Answer:
x = 33.52 amu
Explanation:
It is given that,
Isotope A has a mass of 34 amu and an abundance of 52%, isotope B has a mass of 33 amu and an abundance of 48%.
Let x is the average atomic mass of this element. It can be calculated as follows :

So, the average atomic mass of this element is 33.52 amu.
Answer:
Explanation:
The principle applied is the Markovnikoff's rule which states that when hydrogen chloride adds to a double bond, the hydrogen atoms join to the carbon that already has the most hydrogen atoms bonded to it. The rule wa postulated by a russian chemist known as Vladimir Markovnikoff.
In the markovnikoff's rule, there are sveral conditions that must be met, one of them is that no free radicals must be involved.
The reaction and the structure of the product is as shown in the attachment.
Answer:
Carbohydrates
Explanation:
Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.
In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.
Answer:
5.625 moles of oxygen, O₂.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.
Thus, 5.625 moles of O₂ is needed for the reaction.