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ZanzabumX [31]
3 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]3 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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What is the mass in grams of 9.45*10^24 molecules of methanol (CH3OH)?
Angelina_Jolie [31]
Number of moles:

1 mole ---------- 6.02x10²³ molecules
? moles --------- 9.45x10²⁴ molecules

1 x ( 9.45x10²⁴) / 6.02x10²³ =

9.45x10²⁴ / 6.02x10²³ => 15.69 moles of CH3OH

Therefore:

Molar mass CH3OH = 32.04 g/mol

1 mole ------------ 32.04 g
15.69 moles -----  mass methanol

Mass methanol  = 15.69 x 32.04 / 1 => 502.7076 g


6 0
3 years ago
What is the molarity of 122.5 g of AlCl3 in 1.0 L of solution? (MM = 133 g/mol)
yawa3891 [41]
Answer: Molarity is defined as moles of solute per liter of solution. So, find the moles of solute and divide by the liters of solution.
molar mass AlCl3 = 133g/mole
moles AlCl3 = 127 g x 1 mole/133 g = 0.955 moles
liters of solution = 400 ml x 1 liter/1000 ml = 0.400 liters
Molarity = 0.955 moles/0.400 liters = 2.39 M
Explain: I looked it up on wyzant.com
4 0
3 years ago
Which of the following is not a step in the conversion of 1.3 dam3 to L?
Tanya [424]
First put the choices,and the answer is <span>(10^-3L/1mL)^3 is the correct answer.</span>
3 0
3 years ago
Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate gives, after protonolysis, a good yield of a mixture
tekilochka [14]

Answer:

use google and use the first link

Explanation:

6 0
2 years ago
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
5 0
3 years ago
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