Which element has a greater ionization energy, rubidium or iodine?
1 answer:
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Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
- HA ⇄ H⁺ + A⁻
- Ka = [H⁺][A⁻]/[HA]
We <u>calculate [H⁺] from the pH</u>:
- pH = -log[H⁺]
- [H⁺] =
- [H⁺] = 9.12x10⁻⁴ M
Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
- [HA] * 0.66/100 = [H⁺]
We <u>calculate [HA]</u>:
- [HA] = 0.138 M
Finally we <u>calculate the Ka</u>:
- Ka =
= 6.02x10⁻⁶
Answer:
a. 478.69 K
b. 939.43
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
= 785
= 0.000785
= 400K
= 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 ⋅Pa⋅
⋅
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5 - (34980 + 35.5
)]
83.8 = (0.03 × 35.5) ( - 400K)
83.8 = 1.065 ( - 400)
78.69 = ( - 400)
= 400 + 78.69
= 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
/
=
/
=
/
= (785 × 478.69)/400
= 939.43
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P(-
) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J