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Mrac [35]
3 years ago
14

I need help on my homework I don’t know this answer

Chemistry
1 answer:
Tems11 [23]3 years ago
8 0

Answer:

I believe the answer would be

b.) valence electrons, all thw electrons that surround the nucleus of an atom

Explanation:

You're welcome

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3 years ago
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]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

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

Hence, the freezing point of solution is -0.454°C

3 0
3 years ago
A container of gas is initially at 0.25 atm and 0 ˚C. What will the pressure be at 125 ˚C?
Pani-rosa [81]

Answer:

0.37atm

Explanation:

Given parameters:

Initial pressure  = 0.25atm

Initial temperature  = 0°C  = 273K

Final temperature  = 125°C  = 125 + 273  = 398K

Unknown:

Final pressure  = ?

Solution:

To solve this problem, we use a derivative of the combined gas law;

           \frac{P1}{T1}  = \frac{P2}{T2}

  P and T are pressure and temperature

  1 and 2 are initial and final values

        \frac{0.25}{273}   = \frac{P2}{398}  

         P2  = 0.37atm

3 0
2 years ago
Help me I need to know how do do this or just give me answers lol
Pani-rosa [81]
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5 0
3 years ago
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