Explanation:
The chemical equation is as follows.

And, the given enthalpy is as follows.
;
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.

x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
Answer:
The chemical reaction between the citric acid and sodium bicarbonate
Explanation:
Answer:
Magnesium oxide is a binary compound of magnesium and oxygen while magnesium ribbon consists only of magnesium atoms.
Explanation:
The burning of magnesium in oxygen is a chemical change. It produces magnesium oxide having greater mass than magnesium ribbon. The greater mass results from the fact that the chemical reaction has added another element to the sample- oxygen. The mass of magnesium ribbon is the mass of magnesium atoms alone but in magnesium oxide, we consider the masses of magnesium and oxygen atoms making magnesium oxide heavier than magnesium ribbon.
Explanation:
Yes, the equation is balanced. There are the same number of Hydrogen (H) and Oxygen (O) atoms on both sides of the equation.
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr