Answer:
The standard enthalpy of formation of HgO is -90.7 kJ/mol.
Explanation:
The reaction between Hg and oxygen is as follows.
From the given,
Molar mass of HgO = 216.59 g/mol
Mass of HgO decomposed = 18.5 g
Amount of heat absorbed = 7.75 kJ
From the reaction,
The standard enthalpy of formation = 
During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.
For the formation of 1 mol of HgO , 90.7 kJ of energy is release
Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol
A condensation reaction forming a glycosidic bond. so in other words a monosaccharide joining together to form a disaccharide.
Answer:
B. Valence Electron Pairs
Explanation:
Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.
Answer:- 14.9 M
Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.
Density of the solution is given as 0.90 grams per mL.
From the mass and density we could calculate the volume of the solution as:
= 111 mL
Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.
= 0.111 L
Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.
= 1.65 mole
To calculate the molarity we divide the moles of ammonia by the liters of solution:
= 14.9 M
So, the molarity of the given commercial sample of ammonia is 14.9 M.
The answer is C because it says describe the final sugar and C is unsaturated and in the problem it says as much as sugar will dissovle .
