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3 years ago

Can somebody help with parts c and d

Chemistry

1 answer:

Alecsey [184]3 years ago

6 0

Answer:

13. Na 14. Ne. 15. U. 16. Ca. 17. C. l 18. O. 19. Cl. 20. Si 21. U 22. N.

23. Na 24. Ne. 25. U 26. Sc. 27. N 28. O 29. Cl 30. Si 31. U 32. N.

Explanation:

The general trend as you go to the right in any one period is for the size to decrease.

The general trend is that the size increases as you go down a Group.

The general trend as you go from the bottom left to the top right is a decrease in size.

Ionization energy increases as we move to the right in a period.

It decreases as we move down a group.

Going from the bottom left to top right - the ionization increases.

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A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1

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Answer:

Approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

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It took $2911\; \rm J$ of energy to raise the temperature of this sample by $\Delta T = 92.6\; \rm ^\circ\! C$ . Therefore, raising the temperature of this sample by $1\; \rm ^\circ\! C$ (unit temperature) would take only $\displaystyle \frac{1}{92.6}$ as much energy. That corresponds to approximately $31.436\; \rm J$ of energy.

On the other hand, the energy required to raise the temperature of this material by $1\; \rm ^\circ\! C$ is proportional to the mass of the sample (also assuming no phase change.)

It took approximately $31.436\; \rm J$ of energy to raise the temperature of $57.07\; \rm g$ of this material by $1\; \rm ^\circ C$ . Therefore, it would take only $\displaystyle \frac{1}{57.07}$ as much energy to raise the temperature of $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C\!$ . That corresponds to approximately $0.551\; \rm J$ of energy.

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