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Tpy6a [65]
3 years ago
7

I’m supposed to draw Bohr models for all of them but I have no idea what to even draw-

Chemistry
1 answer:
Setler79 [48]3 years ago
3 0
Honestly you could just google the bohr models for each element
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Mostly located on the bottom of the leaf, open close the stomata ​
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8. Calculate (H^+), (OH^-), pOH and the pH for a 0.00024 M solution of calcium hydroxide. Must show work!
Margaret [11]

Answer:

1. [H⁺] = 2.0×10¯¹¹ M

2. [OH¯] = 4.8×10¯⁴ M

3. pOH = 3.3

4. pH = 10.7

Explanation:

From the question given above, the following data were obtained obtained:

Molarity of Ca(OH)₂ = 0.00024 M

We'll begin by calculating the concentration of the hydroxide ion [OH¯]. This can be obtained as follow:

Ca(OH)₂ (aq) —> Ca²⁺ + 2OH¯

From the balanced equation above,

1 mole of Ca(OH)₂ produced 2 moles of OH¯.

Therefore, 0.00024 M Ca(OH)₂ will produce = 2 × 0.00024 = 4.8×10¯⁴ M OH¯

Thus, the concentration of the hydroxide ion [OH¯] is 4.8×10¯⁴ M

Next, we shall determine the pOH of the solution. This can be obtained as follow:

Concentration of the hydroxide ion [OH¯] = 4.8×10¯⁴ M

pOH =?

pOH = –Log [OH¯]

pOH = –Log 4.8×10¯⁴

pOH = 3.3

Next, we shall determine the pH of the solution. This can be obtained as follow:

pOH = 3.3

pH =?

pH + pOH = 14

pH + 3.3 = 14

Collect like terms

pH = 14 – 3.3

pH = 10.7

Finally, we shall determine the concentration of hydrogen ion [H⁺]. This can be obtained as follow:

pH = 10.7

Concentration of hydrogen ion [H⁺] =?

pH = –Log [H⁺]

10.7 = –Log [H⁺]

Divide both side by –1

–10.7 = Log [H⁺]

Take the antilog of –10.7

[H⁺] = Antilog (–10.7)

[H⁺] = 2.0×10¯¹¹ M

SUMMARY:

1. [H⁺] = 2.0×10¯¹¹ M

2. [OH¯] = 4.8×10¯⁴ M

3. pOH = 3.3

4. pH = 10.7

5 0
2 years ago
Which of the following electron transitions in a hydrogen atom emits light of the shortest wavelength? Circle the correct answer
Lilit [14]

Answer:

Transition from n = 4 to n = 1  corresponds to shortest wavelength.

Explanation:

         Process                   \mid \Delta n\mid =\mid \frac{1}{(n)_{final}^{2}}-(\frac{1}{n_{initial}^{2}})\mid

             A                                         0.9375

             B                                          0.75

             C                                          0.0421

             D                                          0.1875

According to Rydberg equation for electronic transition in H-like atoms:

           \frac{1}{\lambda }=R_{H}\mid (\frac{1}{n_{final}^{2}}-(\frac{1}{n_{initial}^{2}})\mid

where, \lambda is wavelength of light emitted or absorbed, R_{H} is Rydberg constant.

So, higher the value of \mid \Delta n\mid, lower will the corresponding wavelength of light.

Hence process A will be associated with shortest wavelength.

4 0
2 years ago
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