To help you answer this question, you can search a sample reaction in the internet for hydroboration-oxidation. Take a look at the attached picture. The sample reaction is shown at the top. We can deduce that the original compound must be an alkene. Following the pattern, compound a must be 2-methylbutene as shown in the picture.
In balancing nuclear decay types of reaction, the same as balancing a chemical reaction, we use the number and the type of nucleons present for the decay reaction. Regardless of the type of decay, it should be that the total number of nucleons in the shole process should be conserved. For carbon-11, the decay equation would be as follows:
11/6 C --->11/5 B + 0/1β
It is an example of a positron emmision or a positive beta decay. It is a decay for neutron-poor nuclei where a proton is being transformed into a neutron and also emitting a positron that is high in energy.
In the dark waters just below the photic zone, because of cellular respiration, the concentration of dissolved carbon dioxide is higher relative to dissolved oxygen.
In cellular respiration living organisms use oxygen and release carbon dioxide, because there is little or no light there is no photosynthesis and oxygen is little produced.
<span>The expected results of the Rutherford's gold foil experiment were that the relative massive alpha particles (respect to electrons) could go through the gold foil without being deviated of their trajectory or only small deviations due to the uniformly distributed positive charge of the protons. The real results showed that some particles were significantly deviated of the trajectory (large deviation angles and even some particles bounced back to the source). This lead Rutherford to reject the plum pudding model and propose a new one. The new model proposed by Rutherford was that the atom consisted of a small and every dense nucleus (which contained the positive charge, protons) and a vast region, almost empty, but where the electrons were, surrounding the nucleus.</span><span />
<u>Answer:</u> The heat required will be 58.604 kJ.
<u>Explanation:
</u>
To calculate the amount of heat required, we use the formula:
Q= heat gained or absorbed = ? J
m = mass of the substance = 100 g
c = heat capacity of water = 4.186 J/g ° C
Putting values in above equation, we get:
Q = 58604 Joules = 58.604 kJ (Conversion factor: 1 kJ = 1000J)
Thus, heat released by 100 grams of ice is 58.604kJ.
