Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
Answer:
we know, at STP ( standard temperature and pressure).
we know, volume of 1 mole of gas = 22.4L
weight of 1 Litre of hydrogen gas = 0.09g
so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.
similarly,
weight of 2L of a gas = 2.88gm
so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g
hence, molecular weight of the gas = 32.256g
vapor density = molecular weight/2
= 32.256/2 = 16.128g
hence, vapor density of the gas is 16.128g.
Explanation:
Amplitude is the awnser glad I could help you
Answer: 37.0 °C , 42.5 °C
Explanation:
Answer:
1.88 × 10²³ particles
Explanation:
Given data:
Volume of H₂ = 0.7 L
Number of particles at STP = ?
Solution:
First of all we will calculate the number of moles of H₂.
PV = nRT
n = PV / RT
n = 1 atm . 7 L / 0.0821 L. atm. mol⁻¹. k⁻¹ × 273.15 K
n = 7 atm. L / 22.43 L. atm. mol⁻¹
n = 0.312 mol
it is known that,
2 g H₂ = 1 mol = 6.022 × 10²³ particles
For 0.312 mol
0.312 × 6.022 × 10²³ particles
1.88 × 10²³ particles
