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ZanzabumX [31]
4 years ago
9

What is the number of molecules of c2h5oh in a 3m solution that contains 4.00kg h2o?

Chemistry
1 answer:
Veronika [31]4 years ago
3 0
 The number  of C2H5OH  in a 3 m solution that contain 4.00kg H2O is calculate as below

M = moles of the solute/Kg  of water

that is 3M = moles of solute/ 4 Kg
multiply  both side by  4

moles of the  solute  is therefore = 12  moles

by use of Avogadro law constant

1 mole =6.02 x10^23  molecules

what  about 12 moles

=12 moles/1 moles  x 6.02 x10^23 = 7.224 x10^24 molecules
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Which two particles are always found in equal number in neutral atoms
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I hope it helps.

Have a great day.

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Read 2 more answers
The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
STatiana [176]

Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

S_n=\frac{n}{2}(2a+(n-1)d)

Now, it is given that:

For\ n=10,S_n=120\\For\ n=20,S_n=840

Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

4 0
3 years ago
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