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stiks02 [169]
3 years ago
6

True or False? Boiling a solution containing a protein will disrupt all secondary structure but not primary structure.

Chemistry
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

False

Explanation: The structure of denatured protein does not change.The primary structure is the same as the secondary or tertiary, hence the non-polar groups are disrupted at high temperatures.

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Can a element be a metal and a non metal
Dmitry [639]

Answer:

no

Explanation:

because in the periodic table there is a section for metals and a section for non metals. 'non' implies that it is not a metal and having an element be both metal and non metal would be contradictory.

6 0
4 years ago
How many moles of oxygen are in 5.6 moles of al(oh)3?
Mashcka [7]
5.6 Al(OH)3
5.6 Al, 16.8 O, 16.8 H

16.8 mols of oxegyn in 5.6 mols of Al(OH)3
7 0
3 years ago
Solubility is determined by the molecular or ionic structures of the solute and the solvent, and by the _________ that hold toge
Natalija [7]

Answer:

Solubility is determined by the molecular or ionic structures of the solute and the solvent, and by the intermolecular forces that hold together the particles of each substance

Explanation:

- Ionic bonds and covalent bonds are sort of intermolecular forces

-State of matter does not hold together the particles of each substance

4 0
3 years ago
What are the molality and mole fraction of solute in a 22.2 percent by mass aqueous solution of formic acid (HCOOH)?
xxTIMURxx [149]
Since there is no sample, let us assume 100 g of the solution: 
(22.2% of 100 g) / (46.0254 g HCOOH/mol) = 0.48234 mol HCOOH 
(100 g - 22.2 g) = 77.8 g = 0.0778 kg water 
(0.48234 mol HCOOH) / (0.0778 kg) = 6.1997 mol/kg = 6.20 m HCOOH 
(77.8 g H2O) / (18.01532 g H2O/mol) = 4.3185 mol H2O 
(0.48234 mol HCOOH) / (0.48234 mol + 4.3185 mol) = 0.100 [the mole fraction of HCOOH]
3 0
4 years ago
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
trapecia [35]

Answer:

9.9 ml of 0.200M NH₄OH(aq)

Explanation:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

?ml of 0.200M NH₄OH(aq) reacts completely with  12ml of 0.550M FeCl₃(aq)

1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution

1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters  

5 0
3 years ago
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