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Orlov [11]
4 years ago
15

A student carefully pipets 10.00ml of ethanol into a beaker. If ethanol has a density of 0.779g/ml, how many grams of ethanol we

re put into the beaker?
Chemistry
1 answer:
lesantik [10]4 years ago
5 0

Answer: 7.79 grams of ethanol were put into the beaker.

Explanation:

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.779 g/mL

Volume of water = 10.00 mL

Putting values in above equation, we get:

0.779g/mL=\frac{\text{Mass of ethanol}}{10.00mL}\\\\\text{Mass of ethanol}=(0.779g/mL\times 10.00mL)=7.79g

Thus 7.79 grams of ethanol were put into the beaker.

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An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm
frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0
3 years ago
Which structural formula represents an unsaturated hydrocarbon?
bija089 [108]

Answer:

4

Explanation:

your chosen answer is right as unsaturated mean contains double bond and 1st and 3rd is Wrong as Carbon only can have 4 bonds and 2nd is Wrong as it is saturated.

hope this helps :)

5 0
3 years ago
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