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Aliun
1 month ago
9

A typical protein contains 16.2wt%nitrogen. A 0.500 ml aliquot of protein solution was digested, and the liberated NH3 was disti

lled into 10.00 ml of 0.0214 M HCI. The unreacted HCI required 3.26 ml of 0.0198 M NAOH for complete titration. Find the concentration of protein (mg protein /ml) in the original sample.
Chemistry
1 answer:
KiRa [2.9K]1 month ago
3 0

The four corresponding figures and calculations are as follows

Explanation:

[HCL] initial = (10.00ml) (0.02140M) = 0.2140mmol

NaOH = (3.26ml) (0.0198M) = 0.0645mmol

The amount of NH3 generated in HCl will be the difference of 0.2140mmol minus 0.0645mmol = 0.1495mmol

Since 1 mol of N in protein yields 1 mol of NH3, there are 0.1495mmol of N

Molar mass of N = 14.0067mg per mol

(0.1495mmol) (14.0067mg/mL) = 2.094mg N

Given that the protein contains 16.2% N = 2.094mg N / 0.162 mg N = 12.93mg protein

12.95mg protein per 0.500L = 25.85mg protein per liter

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Consider the following system at equilibrium:
VMariaS [2998]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.

6 0
2 months ago
The two-slit diffraction experiment shows how light can be treated as particles and how light waves carry the statistical inform
alisha [2963]

The double-slit experiment serves as a renowned method to exemplify concepts in quantum mechanics. Specifically, it highlights the idea of wave-particle duality. Employing a light wave shows diffraction and interference, which are typical characteristics of wave behavior. Unexpectedly, using an electron beam produces an interference pattern as well, indicating that electrons can exhibit wave-like properties.


Explanation:

The optical phenomenon would nearly resemble, yet be entirely distinct from, that involved with the exploitation of light. Interference and diffraction are the characteristics distinguishing waves from particles: waves can interfere and disperse, whereas particles cannot.

Light curves around obstacles akin to waves, and this bending results in the single-slit diffraction pattern.

5 0
2 months ago
Read 2 more answers
Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^
lorasvet [2795]
Fe 3+ + SCN- --> FeSCN 2+ 

<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>

<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>
5 0
2 months ago
Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g
Anarel [2989]
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.
7 0
2 months ago
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