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
Answer:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
Answer:
The temperature at which the given substance starts to boil is called boiling point.
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Answer:
- 6.38x10²² molecules C₆H₁₂O₆
Explanation:
First we <u>convert the given masses into moles</u>, using the <em>compounds' respective molar mass</em>:
- 64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂
- 83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄
- 19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆
Then we multiply each amount by <em>Avogadro's number</em>, to <u>calculate the number of molecules</u>:
- 2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules
- 0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules
- 0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules
