Answer:
The two atoms will contribute and share two electrons each in a double covalent bond, to form a molecule of the element.
Explanation:
Following the rule which says that an octet structure in the valence electrons shell of an atom, when two atoms of an element with the electronic configuration 1s²2s²2p⁴ comeclose to each other, they will share there electrons in covalent bonding to form a molecule, and thus attain stability by achieving an octet structure.
An atom of the element with electronic configuration 1s²2s²2p⁴, has 6 valence electrons and thus, needs two more electrons in order to achieve an octet. This is achieved by sharing two electrons with another atom of the same element which contributes two electrons as well to form a double covalent bond and a molecule of the element. Therefore, each atom would then have an octet structure in their valence shells.
Answer:
oh i love football. i know this. The guy with 200 kilograms because he is heavier o that means more weight going against the guy. Is this American football. cause i love American football
Explanation:
Answer:
Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole
Explanation:
Using Q = m x Cs x ΔT ................................(1)
where Q = Amount of heat absorbed
m = Mass of solution
Cs = Specific heat capacity of solution
ΔT = Change in temperature
Given Density of solution = 1.20 g/ml
And volume of solution = 150 ml
Mass of solution = density x volume
= 1.2 x 150
= 180 g
From equation (1)
Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj
So, ΔH of the dissolution of the unknown solid =
=
= 4.6 kj/mole
Answer:
- 0.1852
- 0.0947
- 0.7201
- 3.0345 kg CO
/ Kg C
H
- 15.3848 Kg air / kg C
H
Explanation:
Molar masses of each product are :
Butane = 58 kg /kmol
Oxygen = 32 kg/kmol
Nitrogen = 28 kg/kmol
water = 18 kg/kmol
<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>
= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )
= 176 / 950.32
= 0.1852
<em><u>2) Calculate the mass fraction of water </u></em>
= ( 5 * 18 ) / (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))
= 90 / 950.32
= 0.0947
<em><u>3) Calculate the mass fraction of Nitrogen </u></em>
= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))
= 684.32 / 950.32
= 0.7201
<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>
Mco2 = ( 4 * 44 ) / 58 = 3.0345 kg CO
/ Kg C
H
<u>5) Mass of Air required per unit of fuel mass burned </u>
Mair = ( 6.5 * 32 + 24.44 *28 ) / 58 = 15.3848 Kg air / kg C
H