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Veseljchak [2.6K]
2 years ago
15

Which properties of a substance could change during a PHYSICAL change?

Chemistry
1 answer:
olga2289 [7]2 years ago
4 0
It’s appearance
It’s physical form
What it feels like
How hot it gets
If it melts
Shape
Color
You might be interested in
2.(04.01 LC)
Mila [183]

Answer:

2KClO3 —> 2KCl + 3O2

The coefficients are 2, 2, 3

Explanation:

From the question given above, we obtained the following equation:

KClO3 —> 2KCl + 3O2

The above equation can be balance as follow:

There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:

2KClO3 —> 2KCl + 3O2

Now, the equation is balanced.

Thus, the coefficients are 2, 2, 3

8 0
2 years ago
There are ________ σ bonds and ________ π bonds in h3c-ch2-ch=ch-ch2-c≡ch.
FrozenT [24]
The number of sigma and pi bonds are,

          Sigma Bonds  =  16

          Pi Bonds         =   3

Explanation:
                   Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.

6 0
3 years ago
29. What is E for a system which has the following two steps:
jasenka [17]

Answer:

Zero

Explanation:

Recall that;

E = q + w

Where;

q = heat, w = work done

When heat is absorbed by the system q is positive

When heat is evolved by the system q is negative

When the system does work, w is negative

When work is done on the system w is positive

Step 1

ΔE1= 60 KJ + 40 KJ = 100KJ

Step 2

ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ

ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ

4 0
3 years ago
Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate
Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have  CH_3CH_2O^- as nucleophile. Also, this compound is also in excess. So, we will have as solvent CH_3CH_2OH a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0
2 years ago
Explain what helps convert non-rich oxygen blood cells back to oxygen rich cells?
Reil [10]

Answer:

As blood travels through the body, oxygen is used up, and the blood becomes oxygen poor. Oxygen-poor blood returns from the body to the heart through the superior vena cava (SVC) and inferior vena cava (IVC), the two main veins that bring blood back to the heart.

Explanation:

3 0
2 years ago
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