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Fynjy0 [20]
3 years ago
13

Why can you change physical properties and not make a new substance but can't change chemical properties and stay the same?

Chemistry
1 answer:
maxonik [38]3 years ago
8 0

Answer:

Well this has a good chance of being wrong but i wanna say,

When you change a physical property of something it doesn't affect the chemicals that make it up. Like Ice, you can freeze water to make ice. You change a physical property (state of matter) but it's chemical properties don't change because in the end it's still water.

However if you remove a chemical property from something you are changing what made the new substance with will also change the substance along with it.

That's just what I think though

Explanation:

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How are saturated and supersaturated solutions similar?
Serggg [28]

Answer:

They are similar in sense that both cannot dissolve any more solid unless heat or other factors are added. For eg if a solution is saturated it can no longer dissolve the given substance. But if the solution is heated, the solid will dissolved this is now said to be supersaturated.

Explanation:

4 0
2 years ago
Please HELP!!! I DESPERATELY NEED THESE TO BE CORRECT!
kramer

Answer:

i only know 6 whitch is the +ion

Explanation:

6 0
3 years ago
Read 2 more answers
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
Leto [7]

Answer:

0 g.

Explanation:

Hello,

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

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
2 years ago
Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0
nexus9112 [7]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

To calculate the amount of heat absorbed or released, we use the following equation:

q=mc\Delta T    .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of  water = 4.186 J/g ° C      

\Delta T = Change in temperature

  • <u>For Trial 1:</u>

We are given:

m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J

Putting values in equation 1, we get:

q=30g\times 4.186J/g^oC\times 40^oC

q = 5023.2 J

  • <u>For trial 2:</u>

We are given:

m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J

Putting values in equation 1, we get:

q=40g\times 4.186J/g^oC\times 10^oC

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = (5023.2-1674.4)J=3347J

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

4 0
3 years ago
HELP!! ASSIGNMENT DUE TODAY! PLEASE HELP!!! I'LL GIVE 21 POINTS!!!!
ra1l [238]

mitochondria

im not 100% sure but it makes sense

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