A flask containing neon gas is connected to an open-endedmercury manometer. The open end is exposed to the atmosphere, wherethe
1 answer:
Answer: Option (C) is the correct answer.
Explanation:
The given data is as follows.
= 745 torr
= - 50 (the value is negative because on the open side its level is lower than the side connected to the gas)
Now, formula to calculate the pressure of neon gas will be as follows.
Putting the given values into the above formula as follows.
=
= 695 torr
Thus, we can conclude that the neon pressure is 695 torr.
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Answer:
(C) Oxygen because it has the strongest attractive force and the largest
electronegativity
Explanation:
Oxygen will be most reactive
configuration =
As oxygen need only two electrons to complete its octet.
Oxygen will have stronger force than nitrogen as it will tend to pull electrons more to complete its octet. Also along the period electronegativity increases so oxygen has higher electronegativity than nitrogen
The given question is incomplete. The complete question is:
The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:
A.
B.
C.
D.
Answer: A. : decreases
B. : decreases
C. : no change
D. : increases
Explanation:
Entropy is defined as the randomness of the system.
Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.
In reaction , as gaseous reactant is changed to solid product, entropy decreases.
In reaction , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.
In reaction , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.
In reaction , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.
The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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