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Zielflug [23.3K]
3 years ago
14

A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v

olume in Ml of stock solution should the technician use?
Chemistry
1 answer:
mihalych1998 [28]3 years ago
4 0

Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

    Number of moles  = molarity x volume

Parameters given;

        volume of solution = 500mL  = 0.5L

          molarity of solution = 0.5M

   Number of moles  =  0.5 x 0.5  = 0.25moles

Now to know the volume stock to take;

           Volume of stock = \frac{number of moles }{molarity}

  molarity of stock  = 4M

                 volume  = \frac{0.25}{4}   = 0.0625L or 62.5mL

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8 0
2 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

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Answer:

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