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uysha [10]
2 years ago
7

Please help find out code?

Chemistry
1 answer:
Nat2105 [25]2 years ago
4 0
You might need to take more pictures so we can see all the equations clearly
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The solubility of solids______ when the temperature is increased.
Aleksandr-060686 [28]
The answer is increases
4 0
3 years ago
Three substances A, B, and C melt at 00C,500C and -1500C
Drupady [299]
Answer:

That information is better presented and analyzed in a table.

This table shows you all the information and the answers:


Substance         melting point   boiling point    room temperature    conclusion
                                    °C                  °C                      °C                    (state)


A                                  0                    100                    25                    liquid

B                                  50                  200                    25                    solid
C                               -150                   10                     25                    gas

Explanation:

1) Substance A at 25° is above the melting point and below the boiling point, then it is liquid (just like water)


2) Substance B at 25°C is below the melting point, so it is solid.

3) Substance C at 25°C is above the boiling point, so it is gas.
7 0
3 years ago
When energy transforms, which of the following does it produce?
Aliun [14]

motion

Explanation:

it works in our daily life

8 0
2 years ago
If all the deer disappeared from this community, which change would be most
butalik [34]

Answer:c

Explanation:

4 0
2 years ago
Read 2 more answers
What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The
Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}

We are given:

Mass of _{19}^{40}\textrm{K} = 39.963998 u

Mass of _{20}^{40}\textrm{Ca} = 39.962591 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(39.963998-39.962591)=0.001407u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.001407u)\times c^2

E=(0.001407u)\times (931.5MeV)    (Conversion factor:  1u=931.5MeV/c^2  )

E=1.3106MeV

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0
3 years ago
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