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svetoff [14.1K]
3 years ago
11

What is the amount of aluminum chloride produced from 40 moles of aluminum and excess chlorine?

Chemistry
1 answer:
Marrrta [24]3 years ago
6 0

Answer:

is it 26.98

Explanation:

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nadezda [96]

Answer:

JWBVH

Explanation:

NWSBWHSBHS

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3 years ago
How does the structure of amino acids allow them to form a polypeptide
melamori03 [73]
<span>Amino acids which are known to be linked by peptide bonds they form polypeptide chains.

 Proteins are linear polymers are formed by way of linking an a-carboxy group of one amino acid to a-amino of different amino acids which have peptide bond. The formation which results from two amino acids which result in a loss of a water molecule. The best process of the reaction is hydrolysis.</span>
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3 years ago
Hi guys, my question is:
guapka [62]

Answer:

See Explanation Below

Explanation:

A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.  

B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5

C) The rate will increase by 4 since you are doubling both you have to multiply them both.

8 0
3 years ago
12.5 mL of 0.280 M HNO3 and 5.0 mL of 0.920 M KOH are mixed. Is the resulting solution acidic, basic or neutral?
svlad2 [7]

Answer:

The resulting solution is basic.

Explanation:

The reaction that takes place is:

  • HNO₃ + KOH → KNO₃ + H₂O

First we <u>calculate the added moles of HNO₃ and KOH</u>:

  • HNO₃ ⇒ 12.5 mL * 0.280 M = 3.5 mmol HNO₃
  • KOH ⇒ 5.0 mL * 0.920 M = 4.6 mmol KOH

As <em>there are more KOH moles than HNO₃,</em> the resulting solution is basic.

8 0
3 years ago
Explain the difference in the boiling point of 2-methylpropane and 2-iodo-2-methylpropane in terms of both molecular polarity an
Len [333]
The Boiling Point of 2-methylpropane is approximately -11.7 °C, while, Boiling Point of <span>2-iodo-2-methylpropane is approximately 100 </span>°C.

As both compounds are Non-polar in nature, So there will be no dipole-dipole interactions between the molecules of said compounds.

The Interactions found in these compounds are London Dispersion Forces.

And among several factors at which London Dispersion Forces depends, one is the size of molecule.

Size of Molecule:
                          There is direct relation between size of molecule and London Dispersion forces. So, 2-iodo-2-methylpropane containing large atom (i.e. Iodine) experience greater interactions. So, due to greater interactions 2-iodo-2-methylpropane need more energy to separate from its partner molecules, Hence, high temperature is required to boil them.
6 0
3 years ago
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