Answer:
254.5 K
Explanation:
Data Given
initial volume V1 of neon gas = 12.5 L
final Volume V2 of neon gas = 10.5 L
initial Temperature T1 of neon gas = 30 °C
convert Temperature to Kelvin
T1 = °C +273
T1 = 30°C + 273 = 303 K
final Temperature T2 of neon gas = ?
Solution:
This problem will be solved by using Charles' law equation at constant pressure.
The formula used
V1 / T1 = V2 / T2
As we have to find out Temperature, so rearrange the above equation
T2 = V2 x T1 / V1
Put value from the data given
T2 = 10.5 L x 303 K / 12.5 L
T2 = 254.5K
So the final Temperature of neon gas = 254.5 K
As we know that,
1 mm Hg = 1 torr
So,
745 mm Hg = X
X = (745 mm Hg × 1 torr) ÷ 1 mm Hg
X = 745 torr
Also,
1 mm Hg = 133.325 Pa
So,
745 mm Hg = X
X = (745 mm Hg × 133.325 Pa) ÷ 1 mm Hg
X = 99327 Pa
Result:
Option-A (745 torr) is the correct answer.
Answer:
a) 5,3176x10⁻⁴ moles
b) 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch.
d) pH = 4,86. Acidic solution but slighty
Explanation:
a) moles of acetic acid:
9,20x10⁻³L × 57,8x10⁻³M = <em>5,3176x10⁻⁴ moles</em>
<em></em>
b) moles of sodium acetate:
56,2x10⁻³g ÷ 82,0 g/mole = <em>6,85x10⁻⁴ moles</em>
<em></em>
c) The appropriate formula to calculate is Henderson-Hasselbalch:
pH= pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
d) pH= 4,75 + log₁₀ ![\frac{[6,85x10_{-4}]}{[5,3176x10_{-4}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B6%2C85x10_%7B-4%7D%5D%7D%7B%5B5%2C3176x10_%7B-4%7D%5D%7D)
<em>pH = 4,86</em>
<em>3 < pH < 7→ Acidic solution but slighty</em>
I hope it helps!