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Vsevolod [243]
3 years ago
8

What mass of magnesium chloride is needed to make 100.0 mL of a solution that is 0.500 M in chloride ion?

Chemistry
1 answer:
miss Akunina [59]3 years ago
3 0
M = n/V

.5M = n/.100 L

n = .1 L * .5M

n= .05 mols of MgCl2

mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2 

mass of MgCl2 = 4.76 grams

4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused
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A sample of gas occupies 7.80 liters at 425°C? What will be the volume of the gas at 35°C if the pressure does not change?
balandron [24]
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A pure substance made up of molecules of the same kind of atoms is called...
slava [35]

Answer:

All substances that are  made up of the same kind of mole ules are called pure substances.

So what do you mean ?

Explanation:

5 0
3 years ago
Read 2 more answers
What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use
ch4aika [34]

Explanation:

The given data is as follows.

          M_{1} = 10 mM = 10 \times 10^{-3} M

          V_{1} = 750 ml,           V_{2} = 5 ml

          M_{2} = ?

Therefore, calculate the molarity of given NaCl stock as follows.

                  M_{1} \times V_{1} = M_{2} \times V_{2}

                  10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml

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7 0
3 years ago
An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr
Lera25 [3.4K]

Answer:

\sigma_y\ =210.2\ MPa  

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}

\sigma_y\ =Yield\ strength

d = diameter

K =Constant

\sigma_o\ =material\ constant

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

163=\sigma_o+\dfrac{K}{\sqrt 35}      ------------1

d= 17 μm ,yield strength = 192 MPa

192=\sigma_o+\dfrac{K}{\sqrt 17}           ------------2

From equation 1 and 2

192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}

K=394.53

By putting the values of K in equation 1

163=\sigma_o+\dfrac{394.53}{\sqrt 35}

\sigma_o\ =96.31\ MPa

\sigma_y=96.31+\dfrac{394.53}{\sqrt d}

Now when d= 12 μm

\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}

\sigma_y\ =210.2\ MPa

4 0
3 years ago
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