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Gala2k [10]
3 years ago
7

Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most

likely to use?
A reverse osmosis
B osmosis
C filtration
D dilution
Chemistry
2 answers:
BARSIC [14]3 years ago
7 0
Option a is right answer that is reverse osmosis.
Veseljchak [2.6K]3 years ago
4 0

The correct option is A.

Reverse osmosis is a special type of filtration, which involves the passing of a solvent through a semi permeable membrane in the direction that is opposite to that of natural osmosis. Pressure is always used in this process to enforce separation. The method is usually used to remove ions, fine dust particles, molecules and larger particles from solvents. The method is especially popular in water treatment and purification.

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The initial pressure of a gas is 1.58 atm and occupies 1.76 L of space at constant temperature. This gas is compressed so that i
jok3333 [9.3K]

Answer:

Final volume = 0.39 L

Explanation:

V1P1 = V2P2

V2= V1P1/P2

V2= (1.76×1.58)/7.08

V2 = 0.39 L

7 0
3 years ago
A student forgot to read the label on the jar carefully and put potassium chloride in the crucible instead of potassium chlorate
lys-0071 [83]
<span>The object that was trying to be oxidized would end up being reduced. There would be no net reaction otherwise. The KCl would have simply melted after a long enough time and with the application of enough heat to the crucible.</span>
7 0
3 years ago
Read 2 more answers
Can somebody please help me please?
Mrrafil [7]

Answer:

22 in total but round it up to 23

3 0
3 years ago
12. How is the melting point of a substance related to its freezing point? Both indicate the temperature at which the solid and
krok68 [10]
Both indicate the temperature at which the solid and liquid states of a substance are in equilibrium would be your answer.

This is beacause the melting point of a substance is the same as the freezing point of a substance. At this particular temp, the substance can be either a solid or a liquid. 

hope this helps!
8 0
4 years ago
Read 2 more answers
can someone help me with these? I just started chemistry this semester, and since it's involving math I don't see myself doing g
V125BC [204]

Any non zero digits are significant. (120 has 2 sig figs)

Zeroes between significant digits are significant. (103 has 3 sig figs)

Zeroes only count if there is a decimal point and if they are following a significant digit. (.600 has 3 sig figs; .067 has 2 sig figs)

1) <u>1278.50</u> = 6 sig figs

2) <u>12</u>0000 = 2 sig figs

3) <u>90027.00</u> = 7 sig figs

4) 0.00<u>53567</u> = 5 sig figs

5) <u>67</u>0 = 2 sig figs

6) 0.00<u>730</u> = 3 sig figs

That's all I'm gonna do so yeah.

Now for the rounding, it's basically rounding but it must have the x amount of sig figs after rounding. It's pretty hard to explain.

19, I'm not really sure. Might be 120 with a decimal point at the end.

20) 5.457, alright so here, since we are rounding to three significant digits, we look at the third sig fig. The third sig fig here would be 5. Then we just round. Since the number next to 5 is 7, we round 5 up to 6.

So 5.457 rounded to 3 sig figs would be 5.46

21) 0.0008769, the third sig fig here would be 6. You round 6 up to 7 because the number next to it is 9.

So 0.0008769 rounded to 3 sig figs would be 0.000877

So I hope you get it, I'm gonna let you do the rest.

4 0
2 years ago
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