Answer:
Explanation:
1. Molar concentration
Designate chloroform as C and acetone as A.
The molar concentration for C is derived from Moles of C per Litres of solution.
(a) Moles of C
We are assuming there are 0.187 moles of C.
This resolves that step.
(b) Litres of solution
Next, identify 0.813 moles of A.
(i) Mass of each component
(ii) Volume of each component
(iii) Volume of solution
Assuming mixing doesn't alter the total volume.
V = 15.08 mL + 59.70 mL = 74.78 mL
(c) Molar concentration of C
2. Molal concentration of C
Molal concentration is calculated as moles of solute per kilograms of solvent.
Total moles of C = 0.187 mol.
Mass of A = 47.22 g = 0.047 22 kg.
Answer:
The work done in this process will be considered Negative.
Explanation:
The energy transferred by the system to the environment is negative
Therefore, if work is done on the system, it is labeled as positive. Conversely, when work is done by the system, it is regarded as negative.
In this scenario, the argon gas is expanding, and the work is exerted by the system into the surroundings (container), making the sign Negative.
Thus, the result for the work pertaining to this process will carry a Negative sign.
Hello! The mass percent composition of nitrogen in NH₄OH is 14/35×100. To find the percent composition by mass of an element within a chemical compound, divide the atomic mass of that element (AM), which is 14 for Nitrogen, by the entire compound's molar mass (MM) and multiply that by 100. The formula for determining percent composition is as follows: Have a nice day!
Answer:
The process of converting glucose to glucose-6-phosphate is an endergonic reaction, which is coupled with the exergonic hydrolysis of ATP.
Explanation:
Within glycolysis, the phosphorylation of glucose to glucose-6-phosphate occurs first, facilitated by the hexokinase enzyme. This reaction is endergonic. This phosphorylation is a coupled reaction tied to ATP hydrolysis, where the free energy released by ATP hydrolysis drives glucose phosphorylation.
An exponential decay law is generally expressed as: A = Ao * e ^ (-kt) => A/Ao = e^(-kt) Half-life time => A/Ao = 1/2, and t = 4.5 min => 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154. Now substituting k, Ao = 28g, and t = 7 min to determine the remaining grams of Thallium-207 gives: A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g. Final answer is 9.5 g.
