Answer: 0.002 M
Explanation:
The balanced chemical equation for ionization of 
in water is:
According to stoichiometry :
1 mole of ionizes to give =2 mole of 
ions
0.001 mole of ionizes to give = 
mole of ions
Thus of a solution of 0.001 M aqueous sulfuric acid is 0.002 M
155,500
I did this to the best of my ability. I have a hard time comprehending things sometimes so I’m so so so sorry if it’s wrong
Answer:
A. 4-ethyl-hex-3,5-dien-2-ol.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol.
Explanation:
Hello there!
In this case, according to the given problems, it is possible to apply the IUPAC rules to obtain the following names:
A. 4-ethyl-hex-3,5-dien-2-ol because we have an ethyl radical at the fourth carbon and the beginning of the parent chain is on the Me (CH3) because it is closest to first OH.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol: because we start at the alcohol and have a chlorine atom on the second carbon, a methyl radical on the third carbon, a <em>tert</em>-butyl on the fifth carbon and the parent chain is benzene which is phenol as an alcohol.
Regards!
Answer:
8.3334%
Explanation:
You have two masses. To find the percent of sodium chloride in water by mass, you divide the mass of NaCl by water. First, make both units the same. Easiest is to convert kg into g. 1.5kg = 1500g
125g NaCl/1500g H2O = 0.0833333333 ==> 8.3334%
