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tekilochka [14]
3 years ago
8

Tyrosine absorbs maximally at 290 nm. A 0.2162 g sample containing some amount of Tyrosine was placed into a 250.0 mL volumetric

flask and diluted to the mark. This solution had an absorbance of 0.118 (at 290 nm) in a 2.00 cm pathlength cell. A sample of pure Tyrosine (MW - 1818 g/mole) weighing 12.4 mg was dissolved in a 100.0 mL of solution. A 10.0 mL aliquot of this 100.0 mL solution was added to a 250.0 mL flask and diluted to the mark. This last diluted sample had an absorbance of 0.473 at 290 nm in a 1.00 cm cell. Calculate the weight percent of Tyrosine in the sample. Assume no other compounds in the sample, except Tyrosine, absorb at 290 nm.A. 0.14 %
B. 0.069 %
C. 0.028 %
D. 0.040 %
E. 1.80 %
F. 2.63 %

Chemistry
1 answer:
mojhsa [17]3 years ago
7 0

Answer:

B. 0.069 %

Explanation:

It should be initially noted in this answer in particular that, the amino acids tryptophan and tyrosine have a very precise 280 nm absorption rate, which allows a direct A280 size of protein concentration. The 280 nm UV absorbance rate is regularly utilized to approximate protein concentration in laboratories due to its simplistic nature, its affordability and also the ease of usage.

kindly check the attached image below for the solution to the above question.

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How does the circultatory system and muscular system work together?
grin007 [14]

Answer:

d.

Explanation:

i was think b at first but that doesnt really explain how it gets to the circulatory system so d because without the the regulation of the heart rate and oxgen rich blood the muscular system can do absolutely nothing

7 0
3 years ago
What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?
Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

6 0
3 years ago
A gas at 750 mmhg and with a volume of 2. 00 l is allowed to change its volume at constant temperature until the pressure is 600
Gemiola [76]

Answer:

The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.

Explanation:

Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;

P ∝ V

P V = K

P1 V1 = P2 V2

Parameters :

P1 = Initial pressure of the gas = 750 mmHg

V1 = Initial pressure of the gas = 2. 00 Liters

P2 = Final pressure of the gas = 600 mmHg

V2 = Fimal volume of the gas = ? Liters

Calculations :

V2 = P1 V1 ÷ P2

V2= 750 × 2. 00 ÷ 600

V2 = 1500 ÷ 600

V2 = 2.5 Liters.

Therefore, the new volume of the gas is 2. 5 Liters.

8 0
2 years ago
How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain​
11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0
3 years ago
Dubnium chloride chemical formula
cupoosta [38]

Answer:

Db

Dubnium/Symbol

Explanation:

please mark my answer in brainlist plz

4 0
2 years ago
Read 2 more answers
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