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LenKa [72]
3 years ago
13

First to answer first gets 10 pts and brainlesit!

Chemistry
2 answers:
Kryger [21]3 years ago
6 0
A for the first and b for the second!!:))
Sati [7]3 years ago
5 0

Answer:

According to your answers last quarter for number 1

Number 2  waxing gibbous

Explanation:

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Which substance can be broken down by chemical means?
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If 3.00 mL of 0.0250 M CuSO4 is diluted to 25.0 mL with pure water, what is the molarity of copper(II) sulfate in the diluted so
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Answer:

0.00268 M

Explanation:

To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.

<u>Step 1:</u>

3.00 mL / 1,000 = 0.00300 L

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0.0250 M = moles / 0.00300 L

(0.0250 M) x (0.00300 L) = moles

7.50 x 10⁻⁵ = moles

<u>Step 2:</u>

25.0 mL / 1,000 = 0.0250 L

0.0250 L + 0.00300 L = 0.0280 L

Molarity = moles / volume (L)

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1 year ago
The amount of matter in a object is referred as:
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Explanation:

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Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane
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The question is incomplete, the complete question is;

Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.

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E) The positions of the Cl atoms induce the net formation of the terminal alkyne.

Explanation:

In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.

The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.

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