Answer:
C
Explanation:
The detailed solution is found in the image attached. It is necessary to note that the oxidation half equation is multiplied by three to balance electron gain and loss. This is adequately shown in the image below. Inferences are only drawn from balanced redox reaction equation hence the first step is to balance the redox reaction equation.
Answer:
thermal circulation cells
Explanation:
The three cell model of global wind circulation model is the model that give some us the relationship between the wind patterns that is present on our earth to the sunlight (heat) received from the sun.
Now the sunlight (heat) received from the sun is dependent on the earth axis tilt and on the heating of the earth’s surface due to solar heating therefore these two comes under the global circulation model apart from that the movement of air is also dependent on the Coriolis force and hence it also comes under the global wind circulation model but thermal circulation cells are the cells that are described in this model and hence they don’t come under this model therefore thermal circulation cell is the option which is absent in the three cell model of global circulation. Therefore option “c” is correct.
Answer:
C, A, A, B, B
Explanation:
Question 5: C. continental drift
Question 6: A. new rock forms
Question 7: A. happens at deep-ocean trenches
Question 8: B. convection currents in the mantle
Question 9: B. convergent boundary
Answer: Option (d) is the correct answer.
Explanation:
An oxidizing agent is a specie which accepts electrons and gets reduced in a chemical reaction.
In the given reaction,
Oxidation state of Al changes from 0 to +3, therefore, it is reducing agent. Whereas oxidation state of Br is changing from -2 to -3 this means Br is gaining electrons, therefore, it is an oxidizing agent.
Thus, we can conclude that the statement is the oxidizing agent because its oxidation number decreases.
Mass of methane takne = 1.5g
moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles
mass of water = 1000 g
Initial temperature of water = 25 C
final temperature = 37 C
specific heat of water = 4.184 J /g C
1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules
2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J
3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules
This heat is released by 0.094 moles of methane
So heat released by one mole of methane =
- 622851.06 Joules = 622.85 kJ / mole
4) standard enthalpy of combustion = -882 kJ / mole
Error = (882-622.85) X 100 / 882 = 24.84 %