Answer:
Are basic:
[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M
Explanation:
A solution is basic when pH = - log [H₃O⁺] is higher than 7.
It is possible to convert [OH⁻] to [H₃O⁺] using:
[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]
a. [OH⁻] = 3.13x10⁻⁷M
[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]
[H₃O⁺] = 3.19x10⁻⁸M
pH = - log [H₃O⁺] = 7.50
[OH⁻] = 3.13x10⁻⁷M is basic
b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.
This solution is not basic
c. [H₃O⁺] = 9.55x10⁻⁹M
pH = 8.02
This solution is also basic.
Answer: to calculate pH use -log[H+] or - log[OH-]..the solution is basic as the “NaOH” is attached to a hydroxide.Since we need to find the pH (per hydrogen) and not the pOH( per hydroxide) we need to find the pOH of the substance first then we subtract that by 14 so we can arrive at the pH of the substance.
Explanation: So -log( 1 x 10^(-5)) = 5 which is the pOH.Now we subtract that by 14 which gives us -9 and now you’d multiply that by -1 bcuz we can’t have a negative so the pH of the substance is 9
Answer:
20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.
Explanation:
Heat lost by solution ad calorimeter = Q
Heat capacity of solution ad calorimeter = C = 1071 J/°C
Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C
Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J
Moles of sodium nitrate = 
When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.
Then heat absorbed by 1 mole of sodium nitrate :
1 J = 0.001 kJ
20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.
