Ozone which is present in the stratospheric region of atmosphere is helpful for preventing harmful UV rays from reaching the surface of earth. Due to human activity, several compounds (specifically chlorofluorocarbons) are released in atmosphere. Due to inherent chemical stability of these compounds, the remain stable in lower region of atmosphere and slowly diffuse into stratosphere. On reaching the stratosphere, these compounds reacts with ozone and thereby depletes the effective concentration of ozone present in atmosphere. Hence, <span>the Montreal Protocol was signed in 1987 by major countries of the world. This aim of this protocol was to protect the stratospheric ozone layer by phasing out the production and consumption of ozone-depleting substances.</span>
Answer:
-100 kJ
Explanation:
We can solve this problem by applying the first law of thermodynamics, which states that:
where:
is the change in internal energy of a system
Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)
W is the work done by the system (it is positive if done by the system, negative if done on the system)
For the system in this problem we have:
W = +147 kJ is the work done by the system
Q = +47 kJ is the heat absorbed by the system
So , its change in internal energy is:
Answer:
The correct answer is - 800.
Explanation:
Given:
Total amount = ? or assume x
spend in buying birthday item = 3/4 of x
given to sister = 1/5 of x
remaining to mother = 40
solution:
the remaning amount = x- (3x/4+x/5) = 4=
=> x- 19x/20 = 40
=> x = 20*40
=> x = 800
thus, the correct answer is = 800
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
