Answer:
552 g of LiNO₃
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mass of LiNO₃ =?
Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mole of LiNO₃ =?
Molarity = mole /Volume
4 = mole of LiNO₃ / 2
Cross multiply
Mole of LiNO₃ = 4 × 2
Mole of LiNO₃ = 8 moles
Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:
Mole of LiNO₃ = 8 moles
Molar mass of LiNO₃ = 7 + 14 + (16×3)
= 7 + 14 + 48
= 69 g/mol
Mass of LiNO₃ =?
Mole = mass /Molar mass
8 = Molar mass of LiNO₃ /69
Cross multiply
Molar mass of LiNO₃ = 8 × 69
Molar mass of LiNO₃ = 552 g
Thus, 552 g of LiNO₃ is needed to prepare the solution.
Answer:
mass of sodium reacted is 184.1 g
Explanation:
mass Na = X = ?
∴ mass NaCl = 468 g
∴ mass Cl = 0.248 g
∴ molar mass NaCl = 58.44 g/mol
∴ atomic mass Cl = 35.453 a.m.u
∴ atomic mass Na = 22.989 a.m.u
⇒ moles Na = (X gNa)*(mol Na/22.989 g) = X/22.989 mol Na
⇒ mass NaCl = (X/22.989 mol Na)*(mol NaCl/mol Na)*(58.44 gNaCl/mol NaCl) = 468 g NaCl
clearing "X":
⇒ ((58.44)(X))/(22.989) = 468 g
⇒ X = 184.1 g = mass Na reacted
I believe the answer is B??????????? Hope this helps
~Queensupreme
Answer:
4 mL
Explanation:
Use the density formula, d = m/v
Plug in the density and mass to solve for v:
4 = 16/v
4v = 16
v = 4
So, the volume is 4 mL
Explanation:
Since this is an equilibrium problem, we apply le chatelier principle. This principle states that whenever a system at equilibrium is disturbed due to change in several factors, it would move in a way to annul such change.
C2H4(g) + Cl2 ⇔ 2C2H4Cl2(g)
When the concentration of C2H4 is increased, there is more reactant sin the system. In order to annul this change, the equilibrium position will shift to the right favoring product formation.
When the concentration of C2H4Cl2 is increased, there is more product in the system. To annul this change, the equilibrium position will shift to the left, favoring reactant formation.
