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The reaction between 1 mole of NaOH and 1 mole of HCl creates 1 mole of NaCl and 1 mole of water. Meaning that the moles of HCl needs to equal that of NaOH for the solution to be considered equalized. That being said, you first need to find the numbers miles of HCl by multiplying the volume by the molarity to get 0.01 moles HCl. (1Mx0.01L=0.01). That means that you need 0.01 moles of NaOH. I hope that helps. Let me know if anything is unclear.
Answer:
9.80 g
Explanation:
The molecular mass of the atoms mentioned in the question is as follows -
S = 32 g / mol
F = 19 g / mol
The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol
The mass of 6 F = 6 * 19 = 114 g /mol .
The percentage of F in the compound =
mass of 6 F / total mass of the compound * 100
Hence ,
The percentage of F in the compound = 114 g /mol / 146 g / mol * 100
78.08 %
Hence , from the question ,
In 12.56 g of the compound ,
The grams of F = 0.7808 * 12.56 = 9.80 g
The concentration of the HCl solution is 0.72 M.
<h3>How do we calculate the concentration?</h3>
Concentration of the required solution by the use of the known concentration solution will be determine by using the below equation as:
M₁V₁ = M₂V₂, where
- M₁ & V₁ are the molarity and volume of the HCl solution.
- M₂ & V₂ are the molarity and volume of the NaOH solution.
On putting values in the above equation, we get
M₁ = (1)(0.018) / (0.025) = 0.72 M
Hence required concentration of HCl is 0.72M.
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