Answer : The volume of stock solution needed are, 12.5 mL
Explanation :
Formula used :

where,
are the initial molarity and volume of copper (II) chloride.
are the final molarity and volume of stock solution of copper (II) chloride.
We are given:

Putting values in above equation, we get:

Hence, the volume of stock solution needed are, 12.5 mL
Answer:
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<span>Volume = 2.00 L
Molarity = 0.500 M
M = n / V
0.500 = n / 2.00
n = 0.500 * 2.00
n = 1.0 mole KNO</span>₃<span>
Molar mass KNO</span>₃ = <span>101.1032 g/mol
</span>
1 mole ------ <span>101.1032 g
</span>1.0 mole ------ ?
mass = 1.0 * 101.1032 g / 1
mass = 101.1032 g of KNO₃
hope this helps!