200 mL of nitrogen gas at 24K is heated to 47K.
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Answer:
See explanation
Explanation:
The question is incomplete. However, base on the given information, one likely question could be that, we calculate the the volume of the nitrogen gas after heating it.
So, we have that:
Initial Volume, V1 = 200mL
Initial Temperature, T1 = 24k
Final Temperature, T2 = 47k
To solve the final volume of the gas, we apply the following ideal gas law equation
V1 * T1 = V2 * T2
Substitute values for V1, T1 and T2
200mL * 24K = V2 * 47K
200mL * 24 = V2 * 47
4800mL = 47V2
Solve for V2
V2 = 4800/47
V2 = 102.13 mL ------- Approximated
Hence, the volume of nitrogen gas after heating is 102.13mL
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Answers
Answer:
The base-dissociation constant, Kb,for the conjugate base of benzoic acid is :
Explanation:
The product of acid dissociation constant and base dissociation constant is equal to the water dissociation constant.The general formula for the reaction is:
For the acid dissociation reaction:
The conjugate base for the acid is A-
The acid is HA . and its Ka is given.
The value of Kw is fixed at a given temperature , which is equal to:
Answer:
Ka*kb=kw
Explanation:Got it right
Answers
Answers
The concentration of ions in the solution is 0.0063 M
What is pH?
pH is a quantitative measure of the acidity or basicity of aqueous or other liquid solutions.
pH or pOH is the measure of acidity or alkalinity of a solution.
pOH is calculated by taking negative logarithm of hydroxide ion concentration.
pH = -log []
pH + pOH = 14
11.8 + pOH = 14
pOH = 2.2
Putting in the values:
2.2 =-log []
[] =0.0063 M
Hence, the concentration of ions in the solution is 0.0063 M
Learn more about the pH here:
#SPJ2
Answer:
[OH⁻] = 6.31 × 10⁻³ M
SOlution:
Step 1: Calculate pOH as,
pH + pOH = 14
As, pH = 11.8
So,
11.8 + pOH = 14
Or,
pOH = 14 - 11.8
pOH = 2.2
Step 2: Calculate [OH⁻],
pOH is related to [OH⁻] as,
pOH = - log [OH⁻]
Putting value of pOH,
2.2 = -log [OH⁻]
Solving for [OH⁻],
[OH⁻] = 10⁻²'² ∴ 10 = Antilog
[OH⁻] = 6.31 × 10⁻³ M
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(2) 0.480 M (4) 0.801 M
Answers
M( HCl) = ?
V( KOH) = 12.7 mL / 1000 => 0.0127 L
M(KOH) = 0.620 M
Number of moles KOH:
n = M x V
n = 0.620 x 0.0127
n = 0.007874 moles of KOH
number of moles HCl :
HCl + KOH = H2O + KCl
1 mole HCl ------ 1 mole KOH
? mole HCl--------0.007874 moles KOH
moles HCl = 0.007874 * 1 / 1
= 0.007874 moles of HCl
M = n / V
M = 0.007874 / 0.0164
= 0.480 M
Answer (2)
hope this helps!