The total pressure of gases is equivalent to 739mmH divided by the area.
<h3>How to calculate the total pressure of gases?</h3>
To calculate the total pressure of gases, the next formula can be used:
- Total pressure: Force/area
- Total pressure 739 mmH/area
<h3>How to calculate the area?</h3>
The area of a cylinder is equal to 2πrh+2πr².
You can know the radio by checking the measurement of the base of the graduated cylinder. Also, you can determine the height by observing how many mm the gas is occupying. Then follow these steps:
- Find out the area based on the formula provided.
- Replace the area in the pressure formula.
- Solve the formula.
Note: This question is incomplete because the height occupied by the gas and the radio of the cylinder are not provided. Due to this, I answered this question based on general knowledge.
Learn more about cylinder in: brainly.com/question/3692256
<h3>
Answer:</h3>
Limiting reagent: Potassium iodide
Mass of the precipitate (PbI₂) is 4.453 g
<h3>
Explanation:</h3>
We are given;
- 60.0 mL of 0.322 M potassium iodide
- 20.0 mL of 0.530 M lead () nitrate
We are required to identify the limiting reactant and determine the mass of the precipitate formed.
<h3>Step 1: Write the balanced equation for the reaction</h3>
- The balanced equation for the reaction between potassium iodide and lead (II) nitrate is given by;
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂(s)
<h3>Step 2: Determine the number of moles of the reagents</h3>
Moles of KI
Moles = Molarity × volume
Moles of KI = 0.322 M × 0.060 L
= 0.01932 moles
Moles of KNO₃
Moles = 0.530 M × 0.020 L
= 0.0106 M
From the equation;
- 2 moles of KI reacts with 1 mole of Pb(NO)₂
- Therefore; 0.01932 moles of KI will require 0.00966 moles of Pb(NO₃)₂
- This means, KI is the limiting reagent while Pb(NO₃)₂ is the excess reagent.
<h3>Step 3: Determine the mass of the precipitate PbI₂</h3>
2 moles of KI reacts to produce 1 mole of PbI₂
Therefore;
Moles of PbI₂ = Moles of KI ÷ 2
= 0.01932 moles ÷ 2
= 0.00966 moles
But molar mass of PbI² is 461.01 g/mol
Therefore;
Mass of PbI₂ = 0.00966 moles × 461.01 g/mol
= 4.453 g
Therefore, the mass of the precipitate formed (Pbi₂)is 4.453 g
Answer:
838 torr
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
<em>Data:
</em>
p₁ = ?; V₁ = 2.42 L; T₁ = 27.0 °C
p₂ = 754 torr; V₂ = 2.37 L; T₂ = -8.8 °C
Calculations:
(a) Convert <em>temperatures to kelvins
</em>
T₁ = (27.0 + 273.15) K = 300.15 K
T₂ = (-8.8 + 273.15) K = 264.35 K
(b) Calculate the<em> pressure
</em>
p₁ = 754 torr × (2.37 L/2.42) × (300.15/264.35)
p₁ = 754 torr × 0.979 × 1.135
p₁ = 838 torr