Answer:
Explanation:
The change in enthalpy of a substance when heated is given by
ΔH = m x Csp x ΔT
so the enthalpy change is dependent on the specific heat , i.e its capactity to absorb heat, and is not influenced by factors such as being closer to the melting point, volume , potential energy or being a solid or liquid.
Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
Answer:
0.00840
Explanation:
The computation of the mole fraction is as follow:
As we know that
Molar mass = Number of grams ÷ number of moles
Or
number of moles = Number of grams ÷ molar mass
Given that
Number of moles of CaI2 = 0.400
And, Molar mass of water = 18.0 g/mol
Now Number of moles of water is
= 850.0 g ÷ 18.0 g/mol
= 47.22 mol
And, Total number of moles is
= 0.400 + 47.22
= 47.62
So, Molar fraction of CaI2 is
= 0.400 ÷ 47.62
= 0.00840
Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
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