Hot chocolate is stirred with a spoon the spoon gets hot because of

400ml of 2.5M can be deluded to make how many of 2.0 HCI

ahrayia [7]

Answer:

The new volumen of the solution is 500 ml.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

V2= (C1xV1)/C2

V2= (400 ml x 2,5 M)/2,0M

V2= 500 ml

3 0

4 years ago

3) A solution contains 61.3 moles of KC103 in 0.25 L of solution. What is the

marissa [1.9K]

Answer:

The molarity of the solution is 245, 2M.

Explanation:

We calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case KCl03) in 1000ml of solution (1 liter):

0,25 L solution----- 61,3 moles of KCl03

1 L solution----x= (1 L solution x 61,3 moles of KCl03)/0,25 L solution

x=245, 2 moles of KCl03 --> The molarity of the solution is 245, 2M

6 0

3 years ago

Which structure of the cell controls what goes in and comes out of the cell?

LenKa [72]

The Cell Membrane.

The Cell Membrane wraps around the entire cell and only allows certain things to pass through, blocking out potentially harmful things from the organelles within the cell.

Hope this helps!

7 0

3 years ago

What is the percent by mass of oxygen in al2o3 ?

Snezhnost [94]

mr oxygen= 16. 16 x 3= 48
mr Al2O3= 27 + 27 + 48= 102

(48/102) X 100%= 47.05% oxygen

4 0

4 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]

Answer: The freezing point of solution is -0.974°C

Explanation:

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

8 0

3 years ago