The correct answer is this: THE NUCLEUS OF AN ATOM SPLITS INTO FRAGMENTS, RELEASING A LARGE AMOUNT OF ENERGY.
Nuclear fission is the process in which the nucleus of a radioactive element split into two different nucleic of smaller sizes of different elements with a large release of energy. Nuclear fission process is usually used to provide energy for electricity generation.
Answer:
b=200
l=100
Explanation:
so since it is a square all sides would be equal
when you cut the square vertically the dimensions will change
the bredth would be same that is 200
but the length would be half that is 100
Answer:
See explanation
Explanation:
We must first write the equation of the reaction as follows;
C3H8 + 5O2 ----> 3CO2 + 4H2O
Now;
We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles
So;
1 mole of C3H8 yields 3 moles of CO2
3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2
We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles
So;
5 moles of oxygen yields 3 moles of CO2
12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2
We can now decide on the limiting reactant to be C3H8
Therefore;
Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2
Again;
1 moles of C3H8 yields 4 moles of water
3 moles of C3H8 yields 3 × 4/1 = 12 moles of water
Hence;
Mass of water = 12 moles of water × 18 g/mol = 216 g of water
In order to obtain the percentage yield from the reaction, we have;
b) Actual yield = 269.34 g
Theoretical yield = 396 g
Therefore;
% yield = actual yield/theoretical yield × 100/1
Substituting values
% yield = 269.34 g /396 g × 100
% yield = 68%
Answer:
See image attached
Explanation:
a)
The full reaction mechanism of step 1 was obtained from Bartleby and attached to this answer. The steps involved in the reaction are:
1) Loss of Br- and formation of a carbocation
2) Attack of CH3CN on the carbocation
4) Formation of a quaternary nitrogen intermediate
5) Attack of water on the quaternary nitrogen intermediate
6) Loss of the water molecule
5) Formation of the amide product
b)
i) sodium hydroxide
ii) HCl
Explanation:
1. Spontaneous as written at all temperatures
C. When ΔH is negative and ΔS is positive, the sign of ΔG will always be negative, and the reaction will be spontaneous at all temperatures.
2. Spontaneous in reverse at all temperatures
A. When ΔH is positive and ΔS is negative, the sign of ΔG will always be positive, and the reaction can never be spontaneous.
3. Spontaneous as written above a certain temperature
B. ΔH is positive and ΔS is positive - an endothermic reaction (positive ΔH) that also displays an increase in entropy (positive ΔS). It is the entropy term that favors the reaction. Therefore, as the temperature increases, the TΔS term in the Gibbs free energy equation will begin to predominate and ΔG will become negative.
4. Spontaneous as written below a certain temperature
D. ΔH negative and ΔS is negative - When the reaction is exothermic (negative ΔH) but undergoes a decrease in entropy (negative ΔS), it is the enthalpy term which favors the reaction. In this case, a spontaneous reaction is dependent upon the TΔS term being small relative to the ΔH term, so that ΔG is negative. The freezing of water is an example of this type of process. It is spontaneous only at a relatively low temperature.
