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Digiron [165]
3 years ago
15

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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Answer:

The final volume of the balloon is = 28.11 L

Explanation:

Initial pressure P_{1} = 1.03 atm = 104.325 K pa

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We know that

\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }

Put all the values in above formula we get

\frac{(104.325)(22.4)}{299} = \frac{(82)(V_{2} )}{295}

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Explanation:

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iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction \frac{a}{b}

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

\frac{15.39 g  of gold}{2.77 g of chlorine}

The proportion is the equal relationship that exists between two reasons and is represented by:    \frac{a}{b}=\frac{c}{d}

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

\frac{15.39 g  of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}

Solving for the mass of gold gives:

mass of gold=1 g of chlorine*\frac{15.39 g  of gold}{2.77 g of chlorine}

mass of gold= 5.56 grams

So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>

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