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notsponge [240]
3 years ago
13

Retinol is vitamin A and one of the components in multivitamin tablets. A 103.0 mg sample of retinol was dissolved in 1.000 g of

chloroform (Kb = 3.630°C/m), increasing the boiling point of chloroform by 1.305°C.
Chemistry
1 answer:
Eddi Din [679]3 years ago
6 0
The formula for boiling point elevation is:
deltaT = (Kb)(m), where m is molality of moles solute per kilogram solvent
1.305 = 3.630m
m = 0.3595 = moles/kg solvent
Since we have 1 g of chloroform solvent, this is equivalent to 0.001 kg.
The number of moles is 103 mg or 0.103 g, divided by the molar mass (MM).
0.3595 = (0.103/MM) / 0.001
0.0003595 = 0.103/MM
MM = 286.51 g/mol
Therefore, the molar mass of retinol is 286.51 g/mol
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What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the
Maurinko [17]

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

w/w\% = \frac{x}{m}\times 100

3\%=\frac{x}{5.125 g}\times 100

x=\frac{3\times 5.125 g}{100}=0.15375 g

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol

0.004522 moles of hydrogen peroxide molecules are present.

5 0
3 years ago
Why is combustion of fuel necessary for heat engines to do work?
azamat

Because during combustion reaction, heat energy is released and it's this energy that is converted to work

6 0
3 years ago
Read 2 more answers
4. How many moles of KBr are found in 3 Liters of 0.4 M solution?
Allisa [31]

There are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>

The number of moles of a substance can be calculated by multiplying the molarity by the volume.

No. of moles = Molarity × volume

According to this question, 3L of a KBr solution are contained in a 0.4M.

no. of moles = 3L × 0.4M = 1.2moles

Therefore, there are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

Learn more about no. of moles at: brainly.com/question/14919968

4 0
2 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
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