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3 years ago

14

In examining an organic compound, you discover that it contains the following functional group. What class of organic compound i

s it? O H || | R—C—N—R amide amine carboxylic acid ester

2 answers:

ki77a [65]3 years ago

6 0

the correct answer is a 100%

Harman [31]3 years ago

4 0

The answer is A. amide

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What is the temperature of 0.645 mole of neon in a 2.00 L vessel at 4.68 atm?

storchak [24]

Answer:

—96.03°C

Explanation:

We'll begin by writing out the information provided by the question. This includes:

Number of mole (n) = 0.645 mole

Volume (V) = 2.00 L

Pressure (P) = 4.68 atm

Temperature (T) =?

Recall: that the gas constant = 0.082atm.L/Kmol

With the ideal gas equation PV = nRT, the temperature of the gas can be obtained as follow:

PV = nRT

4.68 x 2 = 0.645 x 0.082 x T

Divide both side 0.645 x 0.082

T = (4.68 x 2) /(0.645 x 0.082)

T = 176.97 K

Now, We can also express the temperature obtained in celsius as shown below:

Temperature (celsius) = temperature (Kelvin) - 273

Temperature (celsius) = 176.97 - 273

Temperature (celsius) = —96.03°C

The temperature of the Neon gas is

—96.03°C

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3 years ago

a triangular prism with bases that are right angles measuring 5 inch by 12 inch by 13 in the height of the prism is 2 in what is

valentinak56 [21]

The answer is 1,560 hope this helps if u multiply all of it

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3 years ago

Acidic solitions change blue litmus payer to<br>A.pink <br>B.yellow <br>C.red<br>D.colorless

Ad libitum [116K]

Answer:

D probably

Explanation:

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3 years ago

A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane

zhenek [66]

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

$\frac{1.5 L}{24 hr}$

= 0.0625 L/hr

As, molar mass of He is 4 g/mol and molar mass of $O_{2}$ is 32 g/ mol.

Now,

$\frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}$

= 2.83

or, rate of $O_{2} = \frac{\text{Rate of He}}{2.83}$

= $\frac{0.0625 L/hr}{2.83}$

Rate of $O_{2}$ = 0.022 L/hr.

This means that 0.022 L of $O_{2}$ gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of $O_{2}$ gas is calculated as follows.

$\frac{1.5 L}{0.022 L/hr}$

= 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

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4 years ago

All of the following are evidence of a chemical change except

Norma-Jean [14]

Change of matter
- Know simple science rules and that will help.

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