The closest star to our Sun is Proxima

Chemistry

1 answer:

Aneli [31]1 year ago

8 0

Answer: 40,132,418,653,700 km

Explanation:

1 light year=9460730470000 kilometers

4.242*9460730470000 km=40132418653700 km

You might be interested in

How can you find a tutor??

yanalaym [24]

Depending on what you are doing, virtual, or in school, you can contact your school and you ask them, or you can have your teacher, based on the subject, tutor you.

7 0

3 years ago

Read 2 more answers

What is the difference between theme and plot?

VladimirAG [237]

Answer:

should be D

Explanation:

3 0

2 years ago

Read 2 more answers

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

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 = 2

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

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$