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4 years ago

In a 51.0-g aqueous solution of methanol ch4o the mole fraction of methanol is 0.270. what is the mass of each component?

1 answer:

4 0

Mole fraction is another way of expressing the concentration of a solution or mixture. It is equal to the moles of one component divided by the total moles.

mass CH4O = 0.270 (51) = 13.77 g methanol

water = 51 - 13.77 = 37.23 g water

Hope this answers the question. Have a nice day.

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Determine the mass density, specific volume and specific weight of a liquid whose

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3 years ago

What part of the underside of a geckos feet allows it to stick to surfaces ?

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal depression constant

m = Molality

$\Delta T_f = K_f \times m$

$1.33 = 5.12 \times m$

m = 0.26

Molality = $\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

Mass of solvent (toluene) = 15.0 g = 0.015 kg

$0.26 = \frac{Mole\ of\ compound}{0.015}$

Moles of compound = 0.015 × 0.26 = 0.00389 mol

$Mol = \frac{Mass\ in\ g}{Molecular\ weight}$

Mass of the compound = 1.450 g

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4 0

3 years ago

If atmospheric pressure suddenly changes from 1.00 atm to 0.892 atm at 298 k, how much oxygen will be released from 4.40 l of wa

garri49 [273]

Henry's law constant for oxygen is 0,0013 mol/L·atm. Air has 21,0% oxygen.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 1 atm = 0,000273 mol/l.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 0,892 atm = 0,000243 mol/l.
difference in concentration: 0,000273 - 0,000243 = 0,00003 mol/L.
n(oxygen) = 0,00003 mol/L · 4,40 L = 0,000132 mol.

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4 years ago

Fe2O3(s)+3C(s)→2Fe(s)+3CO(g) how many grams of CO are produced when 32.0 grams of C react

krok68 [10]

Answer:

74.6 g

Explanation:

The computation of the number of grams produced at the time when 32.0 grams of C react is shown below:

As we know that

But before that we need to find out the moles of c i.e carbon which is

$= \frac{Grams }{C\ molar\ mass}$

$= \frac{32\ grams}{12.0107 g/mol}$

= 2.66 mol

As we know that

1 mol of C produces 1 mol of CO

So, for 2.66 mol, the mass of CO is

$= 2.66 \times28.0101 g/mol$

= 74.6 g

Basically we applied the above equation so that we can reach to the answer

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3 years ago