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Reika [66]
3 years ago
10

Calculate the vapor pressure (in torr) at 298 k in a solution prepared by dissolving 15.3 g of the non-volatile non-electrolye u

rea {co(nh2)2} in 107 g of water. the vapor pressure of water at 298 k is 23.76 torr. enter your answer to 2 decimal places
Chemistry
1 answer:
ZanzabumX [31]3 years ago
7 0
In colligative properties like vapor pressure, boiling point, freezing point and osmotic pressure, only the quantity of solute may affect their values. Their identity does not affect these properties. When solute is added to a solution, the vapor pressure decreases, that's why it is called vapor pressure lowering. The equation is

ΔP = xP°, where ΔP is the difference of the vapor pressure of the solution and of the solvent. P° is the vapor pressure of the pure solvent while x is the mole fraction of solute. The molar mass of urea is 60.06 g/mol while that of the water is 18 g/mol.

x =mol urea/total mol = (15.3/60.06)/[(15.3/60.06)+(107/18)]
x = 0.0411

Then,
ΔP = xP°=0.0411*23.76 = 0.976 torr

Therefore, the vapor pressure of the solution is 23.76 - 0.976 = 22.78 Torr.
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Aspartame, phenylalanine, and aspartic acid were examined by thin‑layer chromatography (TLC) using ninhydrin as the visualizatio
Marizza181 [45]

Answer:

Explanation:

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The TLC plate should be laid on a paper towel, to ease picking it from the bench top.

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Spills should be cleaned up and contaminated surfaces should be washed with water.

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4 0
3 years ago
How many grams of glucose, C6H12O6, in 2.47 mole?
statuscvo [17]
First, we need to find the atomic mass of C_{6}H_{12}O_{6}.

According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16

As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of  C_{6}H_{12}O_{6} = 6 * 12.01 + 12 * 1.008 + 6 * 16

The molar mass of  C_{6}H_{12}O_{6} = 180.156 grams/mole

Now that we have the molar mass of  C_{6}H_{12}O_{6}, we can find the grams of glucose by using:

mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)

Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams

Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams

-i
6 0
3 years ago
Read 2 more answers
Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after
Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is 2.83\times 10^{22}

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature of the air = 37^oC=[37+273]K=310K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of air = ?

Putting values in above equation, we get:

1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol

Hence, the number of moles of air present in the RV is 0.047 moles

  • <u>For b:</u>

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.047 moles of air will contain (0.047\times 6.022\times 10^{23})=2.83\times 10^{22} number of gas molecules.

Hence, the number of molecules of gas is 2.83\times 10^{22}

7 0
3 years ago
Wharis the mass in grams of 2.000 mol of oxygen atoms?
Alika [10]

Answer:

32.00 g. Hope this helps! PLEASE GIVE ME BRAINLIEST!!!!! =)

8 0
3 years ago
The elements that contain electrons in an F sublevel near the highest occupied energy level are referred to as?
Mrac [35]

inner transmission levels i took the test

3 0
3 years ago
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