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azamat
1 year ago
12

What do you think is the coldest temperature something can get to? What limits how cold something can get?

Chemistry
1 answer:
damaskus [11]1 year ago
4 0

Answer:

The colder an object it, the slower the atoms are. This is the rule that affects how cold something can get. There is a limit of how cold something can get though. This is called absolute zero. And, actually, we've gotten very close to it. Scientists in Finland have cooled rhodium atoms to a 10th of a billionth of a degree above absolute zero.

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mile

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iVinArrow [24]
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What information did the scientists miss in 1948 and 1966 that caused them to
Debora [2.8K]

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6 0
3 years ago
which of the following amino acids are used to measure the protein concentration by UV? A. His B. Pro C. Trp D. Phe E. Tyr
Goryan [66]

Answer:

C. Trp D. Phe E. Tyr

Explanation:

The concentration of a protein has a direct relation with absorbance of the protein in a UV spectrophotometer. The formula which relates concentration with absorbance is described as under:

                A = ∈ x c x l

where, A = Absorbance

            ∈ = Molar extinction co-efficient

            c  = Concentration of absorbing species i.e. protein

             l  = Path length of light

       

Tryptophan (Trp), phenylalanine (Phe ) and tyrosine (Tyr) are three aromatic amino acids which are used to measure protein concentration by UV. It is mainly because of tryptophan (Trp), protein absorbs at 280 nm which gives us an idea of protein concentration during UV spectroscopy.

The table depicting the wavelength at which these amino acids absorb and their respective molar extinction coefficient is as under:

Amino acid                Wavelength            Molar extinction co-efficient (∈)

Tryptophan                     282 nm                        5690

Tyrosine                          274 nm                        1280

Phenylalanine                 257 nm                        570

In view of table above, we can easily see that Molar extinction co-efficient (∈) of Tryptophan is highest amongst all these 3 amino acids that is why it dominates while measuring concentration.

7 0
3 years ago
A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place
never [62]

Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

Explanation:

<u>Given data;</u>

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

volume of water, v =  109.5 mL = 0.0001095 m³

<u>Known data:</u>

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

specific heat gold is 129 J/kgC

specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

<u>Calculated data</u>

Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge  length forms the <em>hypotenuse side</em>.

L² + L² = 0.0238²

2L² = 0.0238²

L² = 0.0238² / 2

L² = 0.00028322

L = √0.00028322

L = 0.0168

Volume of cube = L³

Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

Gold cube;

let L be the side of the gold cube

L² + L² = 0.0279²

2L² = 0.0279²

L² = 0.0279² / 2

L² = 0.0003892

L = √0.0003892

L = 0.0197

Volume of cube = L³

Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³

Mass of silver cube;

density = mass / volume

mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg

Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be  Q₁

Let the heat lost  by silver cube = Q₂

Let the heat lost  by gold cube = Q₃

Let the final temperature of water = T

Q₁  = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200  (T–19.6)  = 0.0497 x 240 x (81.9–T)  + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04  + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

6 0
3 years ago
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