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enot [183]
3 years ago
11

FIRST ONE TO ANSWER GETS BRAINLIEST!!!!!!!!!!!HURRY!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
Alex777 [14]3 years ago
8 0
I believe that the answer is a
Andreas93 [3]3 years ago
7 0

i believe the answer is a

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How does boiling differ from evaporation? A. Boiling can take place at any temperature, while evaporation takes place at a speci
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<span>Its D :) evaporation happens all the time, and can only occur at the surface (molecules break off and become gases) whereas boiling is when ANY molecule, whether from the surface, the middle or the bottom can just fly off and become a gas.</span>
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4 years ago
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The student plans to conduct a spectrophotometric analysis to determine the concentration of Cu2+(aq) in a solution. The solutio
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Answer:

The wavelength the student should use is 700 nm.

Explanation:

Attached below you can find the diagram I found for this question elsewhere.

Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.

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3 years ago
Write the expression for the equilibrium constant for this reaction. 2N205(g) &lt;==&gt; 4N02(g) + 02(g)
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Answer:

K = [NO₂]⁴[O₂] / [N₂O₅]²

Explanation:

Based on the chemical equilibrium reaction:

2 N₂O₅(g) ⇄ 4NO₂(g) + O₂(g)

The equilibrium reaction is obtained as the ratio between the multiplication of the concentrations of the reactants over the products powered to its reaction coefficient. That is:

<h3>K = [NO₂]⁴[O₂] / [N₂O₅]²</h3>
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3 years ago
A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this r
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M_{Al}=26,98\frac{g}{mol}\\&#10;m=51g\\\\&#10;n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\&#10;M_{NaOH}=39,4\frac{g}{mol}\\&#10;m=84,1g\\\\&#10;n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span>  2mol      :     2mol           :                                  3mol
2,14mol   :     1,89mol      :                                  2,835mol
remains         completely consumed
2,14-1,89=0,25mol


A) Al

B)  

M_{NaOH}=39,4\frac{g}{mol}\\&#10;n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g

C)
n=2,835mol\\&#10;M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g

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3 years ago
33. Which element in Period 5 of the Periodic Table
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B) Ag because it’s in the middle of the periodic table.
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3 years ago
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