Answer:
1.714 M
Explanation:
We'll begin by calculating the number of mole in 46.8 g of NaHCO₃. This can be obtained as follow:
Mass of NaHCO₃ = 46.8 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (3×16)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 46.8 / 84
Mole of NaHCO₃ = 0.557 mole
Next, we shall convert 325 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
325 mL = 325 mL × 1 L / 1000 mL
325 mL = 0.325 L
Thus, 325 mL is equivalent to 0.325 L.
Finally, we shall determine the molarity of the solution. This can be obtained as shown below:
Mole of NaHCO₃ = 0.557 mole
Volume = 0.325 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.557 / 0.325
Molarity = 1.714 M
Therefore the molarity of the solution is 1.714 M
Answer: Option (d) is the correct answer.
Explanation:
An oxidizing agent is defined as the substance that itself gains an electron and helps in oxidation of another substance.
For example,
Here, is the oxidizing agent.
Also, a substance with more positive value of electrode potential will be the strongest oxidizing agent. Whereas a substance with more negative value of electrode potential will be the strongest reducing agent.
Therefore, out of the given options has the highest positive value of electrode potential so, is the strongest oxidizing agent.
Answer:
0.034 M
Explanation:
1/[A] = kt + 1/[A]o
[A] = ?
k= 0.04556
t= 10.0 minutes or 600 seconds
[A]o = 0.50 M
1/[A] = (0.04556 × 600) + 1/0.50
[A] = 0.034 M
Answer:
C. 6 E23 atoms Cl2
Explanation:
S.T.P:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V = 22.4 dm³ = 22.4 L
⇒ mol (n) = PV/RT
⇒ n Cl2(g) = ((1 atm)(22.4 L)/(0.082 atm.L/K.mol)(298 K))
⇒ n Cl2(g) = 0.9166 mol
⇒ atoms Cl2(g) = (0.9166 mol)*(6.022 E23 atoms/mol)
⇒ atoms Cl2(g) = 5.52 E23 atoms ≅ 6 E23 atoms