What is the name of the molecular compound SF

1-Butanol, 1-butene, and di-n-butyl ether are all removed from the product by washing with concentrated sulfuric acid. Explain w

Alecsey [184]

Answer:

1-bromobutane, unlike others, will not be removed because it is an alkane and inert to concentrated sulfuric acid.

Explanation:

Alkanes are saturated organic compounds, primarily made up of hydrogen and carbon atoms. Saturation implies that they posses no double or triple bonds, therefore making them stable and unresponsive to addition reactions and also, concentrated sulfuric acid.

1-Butanol is a primary alcohol. Primary alcohols will react with acids to produce alkyl halides. 1-butene, as an alkene, will react with concentrated sulfuric acid to produce alkyl hyrgensulfate, while di-n-butyl will produce an ether when it reacts with concentrated sulfuric acid.

Therefore, these organic compounds will be washed and converted to other compounds by concentrated sulfuric acid except 1-bromobutane.

3 0

3 years ago

You dissolve 14 g of Mg(NO3)2 in water and dilute to

mart [117]

Answer:

0.127M

Explanation:

Molarity of a solution = number of moles (n) ÷ volume (V)

Molar mass of Mg(NO3)2 = 24 + (14 + 16(3)}2

= 24 + {14 + 48}2

= 24 + 124

= 148g/mol

Using the formula, mole = mass/molar mass, to convert mass of Mg(NO3)2 to mole

mole = 14g ÷ 148g/mol

mole = 0.095mol

Volume = 750mL = 750/1000 = 0.75L

Molarity = 0.095mol ÷ 0.75L

Molarity = 0.127M

3 0

3 years ago

What kind of weather periods are usually followed by cholera and malaria outbreaks?

olga nikolaevna [1]

I would say that it would be excessive rain and heat. Because, after a drought and a flood happen, standing water is home to mosquitos which carry malaria. I hope this helps!

5 0

3 years ago

What is heat of vaporization?

Anestetic [448]

Explanation:

It is the heat required to change a gram of substance from a liquid to a gas.

3 0

3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

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

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

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

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0

3 years ago