Answer:
c
Explanation:
1 calorie = 4.184J/g×°C
This also happens to be the specific heat capacity of water, which is the amount of energy it takes to raise the temperature of 1mL of water by 1°C
Answer:
8.0 moles
Explanation:
Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.
Using the formula: 
Concentration of acid = ?
Volume of acid = 10 mL
Concentration of base = 1.0 M
Volume of base = 40 mL
mole of acid = 1
mole of base = 1
Substitute into the equation:
Concentration of acid = 40/10 = 4.0 M
To determine the number of moles of acid present in 2.0 liters of the unknown solution:
Number of moles = Molarity x volume
molarity = 4.0 M
Volume = 2.0 Liters
Hence,
Number of moles = 4.0 x 2.0 = 8 moles
Answer:
4.86×10^23 molecule of Pb
Explanation:
Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.
So:
2 mol NH3/ 3 mol Pb
Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:
(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb
Then, we just need to use Avagadro's number to get the number of molecules.
(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
