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azamat
2 years ago
6

If 30mL of 0.5M KOH is needed to neutralize 2M HCl, what was the volume of the acid?

Chemistry
1 answer:
defon2 years ago
3 0
The equation for the reaction between KOH and HCl is as follows
KOH + HCl ---> KCl + H2O
the stoichiometry of KOH to HCl = 1:1
the number of KOH moles reacted = 0.5 mol /1000 cm³ * 30 cm³ 
                                                       = 0.015 mol
the number of HCl moles reacted = number of KOH moles reacted 
therefore HCl moles reacted =  0.015 mol
the molarity of HCl is 0.2 mol/dm³
0.2 mol of HCl in - 1000 cm³
Therefore volume required for 0.015 mol = 1000 cm³ / 0.2 mol * 0.015 mol 
                                                                 = 75 cm³
Therefore 75 cm³ of HCl is required

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You have shown no structure, still we can work out for the answer :).

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Third Option: </span>4-methylbutene:<span> I have drawn the structure for this name. But this name is also against IUPAC rules, and its correct name is 1-Pentene.

Fourth Option: </span><span>4,3-methylbutyne: Again incorrect name, 4,3 are two positions, but here only one substituent is given (methyl). I have drawn structure of 3,4-Dimethylbutene, this name is incorrect, and the correct name for this compound is 3-Methyl-1-pentene.

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Although today some professionals do not support that, but rather support a diet with carbohydrates and proteins.

Explanation:

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7 0
3 years ago
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Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

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Best regards!

4 0
3 years ago
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