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anyanavicka [17]

2 years ago

9

Write the structure of all possible peptides containing these amino acids: Asp, Glu, Gln Use single letter abbreviations and cap

ital letters only; i.e. GYR, not Gly-Tyr-Arg. If there are fewer than 6 peptides, leave an appropriate number of answer boxes empty. fill in the blank 1 , fill in the blank 2 , fill in the blank 3 , fill in the blank 4 , fill in the blank 5 , fill in the blank 6

1 answer:

Liula [17]2 years ago

6 0

Answer:

QED, EDQ, DQE, EQD, DEQ, QDE

Explanation:

The structure of all possible peptides that contain the given amino acids are :

QED, EDQ, DQE, EQD, DEQ, QDE

where : Asp is represented by the letter code D

Glu is represented by the letter code E

Gln is represented by the letter code Q

Note : when three amino acids combine they form what is known as tripeptide ( i.e. contains two peptide linkages ) while a peptide linkage is been formed by the combination of a carboxyl group of an amino acid and the amino group of different amino acid

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A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the

DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$ ( $\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$ )

(1/434 × $10^{-9}$ ) = -2.178 × $10^{-18}$ ( $\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$ )

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$ ) $\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

434 × $10^{-9}$ = (1/2.178 × $10^{-18}$ ) $\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }$

434 × $10^{-9}$ × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

⇒ $n_{initial}$ = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

= (hc) ÷ λ

= (6.626 × 10−34 × 3.0 × $10^{8}$ ) ÷ (434 × $10^{-9}$ )

= (1.9878 × $10^{-25}$ ) ÷ (434 × $10^{-9}$ )

= 4.58 × $10^{-19}$ J

= 4.58 × $10^{-22}$ KJ

But 1 mole = 6.02× $10^{23}$ , then;

energy in KJ/mole = (4.58 × $10^{-22}$ KJ) ÷ (6.02× $10^{23}$ )

= 7.61 × $10^{-46}$ KJ/mole

7 0

2 years ago

If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?

kap26 [50]

Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3 1mol CO2
---------------------- x -------------------------
1 84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!

6 0

2 years ago

Read 2 more answers

Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Pepsi [2]

Answer:

$Mass_{chromium}=52.1\ g$

$Mass_{neon}=8.23\times 10^{-7}\ g$

Explanation:

<u>Calculation of the mass of chromium as:- </u>

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1.002\ mol= \frac{Mass}{51.9961\ g/mol}$

$Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g$

<u>Calculation of the mass of neon as:- </u>

Moles = $4.08\times 10^{-8}$ moles

Molar mass of neon = 20.1797 g/mol

Thus,

$1.002\ mol= \frac{Mass}{20.1797\ g/mol}$

$Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g$

6 0

2 years ago

If a car travels 1255 miles how many kilometers is that i mile = 1.609km

ANTONII [103]

Answer:

2,019 km

Explanation:

Step 1: Given data

Distance traveled by the car (D): 1,255 mi

Step 2: Convert the distance traveled by the car to kilometers

To convert one unit into another, we use a conversion factor. In this case, the appropriate conversion factor between miles and kilometers is 1 mile = 1.609 km. The distance traveled by the car, in kilometers, is:

D = 1,255 mi × (1.609 km/1 mi) = 2,019 km

3 0

2 years ago

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

Aleksandr-060686 [28]

To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.

Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed

Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:

Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s

5 0

2 years ago

Read 2 more answers