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____ [38]
3 years ago
7

Which of these scenarios involve a reaction that is at equilibrium?

Chemistry
2 answers:
inessss [21]3 years ago
7 0
The amount of products and reactants is constant.
egoroff_w [7]3 years ago
7 0

Answer:

C

Explanation:

I just took the quizz

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How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
nevsk [136]

Answer:

17.5 g

Explanation:

Given data

  • Mass of solution to be prepared: 50.0 grams
  • Concentration of the salt solution: 35.0%

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl

4 0
3 years ago
Sodium and ____ are the two primary electrolytes.Question 5 options:
Pachacha [2.7K]

C, potassium. Hope this helps.

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3 years ago
What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid?
Shkiper50 [21]
Precipitation and gas evolution
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3 years ago
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4. milk turning sour
sp2606 [1]
I think it’s a physical
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2 years ago
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The Lyman series results from excited state hydrogen atoms transiting to
Nutka1998 [239]

Answer:

I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.

n = 5 4th excited state

n = 4 3rd excited state

n = 3 2nd excited state

n = 2 1st excited state

n = 1 ground state

Here are the possible spectral lines.

n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.

n = 4 to 3, 4 to 2, 4 to 1 = 3 lines

n = 3 to 2, 3 to 1 = 2 lines

n = 2 to 1 = 1 line. Add 'em up. I get 10.

b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.

c.The E for any level is -21.8E-19 Joules/n^2

To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.

So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.

Explanation:

8 0
3 years ago
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