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bekas [8.4K]
2 years ago
6

You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t

o use only 2.5 L of it? *​
Chemistry
1 answer:
solniwko [45]2 years ago
6 0

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

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How many grams of Cl are in 465g of CaCl2
Rama09 [41]
2 ways to do this
a. find %Cl in CaCl2
2 x 35.45g/mole = 70.9g Cl
70.9g Cl / 110.9g/mole CaCl2 = 63.93% Cl in CaCl2
0.6963 x 145g = 92.7g = mass Cl

b. determine moles CaCl2 present then mass Cl
145g / 110.9g/mole = 1.31moles CaCl2 present
2moles Cl / 1mole CaCl2 x 1.31moles = 2.62moles Cl
2.62moles Cl x 35.45g/mole = 92.7g Cl
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3 years ago
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Answer:

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Explanation:

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3 years ago
PLZ HELP FAST
astra-53 [7]
#1
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3 years ago
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Over [174]
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