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3 years ago

What is the density of an object that has a mass of 120 g and volume of 5 ml?

Chemistry

1 answer:

Viktor [21]3 years ago

3 0

The density of an object ρ = 24 g/ml

<h3>Further explanation</h3>

Given

mass of an object = 120 g

volume = 5 ml

Required

The density

Solution

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

Input the value :

ρ = 120 g : 5 ml

ρ = 24 g/ml

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

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

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

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3 years ago

If an acid is added to different samples of buffered water, the sample with the

Helen [10]

Answer:

<h3><u>A). react with acid that is added and make a base.</u></h3>

explanation:

<em>Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added.</em>

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2 years ago

Read 2 more answers

The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arg

almond37 [142]

Answer:

pH 9,8 is likely to work best for this separation

Explanation:

Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).

For lysine, PI is:

$pH = \frac{1}{2} (9,1+10,5) =$ 9,8

For arginine:

$pH = \frac{1}{2} (9,0+12,5) =$ 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, <em>pH 9,8 is likely to work best for this separation</em>

I hope it helps!

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What substance is oxidized in the following reaction? 4HCl + MnO2 → Cl2 + 2H2O + MnCl2

Blizzard [7]

Answer:

Cl⁻ was oxidized.

Explanation:

4HCl + MnO₂ → Cl₂ + 2H₂O + MnCl₂

Oxidation can be defined as the process in which the oxidation number of a substance increases.

On the left side of the equation, Cl has a charge of -1 (in HCl); while on the right side of the equation Cl has a charge of 0 in Cl₂.

Thus, Cl⁻ was oxidized.

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Which planets' orbits is the belt located between the main asteroid belt?

hichkok12 [17]

<span>circumstellar is the answer</span>

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