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2 years ago

What is the ratio of the areas of the signals in the h nmr spectrum of pentan-3-ol?

1 answer:

8 0

The ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.

<h3>What is a NMR spectrum?</h3>

Nuclear magnetic resonance spectroscopy is a spectroscopy that shows the detailed structure and chemical environment of a chemical element.

Pentan-3-ol contain 12 hydrogen atoms. In H-NMR spectra, hydrogen atoms have same environment gives a signal.

There are 4 different kinds of signals due of the 4 different environment experienced by these 12 hydrogens.

Thus, the ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.

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How many subsets of three elements each exist in a set of six elements

OLEGan [10]

Answer:

The answer you're looking for is B.C (6,3)

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3 years ago

Which of the following is necessary to increase the rate of a reaction

frutty [35]

An increase in the frequency of effective collisions because it is these collisions that cause reactants to change to products

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3 years ago

Read 2 more answers

Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at

telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound Pressure at equilibrium

$H_2$ 61.8 atm

$I_2$ 46.5 atm

$HI$ 52.3 atm

Calculate the value of the equilibrium constant $K_p$ for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant $K_p$ for this reaction is, 0.952

Explanation :

The given chemical reaction :

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}$

We are given:

$P_{H_2}=61.8atm$

$P_{I_2}=46.5atm$

$P_{HI}=52.3atm$

Putting values in above equation, we get:

$K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952$

Therefore, the value of equilibrium constant $K_p$ for this reaction is, 0.952

7 0

3 years ago

How many grams of co2 are contained in 550 ml of the gas at stp?

dybincka [34]

V = 550 mL = 0.550 L
V = 22.4 L/mol (STP)
M(CO₂)=44.01 g/mol

n(CO₂)=v/V

m(CO₂)=n(CO₂)M(CO₂)=vM(CO₂)/V

m(CO₂)=0.550*44.01/22.4=1.08 g

4 0

4 years ago

If there are 10 navy beans,3 pinto beans,and 17 lentils in a container,what is the percent composition of the container by bean?

Stells [14]

The percent abundance in the container by bean will be 43.33 % if there are 10 navy beans, 3 pinto beans, and 17 lentils in a container

<h3>What is Percentage composition ?</h3>

The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.

% composition by bean = number of beans / total number of objects in container x 100

Total object = 10 ( navy beans) + 3 (pinto beans) + 17 ( lentils) = 30 objects
Total beans = 10 ( navy beans) + 3 (pinto beans) = 13 beans

hence ;

% composition by beans = 13/30 x 100 = 43.33 %

Therefore, the percent composition of the container by bean is 43.33 %

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2 years ago