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seropon [69]
3 years ago
10

Draw the three alkenes, each of formula C5H10, that will form 2-methylbutane upon hydrogenation?

Chemistry
1 answer:
tiny-mole [99]3 years ago
6 0
Answer is in Word document below.
Hydrogenation is reaction of organic compound with hydrogen. In this chemical reaction, three alkenes react with hydrogen and produce alkanes. Double bond in alkenes breaks up and single bond is formed. Three alkenes are all 2-butene with one methyl group substituent.
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If you start with .235 g nickel, tell me how much of the compound product will be produced.
Yuliya22 [10]

Answer:

Explanation:

There are many compounds that are produced from Nickel.

Molar mass of Ni = 58.7g

So for 0.235g Ni, there are 0.004M.

By conservation of mass, the compound produced will contain 0.04M of Ni.

8 0
3 years ago
A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L
Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l)  ⇄ N₃⁻(aq) + H₃O⁺(aq)          

The pH of this solution is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq)  ⇄  HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles

Now, the number of moles of HN₃ is:

\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles

Then, the pH of the buffer solution after the addition of HCl is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.    

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0
3 years ago
How many grams of sulfur trioxide are produced 18 mol O2 react with sufficient sulfur? Show all your work S8 +12O 2&gt; 8SO3
SOVA2 [1]

Answer:

m_{SO_3}=2.31x10^4gSO_3

Explanation:

Hello!

In this case, since the reaction between sulfur and oxygen is:

S_8 +\frac{1}{2} O_2 \rightarrow 8SO_3

Whereas there is a 1/2:8 mole ratio between oxygen and SO3, and we can compute the produced grams of product as shown below:

m_{SO_3}=18molO_2*\frac{8molSO_3}{1/2molO_2} *\frac{80.06gSO_3}{1molSO_3} \\\\m_{SO_3}=23,057.3gSO_3=2.31x10^4gSO_3

Best regards!

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3 years ago
Any girls got dating advice for me? I am a guy
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Keep ur stuff clean, be respectful and nice
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3 years ago
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PLEASE ROUND TO THE 2ND DECIMAL PLACE!!!
Digiron [165]
I don’t know ask you’re teacher
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3 years ago
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