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stellarik [79]
3 years ago
10

The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c

oncentration in the lab (unless specified otherwise in the protocol), so you will have to dilute the 10x buffer to 1x before use. What is the concentration of Tris and SDS in the 1x buffer
Chemistry
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

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Following laboratory safety guidelines minimizes the chance of lab accidents.

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2 years ago
How many molecules of CO are produced when 2.73 moles of HPO3 react? 4HPO3 + 12C ----&gt; 2H2 + 12CO + P4
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According to the balanced chemical equation:
4 HPO₃ + 12 C → 2 H₂ + 12 CO + P₄
4 moles of HPO₃       ---gives---> 12 moles of CO
2.73 moles of HPO₃  ---gives---> ? moles of CO
so number of moles of CO = \frac{(2.73 * 12)}{4} = 8.19 moles of CO
Number of molecules of CO = number of moles * Avogadro's number
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3 years ago
Mallory combines two chemicals and notices that each chemical retains its original properties. What did Mallory form?
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3 years ago
I’ll mark brainliest &lt;3<br> Can you just check to see if I have this right?
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The specific heat of a certain type of metal is 0.128j/(g.c). What is the final temperature if 305.J of heat is added to 72.7g o
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E answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---&gt; Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
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<span>Hf H2O = -285.83
Now,

</span><span>add them up. 226.77 - 986.2 + (2*285.83) = -187.77 
</span><span>Add back the total enthalpy that is given in the question
 -187.77+127.2 = -60.57 

</span>
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3 years ago
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