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Effectus [21]
2 years ago
8

Can someone explain this to me, please?

Chemistry
1 answer:
zimovet [89]2 years ago
5 0

Answer:

Isotopes have same atomic numbers, no. of protons and no. of electrons. Only their no. of neutrons and atomic mass are changed.

<u>Na - 24:</u>

Atomic Mass = 24

Atomic No. = 11

Hence,

No. of protons in Na-24 = 11

No. of neutrons = Atomic Mass - Atomic Number

No. of neutrons = 24 - 11

No. of neutrons = 13

Atomic Number = 11

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>
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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
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<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

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Answer:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in  H^+ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

Explanation:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

CO_{2} +H_{2} O ⇄ H^+ + HCO^-_{3}

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in  H^+ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

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The pictures to the right, show two different models of the atom. Type in the letter of your answer. Which model best represents
Solnce55 [7]

Answer:

Model B

Explanation:

Just did it on Edgenuity.

6 0
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