Answer:
6.50 g of Hydrogen
Explanation:
We know that in every 20.0g of sucrose, there are 1.30g of hydrogen.
We now have 100.0g of sucrose. 100.0g is 5x larger than the 20.0g sample, which is a 5 : 1 ratio. Applying this ratio to the amount of hydrogen, we would have 5*1.3g of hydrogen in the 100.0g of sucrose.
5*1.3 = 6.5, so our answer is that there are 6.50g of hydrogen in 100.0g of sucrose.
Hope this helps!
It's genre. Look for something either on a final or starting page near the cover, and check to look if it's non fiction, or anything else that could prove it as reliable.
Answer:
1) Separate the aqueous layer from the organic layer using the separation funnel.
2) Treat the aqueous layer to obtain compound A.
3) Distilated the organic layer to obtain compound B.
Explanation:
When <u>NaOH is added</u> to the mixture the acid groups will react to produce a salt and increases the polarity of the compounds due to the net charges generated. (Figure 1).
Therefore, the salt produced by compound A will move to the <u>aqueosus layer</u>. Compound B dont react due to the lack of <u>acid groups</u>. So, this molecule will stay in the <u>organic layer</u>.
When the aqueous layer is separated from the organic layer using the separation funnel we will have a <u>separation</u>. The compound will remain in the aqueous layer and compound b will remain in the organic layer.
Then we aqueous layer can be <u>treated with HCl</u> in order to obtain the initial A molecule, in other words: Undo the ionic form of compound A.
The organic layer can be removed by <u>distillation</u> in order to obtain the pure form of compound B.
Answer:
<h3>An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude.</h3>
Answer:
1.115 g
Explanation:
Applying,
R = R'
................... Equation 1
Where R = original sample of radon-222, R' = sample of radon-222 left after decay, n = Total time, t = half-life.
make R' the subject of the equation
R' = R/()............... Equation 2
From the question,
Given: R = 73.9 g, n = 23 days, t = 3.8 days.
Substitute these values into equation 2
R' = 73.9/(
)
R' = 73.9/
R' = 1.115 g
