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3 years ago

Atomic number of a Carbon atom is 6. How many electrons does it have?

Chemistry

1 answer:

Nostrana [21]3 years ago

6 0

In the first shell there is 2 electrons and on the second shell there is 4 elections.

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The nucleus of the cell

Stells [14]

Answer:

Explanation:

it is c because that is the andwer

3 0

3 years ago

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 3.6-L bulb, then filled

aleksandrvk [35]

Answer:

The gas was N₂

Explanation:

V = 3.6L

P = 2.0 atm

T = 24.0°C = 297K

R = 0.0821 L.atm/K.mol

m = 8.3g

M = molar mass = ?

Using ideal gas equation;

PV = nRT

n = no. Of moles = mass / molar mass

n = m/M

PV = m/M * RT

M = mRT / PV

M = (8.3*0.0821*297) / (2.0*3.6)

M = 28.10

Since X is a diatomic molecule

M = 28.10 / 2 = 14.05 g/mol

M = Nitrogen

X = N₂

5 0

2 years ago

On the basis of which property of the elements are they divided into metals and non metals?

kicyunya [14]

The answer is the elements in a periodic table. On the basis of the elements in the periodic table, they are divided into metals and non metals.

8 0

3 years ago

What is the answers?

ra1l [238]

the numbers are going to be small so like a power but its at the bottom

NH3, H2O2, NHO2

5 0

3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

lions [1.4K]

Answer: The equilibrium concentration of $COF_2$ is 0.332 M

Explanation:

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial: 2.00

At eqllm: 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of $COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M$

Hence, the equilibrium concentration of $COF_2$ is 0.332 M

4 0

3 years ago