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
0.01742919 is the answer because i worked it out
Answer
Na OH reacts with H Cl and forms Na Cl and H₂O
NaOH + HCl → NaCl + H₂O
Here we can see that 1 mole of NaOH reacting with 1 mole of HCl and forming 1 mole of NaCl and 1 mole of H₂O
when NaOH and HCl are added together in equal amount then they will completely neutralize each other but NaOH is hygroscopic in nature which means it can absorb water from air so it will not be weighted accurately.
hence, for neutralization we will take extra NaOH.
Answer:
Conduction directly affects air temperature only a few centimeters into the atmosphere. During the day, sunlight heats the ground, which in turn heats the air directly above it via conduction. At night, the ground cools and the heat flows from the warmer air directly above to the cooler ground via conduction.
Explanation:
According to Avogadro's Law, same volume of any gas at standard temperature and pressure will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).
Data Given:
n = moles = ?
V = Volume = 16.8 L
Solution:
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....
= ( 1 mole × 16.8 L) ÷ 22.4 L
= 0.75 moles
Result:
16.8 L of Nitrogen gas will contain 0.75 moles at standard temperature and pressure.
