol K
Substituting to the given equation:
T = P V / n R
T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm /
mol K)
T = 332.70 K
We can convert K unit to ˚C unit by subtracting 273.15
to Kelvin, therefore
T = 59.55 ˚<span>C</span>
The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol in 1 L solution
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L
therefore volume of acid required = 8 mL
