6000 I pretty sure tell me if it's right
Answer:
Explanation: The strengths of the inter molecular forces varies as follows -

The normal boiling point of CSe2 is 125°C and that of CS2 is 116°C, which explains the trend that as we move down the group, the boiling point of e compound increases as the size increases.
This usually happens because larger and heavier atoms have a tendency to exhibit greater inter molecular strengths due to the increase in size . As the size increases, the valence shell electrons move far away from the nucleus, thus has a greater tendency to attract the temporary dipoles.
And larger the inter molecular forces, more tightly the electrons will be held to each other and thus more thermal energy would be required to break the bonds between them.
Answer:
34.28 L ( 1.5*22.4 L)
Explanation:
Calculation of the moles of aluminum as:-
Mass = 55 g
Molar mass of aluminum = 26.981539 g/mol
The formula for the calculation of moles is shown below:
Thus,

According to the reaction:-

4 moles of aluminum react with 3 moles of oxygen gas
1 mole of aluminum react with
moles of oxygen gas
2.0384 moles of aluminum react with
moles of oxygen gas
Moles of oxygen gas = 1.5288 moles
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K
⇒V = 34.28 L ( 1.5*22.4 L)
Explanation:
there is 2 nitrogen but if you mean nitrate is 6
Answer:
Final temperature of calorimeter is 25.36^{0}\textrm{C}
Explanation:
Molar mass of anethole = 148.2 g/mol
So, 0.840 g of anethole =
of anethole = 0.00567 moles of anethole
1 mol of anethole releases 5539 kJ of heat upon combustion
So, 0.00567 moles of anethole release
of heat or 31.41 kJ of heat
6.60 kJ of heat increases
temperature of calorimeter.
So, 31.41 kJ of heat increases
or
temperature of calorimeter
So, the final temperature of calorimeter = 