The 3 parts are
1) A five carbon ribose sugar
2) A Phosphate molecule
3) The four nitrogenous bases
I hope that's help !
Answer:
The sum would be 69109.4944 ml if that's what you were asking
Hey there mate ;)!,
Given : 100°C
To convert to Kelvin scale unit,
Add the Celsius unit with 273.
→ (100+273) K
→<em><u> </u></em><em><u>373 </u></em><em><u>K</u></em>
<em>Therefore</em><em> </em><em>the </em><em>answer</em><em> is</em><em> </em><u>373 Kelvin</u><em> </em><em>or </em><u>373 </u><u>K.</u>
<em>By </em><em>Benjemin</em> ☺️
Atomic number is 57, and the atomic mass is 138.90547 u + 0.00007 u
Answer:
ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol
Explanation:
Let's consider the combustion of C₈H₁₈.
C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = − 5113.3 kJ
We can calculate the standard enthalpy of formation of C₈H₁₈(g) using the following expression.
ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₈H₁₈(g)) - 25/2 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 25/2 mol × ΔH°f(O₂(g)) - ΔH°rxn
1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × (-393.5 kJ/mol) + 9 mol × (-241.8 kJ/mol) - 25/2 mol × 0 kJ/mol - (− 5113.3 kJ)
ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol
