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
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The answer is D because when a scientist publishes a book it helps other scientists and other scientists can help the scientists that publish the book.
Prions is the answer I took it
М(СО₂)=165 г
н(СО₂)=м(углекислого газа)/М(СО₂)
Н=Nₐ*м(углекислого газа)/М(СО₂)
Н=6.022*1023 * 165/44.01=2.258*10²⁴
Moles XeF6 = 10.0g/ 245.28 g/ mol=0.0408
The ratio between F2 and XeF6 is 3:1
Moles F2 required = 3 x 0.0408=0.122
Mass F2 = 0.122 mol x 37.9968 g/ mol=4.64g
