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4 years ago

What is the mole fraction of NaOH in an aqueous solution that contains 15% NaOH by mass?

Chemistry

1 answer:

zhenek [66]4 years ago

6 0

Answer:

0.074

Explanation:

15% means that in 100 g of solution 15 g sodium hydroxide is present.

Mass of water = 100 - 15 = 85 g

Number of moles of sodium hydroxide:

Number of moles = 15 g/40 g/mol

Number of moles = 0.375 mol

Number of moles of water:

Number of moles = 85 g/18 g/mol

Number of moles = 4.7 mol

Moles fraction of NaOH:

moles of NaOH/ moles of solvent + moles of solute

0.375 mol/ 0.375 mol+4.7mol

0.375 mol / 5.075 mol

0.074

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Alex787 [66]

Answer:

2KBr + MgF₂ –> 2KF + MgBr₂

The coefficients are: 2, 1, 2, 1

Explanation:

KBr + MgF₂ –> KF + MgBr₂

The above equation can be balance as illustrated below:

KBr + MgF₂ –> KF + MgBr₂

There are 2 atoms of F on the left side and 1 atom on the right. It can be balance by writing 2 before KF as shown below:

KBr + MgF₂ –> 2KF + MgBr₂

There 2 atoms of K on the right side and 1 atom on the left side. It can be balance by writing 2 before KBr as shown below:

2KBr + MgF₂ –> 2KF + MgBr₂

Now, the equation is balanced.

The coefficients are: 2, 1, 2, 1

8 0

3 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago

15.0 g of cream at 16.4 °C are added to an insulated cup containing 100.0 g of coffee at 86.5 °C. Calculate the equilibrium temp

Marta_Voda [28]

115 grams total

15/115= 13%

109/115= 87%

.13x 16.4= .277

.87x 86.5= 75.255

75.255+.277= 75.532 deg C
or 75.5 as 3 significant digits

7 0

3 years ago

How do Hydrogen-1, Hydrogen-2, and Hydrogen-3 differ from each other?

Anna71 [15]

Answer:

different from the number of their neutrons

Explanation:

They each have one single proton (Z = 1), but differ in the number of their neutrons. Hydrogen has no neutron, deuterium has one, and tritium has two neutrons. ... Their nuclear symbols are therefore 1H, 2H, and 3H. The atoms of these isotopes have one electron to balance the charge of the one proton.

7 0

3 years ago

D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose?

nataly862011 [7]

Answer:

The projection of the Fisher projection of D-Fructose and D-glucose is that The carbonyl carbon in D-glucose is carbon 1 (aldehyde), whereas in D-fructose, the carbonyl group is on carbon 2 (ketone).

Explanation:

An aldehyde is a compound containing a functional group with the structure −CHO, consisting of a carbonyl center and

A ketone is a functional group with the structure RC(=O)R', where R and R' can be a variety of carbon-containing substituents.

7 0

4 years ago