The volume of 15.7 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid is 0.12 L
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Molarity of stock solution (M₁) = 15.7 M
- Volume of diluted solution (V₂) = 12 L
- Molarity of diluted solution (M₂) = 0.156 M
- Volume of stock solution needed (V₁) = ?
<h3>How to determine the volume of the stock solution needed</h3>
The volume of the stock solution needed can be obtained by using the dilution formula as shown below:
M₁V₁ = M₂V₂
15.7 × V₁ = 0.156 × 12
15.7 × V₁ = 1.872
Divide both side by 15.7
V₁ = 1.872 / 15.7
V₁ = 0.12 L
Thus, the volume of the stock solution needed to prepare the solution is 0.12 L
Learn more about dilution:
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In PV = nRT, n is the number of moles. When 8 grams of oxygen are present,
n = 8/32 = 1/4
So PV = RT/4
Answer:
when the substance is decrease
Explanation:
Answer:
They have fewer hydrogen atoms attached to the carbon chain than alkanes
Explanation:
Let's compare ethane (an alkane) with ethene (an alkene) and ethyne (an alkyne):
- Ethane's formula is C₂H₄, while ethene's is C₂H₄ and ethyne's C₂H₂.
As you can see, alkenes and alkynes have fewer hydrogen atoms attached to the carbon chain due to them having multiple bonds between the carbon atoms.
