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Effectus [21]
2 years ago
5

What is the atomic number of an atom woth six valence electrons?(A) 8(B)10(C) 12​

Chemistry
1 answer:
Alexxx [7]2 years ago
5 0

Answer:

8

Explanation:

The atomic mass of an atom refers to the number of protons in the nucleus of an atom. Hence, an element with 6 valence electron mean that it has 6 electrons in its outermost shell.

The atomic Number of elements with 6 valence electrons cannot necessarily be the same, however, elements with 6 valence electrons belong to group six on the periodic table.

From the options given above, :

Atomic number of 8 :

Configuration : 2, 6

Atomic number of 10 :

Configuration : 2, 8

Atomic number of 12 :

Configuration : 2, 8, 2

Hence, only the element with atomic number of 8 has 6 valence electrons in its outer most shell, hence, the answer.

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8 0
3 years ago
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How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

5 0
2 years ago
Question
VMariaS [17]

Answer:

12.62 L

Explanation:

First, we have to calculate the moles corresponding to 18.0 g of oxygen gas (MW 32.0).

18.0 g × (1 mol/32.0 g) = 0.563 mol

Then, we can find the volume occupied by 0.563 moles of oxygen at STP (273,15 K, 1.00 atm) using the ideal gas law.

P × V = n × R × T

V = n × R × T / P

V = 0.563 mol × 0.0821 atm.L/mol.K × 273.15 K / 1.00 atm

V = 12.62 L

7 0
2 years ago
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