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Aliun [14]
3 years ago
11

PLEASE HELP!!!! WILL GIVE BRAINLIEST!!!!! Is this the only option?

Chemistry
1 answer:
Marta_Voda [28]3 years ago
7 0
I think the first one applys. I may be wrong.
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A sample of water is being heated over a flame. What happens to the water as it is heated? (2 points)
Viefleur [7K]
The answer is D, the kinetic energy of the water molecules begins to increase. When you heat up a pot of water, the water molecules begin to move very fast. Thus, stating that the kinetic energy of water molecules increase since the water molecules go from moving a little bit to moving really fast! I hope this could help!
6 0
3 years ago
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What does the mass defect measure?​
salantis [7]
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A solution has a pH of 4.20. Using the relationship between pH and pOH, what is the concentration of OH−?
frez [133]

Answer : The concentration of OH^- ion is, 1.58\times 10^{-10}M

Solution : Given,

pH = 4.20

First we have to calculate the pOH.

As we know that,

pH+pOH=14

pOH=14-pH

pOH=14-4.20

pOH=9.8

Now we have to calculate the concentration of OH^- ion.

pOH=-\log [OH^-]

9.8=-\log [OH^-]

[OH^-]=1.58\times 10^{-10}M

Therefore, the concentration of OH^- ion is, 1.58\times 10^{-10}M

4 0
3 years ago
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Scientist discover that stars in the universe are mostly composed of which elements?
konstantin123 [22]
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3 0
4 years ago
At what temperature a gas with volume 175 L at 15 oC and 760mmHg will occupy a volume of 198 L at a pressure 640mmHg?
MrRissso [65]

Answer:

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

T

1

P

1

V

1

=

T

2

P

2

V

2

where,

P_1,V_1\text{ and }T_1P

1

,V

1

and T

1

are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2P

2

,V

2

and T

2

are the final pressure, volume and temperature of the gas

We are given:

\begin{gathered}P_1=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K\end{gathered}

P

1

=760mmHg

V

1

=175L

T

1

=15

o

C=[15+273]K=288K

P

2

=640mmHg

V

2

=198L

T

2

=?K

Putting values in above equation, we get:

\begin{gathered}\frac{760mmHg\times 175L}{288K}=\frac{640mmHg\times 198L}{T_2}\\\\T_2=274K\end{gathered}

288K

760mmHg×175L

=

T

2

640mmHg×198L

T

2

=274K

Hence, the temperature when the volume and pressure has changed is 274 K

7 0
3 years ago
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