Answer:
The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons.
Explanation:
Explanation:
The reaction equation will be as follows.

Hence, the expression for
is as follows.
![K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH_%7B2%7DSO%5E%7B-%7D_%7B4%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BH_%7B3%7DAsO_%7B4%7D%5D%7D)
Let us assume that the concentration of both
and
is x.

x = 0.01118034
This means that the concentration of
is 0.01118034.
Since, we know that the relation between pH and concentration of hydrogen ions is as follows.
pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
= 
= 1.958
Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.
Answer:
It is an area that is covered in grasses and wildflowers that receives enough rainfall to support the grassland but not the growth of trees.
Explanation:
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol