2.23 moles of propane react when 294 g of CO₂ is formed .
<h3>What is moles ?</h3>
Moles is a unit which is equal to the molar mass of an element.
A reaction is given
C₃H₈ +50₂ → 3CO₂ + 4H₂O
Grams of CO₂ formed = 294 gm
In moles = 294 /44 = 6.68 moles.
Let x be the moles of C₃H₈ is x
Mole ratio of CO₂ to C₃H₈ = 3 : 1
so
6.68 /x = 3/1
x = 6.68 /3 = 2.23 moles
Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .
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Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>
The process of preparing solutions from stock solutions of higher concentration is known as dilution.
Dilution is done with the aid of the dilution formula given below:
where
- C1 is the concentration of stock solution
- V1 is the volume of stock solution required to prepare a diluted solution
- C2 is the concentration of the diluted solution prepared
- V2 is the final volume of the diluted solution
From the data provided:
C1 is not given
V1 is unknown
C2 = 25%
V2 = 12 mL
- Assuming C1 is 50% solution
Volume of stock, V1, required is calculated as follows:
V1 = C2V2/C1
V1 = 25 × 12 /50
V1 = 6 mL
Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
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The answer is (3) pH 3 to pH 1. The pH is related to the concentration of H3O+ with the relationship: pH = -lg c(H+). So when concentration of H3O+ increase, the pH will decrease. And decrease 2 when when increase hundredfold. Because 100=10^2.
Answer &Explanation:
From Avogadro's lawa equal volume of gas contain equal number of moles
V=N
Hence
13L=965
XL=3.2mol
Hint:as the question state is increase to iteans final mol was 3.2 but if it could state by it would mean initial moles plus adde moles ie 3.2mol)
Cross multiplication
The new volume will be
=(3.2mol×13L÷965mol)
=0.043L
