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Stolb23 [73]
4 years ago
6

1. Write the complete electron configuration for each atom on the blank line. Lithium Fluorine Carbon Argon Sulfur Nickel Rubidi

um Xeon 2. What elements are represented by each of the following electron configurations? a. 1s2 2s2 2p5 b. 1s2 2s2 2p6 3s2 3p6 4s2 c. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p65s2 4d10 5p4 d. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
Chemistry
1 answer:
suter [353]4 years ago
3 0

Answer:

Explanation:

Answer 1:

Lithium : 1s2 2s1  Fluorine: 1s2 2s2 2p5   Carbon: 1s2 2s2 2p2

Argon : 1s2 2s2 2p6 3s2 3p6   Sulphur: 1s2 2s2 2p6 3s2 3p4

Nickel: 1s2 2s2 2p6 3s2 3p6 3d8 4s2  Rubidium: 1s2 2s2 2p6 3s2 3p6 3d10  4s2 4p6 5s1   Xenon: 1s2 2s2 2p6 3s2 3p6 3d10  4s2 4p6 4d10 5s2 5p6

 

Answer 2:  A. Fluorine B. Calcium  

C. It is Tellurium if this was the exact electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4 you intend to write, if not, no element has such electonic configuration.

D. Bromine but the correct electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5

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Answer:

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Oxygen in product = 8

Oxygen in reactant = 8

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1. A student attempts to make a saline solution by adding salt to water. How much water did
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111.1 mL of water

Explanation:

Weight per volume concentration (w/v %) is defined as

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A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
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The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of -6.4^oC and a molal freezing point depression constant K_f=3.96^oC.kg/mol. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at -13.6^oC . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is 1.3\times 10^2g

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}           ......(1)

where,

Freezing point of pure solvent = -6.4^oC

Freezing point of solution = -13.6^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant = 3.96^oC/m

m_{solute} = Given mass of solute (glycine) = ?

M_{solute} = Molar mass of solute (glycine) = 75.07 g/mol

w_{solvent} = Mass of solvent = 950. g

Putting values in equation 1, we get:

-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g

Hence, the mass of glycine that can be dissolved is 1.3\times 10^2g

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