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Svetradugi [14.3K]
3 years ago
7

A student labeled 5 tubes 0, 1, 2, 3, 4, 5. Tube 0 contained undiluted protein lysate. The student used the tube 0 lysate to pre

pared 10-fold serial dilutions in the tubes labeled 1-5. What is the dilution of each of the tubes?
Chemistry
1 answer:
Ivanshal [37]3 years ago
7 0

Answer:

Tube 1: 10⁻¹, tube 2: 10⁻², tube 3: 10⁻³, tube 4: 10⁻⁴, tube 5: 10⁻⁵.

Explanation:

Serial dilutions are dilutions that the concentration decreases by the same quantity in each successive step. It means that the undiluted will be used for the first step, then the first will be used for the second, and successively. So, for a 10-fold, the concentration must decrease 1/10 in each step, it means that the dilution will be 1/10 in the first one (because it's 1 in tube 0).

In tube 1, the dilution is 1/10 = 0.1 = 10⁻¹;

In tube 2, the dilution will decrease more 1/10, so it will be 1/100x10 = 1/100 = 0.01 = 10⁻²;

In tube 3, it will be 1/1000x1/10 = 1/1000 = 0.001 = 10⁻³

In tube 4, it will be 10⁻⁴, and

In tube 5, it will be 10⁻⁵.

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Enter the net ionic equation for this reaction. Express your answer as a net ionic equation. Identify all of the phases in your
Lisa [10]

Answer:

2H+(aq) + 2OH-(aq) → 2H2O(l)

Explanation:

Step 1: The balanced equation

2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)

This equation is balanced, we do not have the change any coefficients.

Step 2: The netionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)

After canceling those spectator ions in both side, look like this:

2H+(aq) + 2OH-(aq) → 2H2O(l)

6 0
3 years ago
Which of the following experimental methods cannot be used to measure the rate of a reaction?
irga5000 [103]

Answer:

I think a is the best answer. but really I can not confirm that.sorry for that.

7 0
3 years ago
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Drawing conclusions can often be difficult for scientists because the data may?
klasskru [66]
Because the data may or may not be true

4 0
4 years ago
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   Answer all parts of the following questions on Particle Stoichiometry.
Ugo [173]
<span>1. Translate, predict the products, and balance the equation above. 

Li + Cu(NO3)2 = Li(NO3)2 + Cu

2. How many particles of lithium are needed to produce 125 g of copper? 

125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles

3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?

</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2
3 0
3 years ago
Calculate the mass of Octane needed to release 6.20 mol Co2
n200080 [17]
The combustion reaction of octane is as follow,

                           C₈H₁₈  +  25/2 O₂     →     8 CO₂  +  9 H₂O

According to balance equation,

8 moles of CO₂ are released when  =  114.23 g (1 mole) Octane is reacted

So,

      6.20 moles of CO₂ will release when  =  X g of Octane is reacted

Solving for X,
                                     X  =  (114.23 g × 6.20 mol) ÷ 8 mol

                                     X  =  88.52 g of Octane
Result:
           88.52 g of Octane is needed to release 6.20 mol CO₂.
8 0
3 years ago
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