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tigry1 [53]
3 years ago
13

Which of the following properties of a protein is least likely to be affected by changes in pH? Tertiary structure Primary struc

ture Secondary structure Net charge
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
7 0

Answer:

Primary structure is the correct answer.

Explanation:

  • The primary structure is the simple level of protein structure.
  • Primary structure is a basic amino acids sequences in a protein.
  • In the primary structure, amino acids are attached together by a covalent bond.
  • Primary structure is when the amino acids are joined together with peptide bonds to produce polypeptide chains
  • Changes in pH are least likely to change the amino acid sequence or disrupt peptide bonds.

Explanation:

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1cm=.01 so it would be 167-34

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If a leaf is fossilized, it would most likely form what type of fossil?
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The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves
adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

<em />

6 0
3 years ago
Select the correct answer.
muminat

Answer:- B. 4.65 g.

Solution:- The given balanced equation is:

2AgNO_3(aq)+Na_2S(aq)\rightarrow Ag_2S(s)+2NaNO_3(aq)

It asks to calculate the mass of silver sulfide formed by when 0.0150 liters of 2.50 M of silver nitrate are used.

Moles of silver nitrate are calculated on multiplying it's liters by its molarity and then on multiplying by mol ratio, the moles of silver sulfide are calculated. These moles are multiplied by the molar mass to convert to the grams.

Molar mass of Ag_2S = 2(107.87)+32.06  = 247.8 g per mol

The dimensional set up for the complete problem is:

0.0150L(\frac{2.50molAgNO_3}{1L})(\frac{1molAg_2S}{2molAgNO_3})(\frac{247.8gAg_2S}{1molAg_2S})

= 4.65gAg_2S

So, the correct choice is B. 4.65 g.


7 0
3 years ago
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