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erica [24]
3 years ago
13

Difference between saturated and supersaturated solutions

Chemistry
1 answer:
Luba_88 [7]3 years ago
5 0
Saturated had as much solute that it can hold, supersaturated holds more than can normally be dissolved.
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Which statement does NOT apply to Alkenes?
nikdorinn [45]

Answer:

They are more stable than alkanes

Explanation:

  • <em><u>Alkenes</u></em><em><u> are a type of unsaturated hydrocarbons </u></em>which means they have a<u> double bond</u> in their structure, or lack maximum number of hydrogen atoms on each carbon.
  • Alkenes have a general formula of CnH2n. They are called <u>unsaturated hydrocarbons</u> since they have a double bond. They are therefore less stable compared to alkanes and also are readily reactive.
6 0
3 years ago
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you have a solution of water and salt. if the total solution weighs 50 grams and the salt weighs 4 grams. what is the percent co
Natali5045456 [20]

Answer:

92%

Explanation:

I believe. Hope this was helpful.

8 0
2 years ago
The gas in a cylinder has a volume of 7 liters at a pressure of 107kPa. The pressure of the gas is increased to 208kPa. Assuming
VashaNatasha [74]
By Boyles Law (P1V1=P2V2), substituting values in and solving for V2, we find that the new volume is 3.6 L<span />
6 0
3 years ago
Why do atoms tend to be bonded to other atoms?
Aleks04 [339]

Answer:

Atoms are often more stable when bonded to other atoms

Explanation:

Like for example let's say ionic bonds..... Since one atom has to lose specific electrons to be stable and the other needs the electrons from the other atom to be stable.....

3 0
3 years ago
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If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain afte
Assoli18 [71]

Answer:

m=0.127g

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g

Best regards.

6 0
3 years ago
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