Salt=compound, soda=liquid solution, aluminum foil=element, milk=colloid,
steel=solid solution
Answer:
Present in both catabolic and anabolic pathways
Explanation:
Glyceraldehyde-3-phosphate abbreviated as G3P occurs as intermediate in glycolysis and gluconeogenesis.
In photosynthesis, it is produced by the light independent reaction and acts as carrier for returning ADP, phosphate ions Pi, and NADP+ to the light independent pathway. Photosynthesis is a anbolic pathway.
In glycolysis, Glyceraldehyde-3-phosphate is produced by breakdown of fructose-1,6 -bisphosphate. Further Glyceraldehyde-3-phosphate converted to pyruvate and pyruvate is further used in citric acid cycle for energy production. Therefore, it is used in catabolic pathway too.
Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because it is present in both catabolic and anabolic pathways.
First, find moles of gold given the mass of the sample:
(35.9g Au)/(197.0g/mol Au) = 0.182mol Au
Second, multiply moles of Au by Avogrado's number:
(0.182mol)(6.02 x10^23)= 1.10x10^23 atoms Au
We can rephrase the statement with a little more specificity in order to understand the answer here.
The mass of the products can never be more than the The mass that is expected.
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{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 
.
