H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.
Answer:
The correct answer to the following question will be "Arrhenius base".
Explanation:
- An Arrhenius base seems to be a material that raises the ion concentration (hydroxide) when exposed to water and thereby reduces the concentration of ions (hydronium).
- This acid, as well as base model, claims an acid is indeed a material that incorporates hydrogen or ionizes protons throughout aqueous, while a base would be a material that comprises hydroxide while releases everything in a that solution
So that the above is the right answer.
Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode):
Reduction half reaction (cathode):
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature =
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
=
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Scientific questions and hypotheses come up frequently while one is engaged in investigating a scientific phenomenon such as natural geological phenomena as may occur in geological mapping in the field. For example, there may be a question does this canyon or deeply incised valley which is quite straight follow a weakness in the earth's crust like a major fault or the direction of bedding in well bedded sedimentary rocks. In a particular topographic area, some hypotheses which may be developed is that valleys follow geological structure whereas ridges follow resistant rocks like quartzites or quartz sandstones or in the ocean, points or capes may represent resistant quartz sandstones and bays may represent weak soft shales recessively weathering