Answer:
clf3
Explanation:
it occupied more than 8 valence electrons
I have provided the steps to figure out the number of atoms in the chemical compound.
1. What is a property of a base? You should N-O-T taste laboratory chemicals!!! I don't know why textbooks emphasize the taste of acids and bases. But that is the answer.
<span>2. In the reaction of aluminum bromide with ionized sodium bromide, which compound is the Lewis acid? </span>
<span>What reaction??? </span>
<span>3 In a neutral solution the [H^+] is ____. </span>
<span>At 25C a solution is said to be neutral when the hydrogen ion concentration is 1.00x10^-7M. </span>
<span>4 With solutions of strong acids and strong bases, the word strong refer to ____. </span>
<span>The strength of electrolytes, including acids and bases, describes the degree to which the substance ionizes. Strong acids and bases ionize completely in water. </span>
<span>5 Which of the following pairs consists of a weak acid and a strong base? </span>
<span>a. sulfuric acid, sodium hydroxide == strong, strong </span>
<span>b.acetic acid, ammonia == weak, weak </span>
<span>c. acetic acid, sodium hydroxide* == weak, strong </span>
<span>d. nitric acid, calcium hydroxide == strong, strong </span>
<span>6. The ionization constant (K^a) of HF is 6.7 x 10^-4. Which of the following is true in a 0.1M solution of this acid? </span>
<span>a. [HF] is greater than [H^+][F^-].* == Yep </span>
<span>b. [HF] is less than [H^+][F^-]. == Nope </span>
<span>c. [HF] is equal to [H^+][F^-]. == if K=1 </span>
<span>d. [HF] is equal to [H^+][F^2-] == nonsense </span>
<span>7. The process of adding a known amount of solution of known concentration to determine the concentration of another solution is called ____. </span>
<span>The process of finding the concentration of an acid or base by neutralizing it with a known concentration of a known volume is a titration.</span>
Answer:
The reaction when the Borane (BH3) is add to an alkene and form an alkylborane is shown below.
Explanation:
The boron of the borane does not have extra electron pairs, in this way the double bond of the alkene attacks the boron and the hydrogen belonging to the borane adheres to the carbon that is more substituted, thus forming an alkyl borane.