The answer is 0.2 M.
Molarity is a measure of the concentration of solute in a solution
c = n ÷V<span>
c - concentration of solute,
n - number of moles of solute
V - volume of solution
</span>
We know:
V = 50.00 mL = 0.05 L
c = ?
n = ?
Let's calculate concentration:
c = m/(V * Mr)
The molar mass of <span>(NH4)2SO4 is the sum of atomic masses (Ar) of its elements):
Mr (</span><span>(NH4)2SO4) = 2Ar(N) + 8 Ar(H) + Ar(S) + 4Ar(O)
= 2 * 14 + 8 * 1 + 32 + 4 * 16 =
= 28 + 8 + 32 + 64 =
= 132 g/mol
c = 26.42 g/ (0.05 L * 132g /mol) =26.42 g/ 6.6 L*g/mol = 4 mol/L = 4 M
Now, calculate molarity:
</span>c = n ÷V
n = c * V
n = 4 mol/L * 0.05 L
n = 0.2 mol
Answer:
N₂ = 6.022 × 10²³ molecules
H₂ = 18.066 × 10²³ molecules
NH₃ = 12.044 × 10²³ molecules
Explanation:
Chemical equation;
N₂ + 3H₂ → 2NH₃
It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
Number of molecules of nitrogen gas:
1 mol = 6.022 × 10²³ molecules
Number of molecules of hydrogen:
3 mol × 6.022 × 10²³ molecules/ 1 mol
18.066 × 10²³ molecules
Number of molecules of ammonia:
2 mol × 6.022 × 10²³ molecules/ 1 mol
12.044 × 10²³ molecules
Answer:
1.89 V
Explanation:
To calculate the standard cell potential, subtract the reduction potential of the anode from the reduction potential of the cathode.
So for your calculation,
-0.22 V - (-2.11 V)= 1.89 V
