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What is the general formula for a secondary amine? rnh2 h2nrnh2 rn2h r2nh

zvonat [6]

The general fomula of the amine group ought to be R2NH because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

<h3>What is a secondary amine?</h3>

A secondary amine is one in which two alkyl groups are subtituted on the nitrohen atom of ammonia.

This implies that the general fomula of the amine group ought to be R2NH. This is because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

Learn more about secondary amine: brainly.com/question/12682525

4 0

2 years ago

What is the hydrogen ion concentration of a solution with pH = 8.75 ?

Tom [10]

Answer:

The answer is 1.778*10^-9

Explanation:

Using the formula

pH = -log[H^+]

Since the pH of the solution has been given to be 8.75,

Therefore,

8.75= -log [H^+]

Let's take H^+ to be x

8.75= -log x

Let's look for x, since the log is in base 10

X = 10^-8.75

X = 1.778 * 10^ -9

Therefore, H^+ =1.778 *10^-9

4 0

4 years ago

the pH of a solution is a measure of the acidity or alkalinity of a solution. Water has a pH of 7 and, in general, acids have a

Sliva [168]

Answer:

7.1

Explanation:

equation to calculate pH is

$pH=-Log_{10}(a_{H^+})=-Log_{10}(8*10^-8) = -Log_{10}(8)-Log_{10}(10^{-8})=-0.9+8=7.1$

6 0

3 years ago

The nonvolatile, nonelectrolyte urea, CH4N2O (60.1 g/mol), is soluble in water, H2O. How many grams of urea are needed to genera

Sergeeva-Olga [200]

<u>Answer:</u> The mass of urea needed is 12.89 grams

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 24.3 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of urea = ? g

$M_{solute}$ = molar mass of urea = 60.1 g/mol

$V_{solution}$ = Volume of solution = 216 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$24.3atm=1\times \frac{m_{solute}\times 1000}{60.1\times 216}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\m_{solute}=\frac{24.3\times 60.1\times 216}{1\times 1000\times 0.0821\times 298}=12.89g$

Hence, the mass of urea needed is 12.89 grams

3 0

3 years ago

44. If there are 2.54 cm in an inch and 12 inches in a foot, how many cm are in 1.50 feet

lutik1710 [3]

Answer:

45.72 cm

Explanation:

We are told that there are 2.54 cm in an inch and 12 inches in a foot.

Since 1 inch = 2.54 cm

Then, 12 inches = (12 × 2.54)/1 = 30.48 cm

Now, 12 inches makes 1 foot.

Thus, 1 ft = 30.48 cm

Thus, 1.5 ft = (1.5 × 30.48)/1 = 45.72 cm

Thus, 45.72 cm will make 1.5 feet

7 0

4 years ago