No, don't try, it will explode close to 187 kPa
<h3>Answer:</h3>
The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).
<h3>Solution:</h3>
According to Boyle's Law, " <em>The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure</em>". Mathematically, the initial and final states of gas are given as,
P₁ V₁ = P₂ V₂ ----------- (1)
Data Given;
P₁ = 500 mmHg
V₁ = 9.0 mL
P₂ = 750 mmHg
V₂ = ??
Solving equation 1 for V₂,
V₂ = P₁ V₁ / P₂
Putting values,
V₂ = (500 mmHg × 9.0 mL) ÷ 750 mmHg
V₂ = 6.0 mL
<h3>Result:</h3>
The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).
Nitrogen is more reactive than oxygen and oxygen than chlorine
First, we will get the average pH of the two given values:
average pH = (6.4+8) / (2) = 7.2
At this average pH, the concentration of the acid from the phenol red is equal to the concentration of the base.
pH = 7.2
[H+] = 10^(-7.2) = 6.3 * 10^-8
Phenol red has the general formula HA, this gives us:
HA <.......> H+ + A-
At pH = 7.2, [H+] = [A-]
<span>Ka = [H+][A-]/ [HA]
</span>Ka = [H+] = <span>6.3 x 10^-8</span>
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
