Answer:
0.373M
Explanation:
Given parameters:
Mass of NaOH given = 112g
Volume of solvent = 0.75L
Solution:
1....To solve this problem, we first find the number of moles of the solute using the expression below:
Number of moles = mass/molar mass
Solving:
Molar mass of NaOH
Atomic mass of Na = 23g
Atomic mass of O = 16g
Atomic mass of H = 1g
Molar mass = 23 + 16 + 1 = 40g/mol
Number of moles = 112/40 = 2.8mol
2..... From the number of moles of solute, we then evaluate the molarity of the solution that would be produced using the expression below:
Molarity = number of moles/Volume
Molarity = 2.8/0.75 = 0.373M
I can't answer this question if the structural formula is not given. However, I found a similar problem in terms of wording. Taking this problem to be solved, let's take a look at the structural formula as shown in the second picture. First, you must know the parent chain, which is the longest chain. This is a trial-and-error process. The longest chain which has a branching group that is nearest to the head is the correct numbering. In this case, the longest chain has 8 carbon atoms. Thus, the base of the name if octane. Because a 3-carbon chain is branching from the 4th carbon, the IUPAC name of the compound shown is:
<em>4-propyloctane.</em>
Answer:
(a) PbSO4 anode is reduced to Pb.
Explanation:
Cells which can be recharged are known as secondary cells. They can be recharged by passing a direct current through them. An example of a secondary cell is the lead storage battery used in motor cars.
During the process of recharging, the electrodes in the cell are reversed from what we have know to occur during discharging because it is electrolized.
At the anode: The lead (II) tetraoxosulphate (VI) absorbs two electrons to become reduced to Pb. Thus at the anode PbSO₄ is reduced to Pb.
![PbSO_4+2e^-\rightarrow Pb+SO_4^{2-}](https://tex.z-dn.net/?f=PbSO_4%2B2e%5E-%5Crightarrow%20Pb%2BSO_4%5E%7B2-%7D)
At the cathode:
The electrons from the anode are accepted at the cathode where the lead (iV) oxide is oxidised into lead (II) ions.
![PbSO_4+2H_2O\rightarrow PbO_2+SO_4^{2-}+4H^++2e^-](https://tex.z-dn.net/?f=PbSO_4%2B2H_2O%5Crightarrow%20PbO_2%2BSO_4%5E%7B2-%7D%2B4H%5E%2B%2B2e%5E-)
Answer:
The addition of inorganic phosphate helps to sustain the reaction by helping the reaction to occur in a thermodynamically unfavorable manner.
Explanation:
The reaction of inorganic phosphate with glucose is an endergonic reaction which is thermodynamically unfavorable as a result reaction need coupling of ATP hydrolysis to take place in biological system because ATP hydrolysis releases huge amount of free energy as a result the overall free energy change of this reaction becomes negative which helps the reaction to proceed in a thermodynamically favorable manner.
That"s why inorganic phosphate can be used to sustain the rapidly proceeding glycolysis pathway. The addition of inorganic phosphate makes the reaction thermodynamically unvorable.