Their are natural since they come from trees which is natural
Answer:
<h2>1.505 × 10²⁴ particles</h2>
Explanation:
The number of particles in iron (II) chloride can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 2.5 × 6.02 × 10²³
We have the final answer as
<h3>1.505 × 10²⁴ particles</h3>
Hope this helps you
Answer:
D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.
Explanation:
Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.
<u>(1) Preparatory phase</u>
During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.
When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.
<u>(2) Pay-off phase</u>
During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.
It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.
Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, 
.
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
The given reaction equation is as follows.
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, .
Answer:
and 
Explanation:
Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine (
) and <u>oxygen</u> (
). So, we can start with the <u>reaction</u> between these compounds:
Now we can <u>balance the reaction</u>:
In the problem, we have the values for both reagents. Therefore we have to <u>calculate the limiting reagent</u>. Our first step, is to calculate the moles of each compound using the <u>molar masses values</u> (32.04 g/mol for and 31.99 g/mol for ):
In the balanced reaction we have 1 mol for each reagent (the numbers in front of and are 1). Therefore the <u>smallest value would be the limiting reagent</u>, in this case, the limiting reagent is .
With this in mind, we can calculate the number of moles for each product. In the case of 
we have a <u>1:1 molar ratio</u> (1 mol of is produced by 1 mol of ), so:
We can follow the same logic for the other compound. In the case of 
we have a <u>1:2 molar ratio</u> (2 mol of is produced by 1 mol of ), so:
I hope it helps!
