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xeze [42]
2 years ago
12

A student made an initial 1:5 dilution of protein lysate. Then 2mL of that was added to 8mL of water. Lastly, the student made a

1:20 dilution of the second tube. What is the final dilution of protein lysate.
Chemistry
1 answer:
Licemer1 [7]2 years ago
3 0

Answer:

The final dilution is 1:400

Explanation:

Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.

So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400

This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.

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The instruction booklet for your pressure cooker indicates that its highest setting is 11.8 psipsi . You know that standard atmo
zavuch27 [327]

<u>Answer:</u> The temperature at which the food will cook is 219.14°C

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.8)psi=26.5psi\\T_2=?

Putting values in above equation, we get:

\frac{14.7psi}{273K}=\frac{26.5psi}{T_2}\\\\T_2=\frac{26.5\times 273}{14.7}=492.14K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

492.14=T(^oC)+273\\\\T(^oC)=219.14^oC

Hence, the temperature at which the food will cook is 219.14°C

7 0
3 years ago
How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0
3 years ago
Describe how stirring, surface area and temperature affect the rate of dissolving.
denis23 [38]

Answer:

friction

Explanation:

since it has a high tempature the friction increases like blowing air in a furnace

7 0
2 years ago
The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions
musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,  

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,  

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.  

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,  

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,  

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,  

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,  

x as 0.115 and y as 0.165

Now the mass of AgCl will be,  

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,  

278.1 × 0.165 = 45.88 grams.  

3 0
3 years ago
What is the systematic name for TiCl2​
Anarel [89]

Answer:

66228

Molecular Formula: TiCl2 or Cl2Ti

Chemical Names: Titanium chloride (TiCl2) 10049-06-6 TiCl2 Titanium(II) chloride dichlorotitanium More...

Molecular Weight: 118.77 g/mol

Dates: Modify: 2019-08-10 Create: 2005-03-26

Explanation:

66228

Molecular Formula: TiCl2 or Cl2Ti

Chemical Names: Titanium chloride (TiCl2) 10049-06-6 TiCl2 Titanium(II) chloride dichlorotitanium More...

Molecular Weight: 118.77 g/mol

Dates: Modify: 2019-08-10 Create: 2005-03-26

8 0
3 years ago
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