Answer:
Al is oxidized while Ag is reduced.
Explanation:
The complete molecular equation is;
3Ag2S + 2Al --> 6Ag + Al2S3
Oxidation half equation;
2Al ------> 2Al^3+ + 6e
Reduction half equation;
6Ag^+ + 6e -------> 6Ag
Overall redox reaction equation;
2Al + 6Ag^+ ----->2Al^3+ + 6Ag
Hence; Al is oxidized while Ag is reduced.
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 pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
Basically since there’s 2 hydrogen’s there will be two H’s on either side of your other element. And Se is in group 6 which means it has 6 valence electrons. When you combine 6 and 2 from the hydrogen you get 8. You then should place 8 dots around Se two on each side.
So something like H Se(with 8 dogs around it) H
For water to freeze it would need to be 32 degrees Fahrenheit or 0 degrees Celsius.
