Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ.
Answer:
D. Nuclei with small masses combine to form nuclei with larger masses.
B. A small amount of mass in the nuclei that combine is converted to energy
Explanation:
A nuclear fusion, in contrary to fission, is the process by which the nuclei of two atoms combine to form a much larger atom with a large nuclei. Likewise, during a fusion reaction, a large amount of energy is released from the small amount of mass in the nuclei (two) that combines.
According to this question, the following are true of a fusion reaction:
- Nuclei with small masses combine to form nuclei with larger masses.
- A small amount of mass in the nuclei that combine is converted to enormous energy.
The cell notation is:
here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell
in anodic half cell : oxidation takes place [loss of electrons]
in cathodic half cell: reduction takes place [gain of electrons]
1) this is a galvanic cell
2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode
Therefore
3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative
Δ
As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.
To solve this problem,
we can use the Henderson-Hasselbalch Equation which relates the pH to the measure
of acidity pKa. The equation is given as:<span>
<span>pH = pKa + log ([base]/[acid]) ---> 1</span></span>
Where,
[base] = concentration
of C2H3O2
in molarity or moles
<span>[acid] = concentration of HC2H3O2 in molarity or moles</span>
For the sake of easy calculation, let us assume that:
[base] = 1
[acid] = x
<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x)
<span>log (1 / x) = - 0.5
1 / x = 0.6065 </span></span>
x =
1.65<span>
The required ratio of C2H3O2 /HC2H3O2 <span>
is 1:1.65 or 3:5. </span></span>
