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Irina-Kira [14]
3 years ago
10

How could I study for my science test on science inquiry?

Chemistry
1 answer:
Debora [2.8K]3 years ago
7 0

You could make flashcards. Or review previous lessons about the subject.

Hope this helped! :D

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Write formulas and names for two binary molecular compounds
lyudmila [28]

Answer:

NaCl sodium chloride

HF hydrogen flouride

Explanation:

7 0
3 years ago
A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A
anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

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3 years ago
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What are some physical properties of a candle
Troyanec [42]
The wick and the wax

Sorry if that was useless, I'm not sure how generalized you were being

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3 years ago
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Calculate the number of atoms of argon in 75.0 L of argon at STP.<br> ?????
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3 years ago
Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me
LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0
3 years ago
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