Answer: 25,200.
Explanation:
1) Given: 4.659 × 10⁴ - 2.14 × 10⁴
2) You have to deal with significant figures.
Since, the powers are the same (10⁴) you can directly subtract the decimal numbers. But first analyze the significant figures and the number of decimal digits.
3) The number 4.659 × 10⁴ has four significant figures (4, 6, 5, and 9), while the number 2.14 × 10⁴ has three significan figures (2, 1, and 4).
4) When you add or subtract numbers with diferent amount of decimal digits, the result must show the same number of decimal digits as the term with the least number of decimal digits.
5) Before subtracting, you must round all the terms to the least number of decimal digits. So, since 2.14 has two decimal digits and 4.659 has three decimal digits, you shall round 4.659 to 4.66.
6) Now you subtract 4.66 - 2.14 = 2.52
7) Multiply by the power of 10: 2.52 × 10⁴ = 25,200. And that is the answer.
We calculate for the amount of chromium metal in the reactant by,
= 350 x (mass of Cr2/mass of Cr2O3)
= 350 x (104/152)
= 239.47 grams
The amount of Cr metal in the product is only 213.2 grams. Thus, the percent yield.
percent yield = (213.2 grams/239.47) x 100%
= 89%
Answer:
1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).
2. The difference in pH values is 4.95.
Explanation:
1. The pH of a compound can be found using the following equation:
![pH = -log([H_{3}O^{+}])](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20)
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
<u>Trimethyl ammonium</u>:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
![Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5B%28CH_%7B3%7D%29_%7B3%7DN%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5B%28CH_%7B3%7D%29_%7B3%7DNH%5E%7B%2B%7D%5D%7D)
By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
![pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%280.097%29%20%3D%201.01%20)
<u>Phenol</u>:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
![Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DOH%5D%7D)
Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
![pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%289.99%20%5Ccdot%2010%5E%7B-6%7D%29%20%3D%205.00%20)
Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.
2. The difference in pH values for the two acids is:
Therefore, the difference in pH values is 4.95.
I hope it helps you!
The correct answer is B. H2SO4 + B(OH)3 B2(SO4)3 + H2O
Hope this helps!
<span>the answer is
1A = 10^-10 m
so </span>1.61Å = 1.61 x 10^-10 m
he distance between the atoms of H−I is 1.61 x 10^-10 m
