Stable isotopes are not radioactive nor dangerous.
Unstable isostopes are radioactive and have different mass of the element.
Answer:
0.46 grams (C₆H₅)₂CO
Explanation:
To find the mass of benzophenone ((C₆H₅)₂CO), you need to (1) convert mmoles to moles and then (2) convert moles to grams (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units. The final answer should have 2 sig figs to match the sig figs of the given value (2.5 mmoles).
Molar Mass ((C₆H₅)₂CO): 13(12.011 g/mol) + 10(1.008 g/mol) + 15.998 g/mol
Molar Mass ((C₆H₅)₂CO): 182.221 g/mol
2.5 mmoles (C₆H₅)₂CO 1 mole 182.221 g
----------------------------------- x ------------------------ x ------------------- =
1,000 mmoles 1 mole
= 0.46 grams (C₆H₅)₂CO
Answer:
<h3>The answer is 32 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 768 g
volume = 24 cm³
We have

We have the final answer as
<h3>32 g/cm³</h3>
Hope this helps you
Answer:
What will determine the number of moles of hydronium in an aqueous solution of a strong monoprotic acid? The amount of acid that was added.
Explanation:
The dissociation of formic acid is:

The acid dissociation constant of formic acid,
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)


Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is
.