Answer:
P(total) = 1110 mmHg
Explanation:
According to the Dalton law of partial pressure,
The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.
P(total) = P1 + P2 + P3+ .....+ Pn
Given data:
Sample A = 740 mmHg
Sample B = 740 mmHg
Sample C = 740 mmHg
Total pressure = ?
Solution:
<em>Sample A:</em>
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 740 mmHg × 2L/4L
P₂ = 370 mmHg
<em>Sample B:</em>
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 740 mmHg × 2L/4L
P₂ = 370 mmHg
<em>Sample C:</em>
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 740 mmHg × 2L/4L
P₂ = 370 mmHg
Total pressure:
P(total) = P1 + P2 + P3
P(total) = 370 mmHg + 370 mmHg+ 370 mmHg
P(total) = 1110 mmHg
Answer:
The molarity of a sugar solution is 2 M.
Explanation:
Molarity is a concentration measure that expresses the moles of solute per liter of solution. In this case it is calculated with the simple rule of three:
4 L of solution--------8 moles of sugar
1 L of solution ------x= (1 L of solution x 8 moles of sugar)/4 L of solution
x=2 moles of sugar---> <em>The solution is 2M</em>
<span>Displaced volume :
</span>Final volume - <span>Initial volume
</span>13.45 mL - 12.00 mL => 1.45 mL
Mass = 4.50 g
Therefore:
density = mass / volume
D = 4.50 / 1.45
<span>D = 3.103 g/mL </span>
After some thinking I have come to the conclusion that the answer is C.
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
<em />
Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
<em />
The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
<em />