There are six electrons in the covalent bonds.
Two N atoms would be :N:· + ·:N:
An N₂ molecule would be :N:::N: or :N≡N:
This gives each N atom an octet of eight electrons in its valence shell.
Answer:
8.99×10^-7m
Explanation:
The wavelength can be calculated using the expression below
E=hcλ
Where E= energy= 2.21 x 10^-19 J.
C= speed of light= 3x10^8 m/s
h= planks constant= 6.626 × 10^-34 m2 kg / s
E=hcλ
λ= E/(hc)
Substitute for the values
λ=( 2.21 x 10^-19 )/(6.626 × 10^-34 × 3x10^8 )
= 8.99×10^-7m
<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
Ionic: transfer of electrons
Covalent: sharing of electrons
Metallic: sharing of free electrons in a structure of cations
