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Ludmilka [50]
2 years ago
9

How many mL of a 0.300 M AgNO3 solution will it take to make 500 mL of a 0.100 M AgNO3 solution?

Chemistry
2 answers:
AfilCa [17]2 years ago
7 0

Answer: We need 166, 7 mL of AgNO3 [0.300M] to prepare 500mL of AgNO3[ 0.100M]

Explanation:

When calculating the amount of volumen from an initial solution in order to prepare other one. we always use this formula for dilutions.

C1V1=C2V2

  • First we gather the information that will be replaced in our equation.

Inicial conditions that will be represented by  number 1

C1= 0.300M AgNO3

V1=  ?

Final conditions that will be represented by  number 2

C2= 0.100M AgNO3

V2=  500mL

  • Second, we replace the parameters giving in the equation and find out the value for V1.

V1= C2. V2 / C1

V1= 0.100M. 500mL /0.300M

V1= 166.7 mL

sergeinik [125]2 years ago
3 0
The answer is 166 mL. You use the formula V1*M1=V2*M2. For what you're given you rearrange the formula to V1=M2*V2/M1 so V1= (0.500 mL)(0.100 M AgNO3)/(0.300 M AgNO3) = 166 mL
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Answer:

19,26 kJ

Explanation:

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W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×\frac{1molZn}{65,38g}×\frac{1molH_{2}(g)}{1molZn} = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

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V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

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W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×\frac{101,325}{1atmL} = <em>19,26 kJ</em>

I hope it helps!

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2 years ago
An electrically neutral atom of gallium has 31 electrons and 39 neutrons. What is the mass number for an atom of gallium?. A. 31
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Answer:

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Explanation:

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