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Sphinxa [80]
3 years ago
6

List the symbols for all the elements with electron configurations that end as follows. Each n represents an energy level.

Chemistry
1 answer:
Gwar [14]3 years ago
6 0

List the symbols for all the elements with electron configurations that end as follows. Each n represents an energy level.

a. ns1

b. ns2np4

Answer: a) H, Li, Na, K, Rb, Cs, and Fr

b) O, S, Se, Te, Po

Explanation:

a. Determining the elements correlating to the electron configuration of "ns1" , you should understand that this suggests that the elements are in Group 1A.

This is because that there is one electron in an s-orbital. The orbital is located in the first column of the periodic table.

Thus, the elements that have this electron configuration are H, Li, Na, K, Rb, Cs, and Fr.

b. Determining the elements correlating to the electron configuration of ns^2np^4 characterizes group 6A of the periodic table.

This is because that there is have six valence electrons in their highest-energy orbitals (ns2np4). The orbital is located in the first column of the periodic table.

Thus, the elements that have this electron configuration are the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po).

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Which statement is true about the radii of Na and Na+?
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Na has a larger atomic radius than Na+.

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2 years ago
Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH
uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
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Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

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where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

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So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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Yes, because you made a new substance to make a gas.

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