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nlexa [21]
3 years ago
6

15. If you dilute a 6 M solution of HCl from 5 mL to 50mL, what is the concentration of this new solution? (M1V1 = M2V2)

Chemistry
1 answer:
Andreyy893 years ago
6 0

Answer:

B) 0.6M

Explanation:

               I apologize in advance if it is not correct :l

The (M1V1= M2V2) is given for you to plug in the correct numbers so let's jot this down.

                    (M1*V1= M2*V2)

so they give us 6M which would be our (M1), from this we can also conclude that 5mL is also V1; ( if you notice the M1's and V1's are always found next to eachother). This leads us to our 50mL, this would be our V2 because the volume went from 5mL to 50mL. Now lets put this in order based on what we know.

M1= 6M                                (M1*V1= M2*V2)

V1= 5mL

M2= ?

V2= 50mL

now we plug in what we know into the equation to find the unknown (M2)

                   (6M*5mL= M2*50mL)

now we could do the long math, but I don't think your on brainly to do the hard way. so lets keep it simple!

                   We are going to put the 50mL under the (6M*5mL) for division.

                             \frac{(6M*5mL)}{(50mL)} This is honestly MUCH easier, than manually answering. you just put that in the calculator and it'll give you B) 0.6M

                 

   honestly though I might not know what I'm doing cuz im currently doing my test and decided to answer this question ;)

        Good Luck!

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The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

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37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

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y = 0.241 = 24.1 %

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37Cl = 36.96590258 amu = 24.1 %

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