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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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Determine the net change in NADHNADH , FADH2FADH2 , and GTPGTP from the passage of two acetyl-SCoA molecules through the citric
CaHeK987 [17]

Answer:The citric acid cycle is the final common pathway for the aerobic oxidation of fuel molecules. Moreover, as we will see shortly (Section 17.3) and repeatedly elsewhere in our study of biochemistry, the cycle is an important source of building blocks for a host of important biomolecules. As befits its role as the metabolic hub of the cell, entry into the cycle and the rate of the cycle itself are controlled at several stages.

Explanation:

8 0
3 years ago
Which of the following would represent a single displacement reaction between Potassium Bromide (K^+Br^-) and Iodine (I), which
ra1l [238]

Answer:

Option : KBr + I -> KBr+I

Explanation:

Single-replacement reaction or single displacement reactions are a type of chemical reactions in which a whole compound reacts with an element in such a way that the element takes place of one of the compound's own elements and sets it free.

If we talk about KBr and I displacement reaction is not possible among these because Iodine is less reactive than Bromine that is why it will not react with KBr or replace Br.

                       KBr + I -> KBr+ I


Potassium Bromide + Iodine -> Potassium bromide + Iodine


Hope it help!

5 0
3 years ago
The closest hominin to a current day human is?
bearhunter [10]
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8 0
3 years ago
A) explain why caesium is more reactive than potassium.
Kamila [148]

Answer:

Explanation:

(a) Firstly, caesium abd potassium are both in Group 1 of the periodic table. Group 1 metals (also called alkali metals) are the most reactive metals of the periodic table. Caesium is more reactive than Potassium because it has a higher electropositivity than Potassium. Electropositivity is the tendency of a metal to donate electron(s) to form a cation. Electropositivity increases down the group; this is because it is easier for atoms to loose electrons on the outermost shell that are far away from the central nucleus as against atoms whose outermost electrons are closer to the central nucleus. <u>Thus, the more "bulky" an atom is, the farther it's outermost electrons (valence electrons) get from the central nucleus and the easier it is to lose the outermost electron(s). And the easier it is for the valence electron(s) to be removed, the more reactive the atom would be and vice-versa.</u>

Caesium is more reactive than potassium because it is more bulky than potassium, with an atomic number of 55, while potassium has an atomic number of 19.  

NOTE: The closer an electron is to the nucleus, the more difficult it is to be removed from it's shell.

(b) i. Formula for Caesium Nitrate:

Symbol for Caesium is Cs and Nitrate is NO₃⁻.

Cs⁺ + NO₃⁻  ↔  CsNO₃

Formula for Caesium Nitrate is  CsNO₃

ii. Formula for Caesium sulphate

Symbol for caesium is Cs and Sulphate is SO₄²⁻

Cs⁺ + SO₄²⁻  ↔   Cs₂SO₄  

Formula for Caesium sulphate is Cs₂SO₄

NOTE: When writing the formulae, the charges would be exchanged to form the subscript as seen on the product sides above.

7 0
2 years ago
Limiting Reactant
liraira [26]

Answer:

(1) Cl₂ is the limiting reactant.

(2) 8.18 g

Explanation:

  • 2Na(s) + Cl₂(g) → 2NaCl(s)

First we <u>convert the given masses of reactants into moles</u>, using their <em>respective molar masses</em>:

  • Na ⇒ 12.0 g ÷ 23 g/mol = 0.522 mol Na
  • Cl₂ ⇒ 5.00 g ÷ 70.9 g/mol = 0.070 mol Cl₂

0.070 moles of Cl₂ would react completely with (2 * 0.070) 0.14 moles of Na. There are more Na moles than that, so Na is the reactant in excess while Cl₂ is the limiting reactant.

Then we <u>calculate how many moles of NaCl are formed</u>, <em>using the limiting reactant</em>:

  • 0.070 mol Cl₂ * \frac{2molNaCl}{1molCl_2} = 0.14 mol NaCl

Finally we <u>convert NaCl moles into grams</u>:

  • 0.14 mol NaCl * 58.44 g/mol = 8.18 g
3 0
3 years ago
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