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2 years ago

14

How do you determine the group and period of an element based on its electron configuration?

1 answer:

il63 [147K]2 years ago

4 0

Answer:

If you are given with the atomic number of an element you can find it's period number and group number. The period number is related to the number of electron occupied shells in the element and the period number is linked to its valence electrons.

Explanation:

hope it helps luv <3

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Which is the average kinetic energy of the individual particles of a substance?

zhannawk [14.2K]

Answer: Temperature

Explanation: Temperature is a measure of average kinetic energy of particles in an object. The hotter the substance, higher is the average kinetic energy of its constituent particles. When we heat a substance, the particles that constitute the substance gain some energy and begin to move faster.

6 0

2 years ago

A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem

Leto [7]

Answer : The final temperature of the copper is, $87.29^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat gained = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = $0.0923cal/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC$

Now put all the given values in the above formula, we get the final temperature of copper.

$299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)$

$T_{final}=87.29^oC$

Therefore, the final temperature of the copper is, $87.29^oC$

4 0

3 years ago

Which of the following has the electron configuration [ar] 4s23d5 ?

nignag [31]

Answer:

Mn Manganese

Explanation:

5 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

In 1953, who devopled the model that is shown below

PilotLPTM [1.2K]

What model? can you screenshot it or send a link?

7 0

2 years ago