Answer:
294.87 gm CaCl_2
Explanation:
The computation of the mass of calcium chloride is shown below:
But before that following calculations need to be done
Number of moles of chlorine atom is
= 3.20 × 10^24 ÷ 6.022 × 10^23
= 5.314 moles
As we know that
1 mole CaCl_2 have the 2 moles of chlorine atoms
Now 5.341 mole chloride atoms would be
= 1 ÷ 2 × 5.314
= 2.657 moles
Now
Mass of CaCl_2 = Number of moles × molar mass of CaCl_2
= 2.657 moles × 110.98 g/mol
= 294.87 gm CaCl_2
Answer:
Activation energy of phenylalanine-proline peptide is 66 kJ/mol.
Explanation:
According to Arrhenius equation- 
, where k is rate constant, A is pre-exponential factor, 
is activation energy, R is gas constant and T is temperature in kelvin scale.
As A is identical for both peptide therefore-
![\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7Bala-pro%7D%7D%7Bk_%7Bphe-pro%7D%7D%3De%5E%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-E_%7Ba%7D%5E%7Bala-pro%7D%5D%7D%7BRT%7D)
Here 
, T = 298 K , R = 8.314 J/(mol.K) and 
So, ![\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.05%7D%7B0.005%7D%3De%5E%7B%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-%2860000J%2Fmol%29%5D%7D%7B8.314J.mol%5E%7B-1%7D.K%5E%7B-1%7D%5Ctimes%20298K%7D%7D)
(rounded off to two significant digit)
So, activation energy of phenylalanine-proline peptide is 66 kJ/mol
Answer:
the answer for this question is true
Answer:
The content that follows is the substance of lecture 14. In this lecture we cover the use of balanced chemical equations to make mole and mass predictions in reaction. Reaction stoichiometry is a fancy way of describing mole relations in a reaction. When we read a balanced equation we relate every compound in the reaction to the others by means of the mole coefficients in the reaction. For example: HCl + NaOH → NaCl + H2O This equation is all 1:1 relationships, 1 mole of NaCl is made for every 1 mole of HCl or NaOH etc. In a previous lecture I introduced the following analogy: What can we do with a balanced equation? A chemical reaction equation is essentially a table of conversion factors that we can use to predict amounts of products that can be made, reactants needed to make a specific amount of product or exact amounts of reactants needed to completely consume another reactant (eg. acid and base neutralization). While the use of the stoichiometry (fancy words for mole relationships in a reaction equation) may seem difficult, it really isn't. Let's use an example that you all can understand first: 1 slice bologna + 2 slices of bread → 1 Sandwich If I asked you how many sandwiches you could make with 12 slices of bread, you would immediately say 6, right? How did you know this? Well, obviously based on the equation, you make 1 sandwich for every 2 slices of bread you have.12 slices bread x 1 sandwich/2 slices bread = 6 sandwiches If I asked you how many slices of bologna and bread you need to create 10 sandwiches, you would immediately say 10 slices of bologna and 20 slices of bread, right? Again the relationship given in the equation tells you how much is needed. 10 sandwiches x 1 slice bologna/1 sandwich = 10 slices bologna and 10 sandwiches x 2 slices bread/1 sandwich = 20 slices bread Finally if I asked you how many sandwiches you could make with 5 slices of bologna and 8 slices of bread? Which of the two sandwich makings would run out first and how much of the excess makings would be left over? Well, the answers are 4 sandwiches could be made and 1 slice of bologna is left over. 5 slices of bologna x 1 sandwich/1 slice bologna = 5 sandwiches 8 slices bread x 1 sandwich/2 slices bread = 4 sandwiches 8 slices bread x 1 slice bologna/2 slices bread = 4 slices of bologna used 5 slices bologna - 4 slices used = 1 slice left over
EZ U-U
Explanation:
