PLEASE HELP!!

If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of

irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of $C_{18}H_{21}NO_3$ = 46.85 g

Molar mass of $C_{18}H_{21}NO_3$ = 299.4 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{46.85\ g}{299.4\ g/mol}$

$Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol$

Given, Mass of $C_{2}H_{5}OH$ = 125.5 g

Molar mass of $C_{2}H_{5}OH$ = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{125.5\ g}{46.07\ g/mol}$

$Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}$

$Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22$

4 0

3 years ago

You throw a ball straight up in the air trying to throw it as high as possible. What do you think happens to the balls kinetic e

mote1985 [20]

With the info given i would have to say their is no kinetic energy, it's all potential energy.

6 0

3 years ago

The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$

Where

xB = mole fraction of solute=?

$p^{0}=23.8torr$

p = 22.8 torr

$x_{B}=\frac{23.8-22.8}{23.8}=0.042$

mole fraction is ratio of moles of solute and total moles of solute and solvent

moles of solvent = mass / molar mass = 500 /18 = 27.78 moles

putting the values

$molefraction=\frac{molessolute}{molesolute+molessolvent}$

$0.042=\frac{molessolute}{27.78+molessolute}$

$1.167+0.042(molesolute)=molessolute$

$molessolute=1.218$

mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams

3 0

3 years ago

A molecule tends to attain A geometry where the repulsive interactions with in the molecule are?

Ivenika [448]

The shape of a molecule is determined by the location of the nuclei and its electrons. ... Molecular geometry, on the other hand, depends on not only on the number of electron groups, but also on the number of lone pairs. When the electron groups are all bond pairs, they are named exactly like the electron-group geometry.

- BRAINLIEST answerer ❤️✌

3 0

2 years ago

Magnalium alloys contain 70% Al and 30.0% Mg by mass. How many grams of H2(g) are produced in the reaction of a 0.710g sample of

Elis [28]

Answer:

H2 produced = 0.4235g

Explanation:

Equations for reaction with Hcl

Aluminium

2Al + 6 Hcl ------------ 2Alcl3 + 3H2

Magnesium

Mg + 2Hcl -------- Mgcl2 + H2

Aluminium = 70% of 0.710g of sample

sample contains (70 x 0.710)/100 = 0.497g of aluminium

Magnesium = 30% of 0.710g of sample

sample contains(30 x 0.710)/100 = 0 .213g of magnesium

Moles = mass/ molecular mass

Moles of aluminium in sample = 0.497/27 = 0.0184

2 moles of aluminium gives 3 moles of H2

No of moles of H2 from reaction with aluminium = (2 x0.0184)/3

= 0.0123 moles

1 mole of H2 = 2g therefore mass of H2 produced = 0.0123 x 2 = 0.0246g

Moles of magnesium in sample = 0.213/24 = 0.008875

1 mole of mg gives 1 mole of H2

No of moles of H2 from reaction with magnesium = 0.008875 x 1

= 0.008875

1 mole of H2 = 2g therefore mass of H2 produced = 0.008875 x 2

= 0.01775g

Ttal mass of H2 = 0.0246 +0.01775 = 0,04235g

8 0

3 years ago