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3 years ago

How many electrons will be found in the “p” orbitals in the ground state of chlorine atom?

1 answer:

7 0

Chlorine, Cl , is located in period 3, group 17 of the periodic table, and has an atomic number equal to 17 . This tells you that a neutral chlorine atom will have a total of 17 electrons surrounding its nucleus. Now, notice that the first energy level doesn't not contain a p-subshell,

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The __________ is the part of the solution that is in the greater quantity, and the __________ is the smaller portion of the sol

nikklg [1K]

Answer:

its c

Explanation:

solvent is the part of the solution that is in the greater quantity, and solute is the smaller portion of the solution

hope this helps :)

7 0

3 years ago

Read 2 more answers

at what height is an object that has a mass of 16kg, if it’s gravitational potential energy is 7500J?

dolphi86 [110]

Answer:

46.875

Explanation:

P.E=MGH

7500=16*10*h

7500=160h

h=7500/160

h=46.875

5 0

3 years ago

1. In this experiment, the procedure instructs you to dissolve solid potassium hydrogen tartrate (KHT) in two different solvents

GuDViN [60]

Answer:

Water

Explanation:

Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.

The solvent for KHT is water.

3 0

3 years ago

In the mass spectrum of the molecule phenol, C6H5OH, the approximate intensity of the peak at m/z 95, relative to the molecular

klasskru [66]

The correct answer to thi

6 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

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

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago