19.6 is your answer hope this is right
Answer:-
95 grams
Explanation:-
Let the mass of water to be added be M
So total mass = 5 + M
So 5% of this solution has 5g of NaCl by mass.
∴ (M+5) x (5/100) = 5
M+5 = 5 x 100/5
M+5=100
M= 100-5
95
So amount of water to be added is 95 gram
I'm not sure but it should be 7.31 times 2 times 3.14 or pi
7.31 x 2 x 3.14 = 45.9068
If you need to round to the nearest tenth it will be 45.9
If you need to round to the nearest hundredth it will be 45.91
A neutral carbon doesn't lack any electrons. It has exactly the same number of electrons as it has neutrons.
However, it has four electrons in its outer shell in comparison with eight electrons for a noble gas.
In that sense, it needs four electrons to complete its second shell.<span />
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
