Answer:
The oxidation number of oxygen is -2 and hydrogen is +1
Hope this helps ^-^
Explanation:
I attached the working and the answer to the question below.
<span>Note that in the formula, C = speed of light, ν = frequency, λ= wavelength
</span><span>
The wavelength of a 4.3 x 10</span>¹⁵<span> /s wave is
6.98 </span>
× 10⁻⁶.
Answer: from the lies this is synthesis reaction
Explanation:
It's also oxidation-reduction reaction. Li is oxidised and
H is reduced
Answer:
1.75M
Explanation:
The problem deals with finding the molarity of the given compound.
The compound is:
Na₂SO₄.10H₂O
Mass of the compound = 10.5kg = 10500g
Volume of the compound = 18.6L
Molarity is the number of moles of solute in a solution;
Molarity =
Number of moles =
Molar mass of Na₂SO₄.10H₂O = 2(23) + 32 + 4(16) + 10[2(1) + 16]
= 322g/mol
Now;
Number of moles =
= 32.6mole
So;
Molarity =
= 1.75M
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05