Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
Answer:
0.54 mole
Explanation:
CH3COOH CH3CH2OH CH3COOCH2CH3 H2O
Initial concentration 1.0 mole 1.0 mole 0 mole 1.0mol
Change - x - x + x + x
Equilibrium (1.0 - x) (1.0 - x) x (1.0 + x)
K = [CH3COOCH2CH3]*[H2O]/[CH3COOH]*[CH3CH2OH]
x*(1.0+x)/(1.0-x)(1.0-x) = 4.0
x+x²=4*(1-x)²
x+x² = 4(1² - 2x + x²)
x + x² = 4 - 8x + 4x²
4 - 8x + 4x²- x² - x= 0
3x² - 9x + 4 = 0
x=2.5 , x=0.54
2.5 mole of acid cannot be esterified, because there is only 1.0 mole of acid,
so answer is 0.54 mole.
We have 25cm^3 of 0.1mol AgNO3.
25cm^3 = 0.025L, so we have 0.025 x 0.1 = 0.0025mol AgNO3, so
0.0025AgNO3 + 0.0025NaCl = 0.0025AgCl + 0.0025NaNO3
Change in Free Energy: ΔG(20C) = -0.064kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -0.110kJ (negative, so the reaction is exothermic)
This reaction produces 0.358g of AgCl and 0.213g of NaNO3
Les Mclean PhD
2C4H10 + 13O2 -> 8CO2 + 10H2O
