<span>The product would be :
H3O+ Benzaldehyde -----> C6H5CHO.
It will form a neutral molecule. So the charge is neutral.
I hope it helped.</span>
Answer:
approximately 15.1 grams.
Explanation:
The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.
change 25 grams of potassium chlorate to moles.
calculate the gram molecular mass of potassium chlorate.
Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.
1 K x 39 grams/mole
+1 Cl x 35.4 grams/ mole
+3 O x 16 grams/ mole
= 122.4 grams / mole Potassium Chlorate
25
122.4
= moles.
2.05 moles of Potassium Chlorate.
There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.
2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.
Find the gram molecular mass of Potassium Chloride.
1 K x 39 = 39
+1 Cl x 35.4 = 35.4
= 74.4 grams / mole.
2.05 moles x 74.4 grams/ mole = 15.2 grams
hi im breanna
Answer:
The mole is simply a very large number that is used by chemists as a unit of measurement.
Explanation:
The mole is simply a very large number,
6.022
×
10
23
, that has a special property. If I have
6.022
×
10
23
hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have
6.022
×
10
23
H
2
molecules, I have a mass of 2 gram of hydrogen molecules. If I have
6.022
×
10
23
C
atoms, I have (approximately!) 12 grams.
The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of
C
, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of
O
2
are required to react with the
C
, which I could measure by mass or by volume.