A gas has a pressure of 1.00 atm and a volume of 500 mL. When its pressure is doubled, at constant temperature, to 2.00 atm, its
1 answer:
You might be interested in
- E(Bonds broken) = 1371 kJ/mol reaction
- E(Bonds formed) = 1852 kJ/mol reaction
- ΔH = -481 kJ/mol.
- The reaction is exothermic.
2 H-H + O=O → 2 H-O-H
There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb
- E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
- ΔH(Breaking bonds) = +1371 kJ/mol
Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release
- E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
- ΔH(Forming bonds) = - 1852 kJ/mol
Heat of the reaction:
is negative. As a result, the reaction is exothermic.
Answer:
The answer to your question is: T2 = 235.44 °K
Explanation:
Data
V1 = 3.15 L V2 = 2.78 L
P1 = 2.40 atm P2 = 1.97 atm
T1 = 325°K T2 = ?
Formula
Process
T2 = (P2V2T1) / (P1V1)
T2 = (1.97x 2.78x 325) / (2.40 x 3.15)
T2 = 1779.895 / 7.56
T2 = 235.44 °K
Answer:
Option "B" is correct.
Explanation:
According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In and
, the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.