Answer:
Explanation:
Hello,
In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:
We can represent the heats in terms of mass, heat capacities and temperatures:
Thus, we solve for the mass of platinum:
Next, by using the density of platinum we compute the volume:
Which computed in terms of the edge length is:
Therefore, the edge length turns out:
![a=\sqrt[3]{180cm^3}\\ \\a=5.65cm](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B180cm%5E3%7D%5C%5C%20%5C%5Ca%3D5.65cm)
Best regards.
Answer:
Dipole-dipole interactions
Step-by-step explanation:
Each molecule consists of <em>two different elements</em>.
Thus, each molecule has permanent <em>bond dipoles</em>.
The dipoles do not cancel, so the attractive forces are dipole-dipole attractions.
"Covalent bonds" is <em>wrong,</em> because there are no bonds between the two molecules.
There are dipole-induced dipole and London dispersion forces, but they are much weaker than the dipole-dipole attractions.
The law of conservation
Hope this helps
~ Jordan ~
Answer:
The correct answer is: Serine, Histidine, Aspartate
Explanation:
The catalytic triad of an enzyme is composed of three aminoacid residues which are the most important for its catalytic activity. They are located in the catalytic site of the enzyme. In the case of chymotrypsin- a serine protease, the catalytic triad is composed by serine, histidine and aspartate (Ser-His-Asp). Serine proteases hydrolyse peptidic bonds in proteins and peptides. To do that, the histidine-which interacts with the aspartate by a hydrogen bond so its pKa increases- take a proton from the serine. Thus, deprotonated serine is able to attack the peptide bond and to perform hydrolysis.
<h3>
Answer:</h3>
81.9 grams
<h3>
Explanation:</h3>
From the question we are given;
- Half-life of C-14 is 5730 years
- Original mass of C-14 (N₀) = 150 grams
- Time taken, t = 5000 years
We are required to determine the mass left after 5000 years
- N = No(1/2)^t/T, where N is the remaining mass, N₀ is the original mass, t is the time taken and T is the half-life.
t/T = 5000 yrs ÷ 5730 yrs
= 0.873
N = 150 g ÷ 0.5^0.873
= 150 g × 0.546
= 81.9 g
Therefore, the mass of C-14 left after 5000 yrs is 81.9 g
