The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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Answer:
C
Explanation:
Large chlorine atoms can not fit within the atoms of boron
<u>Answer:</u> The percentage abundance of 
and 
isotopes are 77.5% and 22.5% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of isotope be 'x'. So, fractional abundance of isotope will be '1 - x'
- <u>For isotope:</u>
Mass of isotope = 35 amu
Fractional abundance of isotope = x
- <u>For isotope:</u>
Mass of isotope = 37 amu
Fractional abundance of isotope = 1 - x
Average atomic mass of chlorine = 35.45 amu
Putting values in equation 1, we get:
![35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775](https://tex.z-dn.net/?f=35.45%3D%5B%2835%5Ctimes%20x%29%2B%2837%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.775)
Percentage abundance of isotope = 
Percentage abundance of isotope = 
Hence, the percentage abundance of and isotopes are 77.5% and 22.5% respectively.
Answer:
40.8g of sodium sulfate must be added
Explanation:
The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:
Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)
That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:
<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>
75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄
<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>
0.287 moles * (142.04g / mol) =
<h3>40.8g of sodium sulfate must be added</h3>
