You take a small pebble and want to find the volume. You fill your

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing point depression of

sergij07 [2.7K]

From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent.
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K

7 0

3 years ago

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

IceJOKER [234]

Answer:

$1.75\cdot 10^{-4} M$

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

$S = k_H p^o$

Where $k_H$ is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

$S_1 = 2.67\cdot 10^{-4} M$

Also, at sea level, we have an atmospheric pressure of:

$p = 1.00 atm$

Given mole fraction:

$\chi_{O_2} = 0.209$

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

8 0

3 years ago

NaOH+HCl->............

Oxana [17]

Answer:

NaCl+H20

Explanation:

It is a neutralisation reaction in which NaOH is a base and HCl is an acid. On reaction it forms salt and water.

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3 0

3 years ago

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The human body excretes nitrogen in the form of urea, NH₂CONH₂. The key step in its biochemical formation is the reaction of wat

Misha Larkins [42]

Mass perventage of nitrogen in urea, arginine, ornithine is 46.67%, 32.2%,21% respectively.

What is Mass perventage ?

Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100.

Molecular mass of urea (NH2CONH2) is= 2× (Atomic mass of N ) + 4 × (Atomic mass of H) + (Atomic mass of C) (Atomic mass of O) = 60

Mass percentage of N = $Total mass of N atoms in the compoundMass of compound×100=2860×100=46.67%$

chemical formula for arginine is C6H14N4O2

molar mass of C6H14N4O2=174g/mol

moles of N atoms in C6H14N4O2=4

mass of N atoms=14*4=56g

mass percent of N in C6H14N4O2=(56/174)*100=32.2%

the chemical formula for ornithine is C5H12N2O2

molar mass of C5H12N2O2=132g/mol

moles of N atoms in C5H12N2O2=2

mass of N atoms in C5H12N2O2=2*14=28

mass percent of N in C5H12N2O2=(28/132)*100=21%

To learn more about urea click on the link below:

brainly.com/question/17812875

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7 0

1 year ago

Aleks [24]

Electron microscopes differ from light microscopes in that they produce an image of a specimen by using a beam of electrons rather than a beam of light. Electrons have much a shorter wavelength than visible light, and this allows electron microscopes to produce higher-resolution images than standard light microscopes

3 0

3 years ago