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Vladimir79 [104]
2 years ago
13

Which functional group is within the compound shown below?

Chemistry
1 answer:
mamaluj [8]2 years ago
4 0

Answer:

A. Amino

example phenylalanine

Explanation:

B. Ester is R-CO-OR

example nitroglycerin

C. Hydroxyl R-OH

example ethanol

D. Carbonyl R-(C=O)-R

examples formaldehyde

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What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention
Elena L [17]

Answer: Lead(II) nitrate but idk the rest

Explanation:

5 0
2 years ago
What kind of intermolecular force is between hydrogen sulfide and hydrogen iodide?
RUDIKE [14]
I will have to go with carbon monoxide
8 0
3 years ago
Year/ Distance (cm)
luda_lava [24]

Answer:

the answer is C

Explanation:

6.7 to 13.2 then look at the numbers they go up but not a lot each time

5 0
3 years ago
Which statement best explains why a carbon atom can bond easily with other atoms?
Kruka [31]
The best and most correct answer among the choices provided by the question is the second choice.

Carbon can easily bond with other atoms because it is an organic element.

I hope my answer has come to your help. God bless and have a nice day ahead!
3 0
2 years ago
Read 2 more answers
2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0
2 years ago
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