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Marianna [84]
3 years ago
6

If a object has a density of 0.55 g/mL , what is it density in cg/L

Chemistry
2 answers:
chubhunter [2.5K]3 years ago
5 0

Answer : The density of a object in cg/L is, 55000 cg/L

Explanation :

As we are given that the density of an object is, 0.55 g/ml.

We have to convert g/ml into cg/L.

As, 1 gram = 100 centigram

As, 1000 milliliter = 1 liter

0.55 g/ml = \frac{0.55g}{1ml}\times \frac{100cg}{1g}\times \frac{1000ml}{1L}=55000cg/L

Therefore, the density of a object in cg/L is, 55000 cg/L

Tpy6a [65]3 years ago
3 0

Answer : The density in cg/L is 0.55\times 10^5cg/L.

Solution :

Conversions :    1g/ml=10^{5}cg/L

Conversion of 'g' to 'cg' :    1g=100cg

Conversion of 'L' to 'ml' :    1L=1000ml

Given density = 0.55g/ml

After conversion,

Density = 0.55\times 10^5cg/L

Therefore, the density in cg/L is 0.55\times 10^5cg/L.


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A cube has sides that are 0.03 m . What is the volume of the cube in liters?
Marizza181 [45]

Answer:

0.027 litres

Explanation:

volume of cube = length × base area

volume of cube = 0.03m ×( 0.03m × 0.03m )

volume of cube = 0.03m × ( 0.0009m^2 )

volume of cube = 0.000027m^3

1 cubic metre = 1000 litres

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8 0
3 years ago
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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Answer:

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RFM 24 RFM 95

moles 2.708 moles 2.708

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Natali5045456 [20]

Answer:

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8 0
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