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
I would say A but then again im not too sure so hope that makes it easier to somehow
<u>Answer:</u> The number of moles of gas present is 0.276 moles
<u>Explanation:</u>
To calculate the number of moles of gas, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of the gas = 725 mm Hg
V = Volume of the gas = 7.55 L
n = number of moles of gas = ?
R = Gas constant = 
T = Temperature of the gas = 
Putting values in above equation, we get:
Hence, the number of moles of gas present is 0.276 moles
Answer:
3. be
Explanation:
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