Answer:
Prefix used for a molecule with two carbon atoms is (Eth-).

Explanation:
Those organic compounds which contain two carbon atoms are specified by using a prefix (Eth-).

Examples:

An Alcohol with two carbon atoms = Ethanol

An Alkyne with two carbon atoms = Ethyne

An Alkene with two carbon atoms = Ethene

A Carboxylic Acid with two carbon atoms = Ethanoic Acid

An Aldehyde with two carbon atoms = Ethanall

5 0

3 years ago

Any help?

lutik1710 [3]

Answer:

pH = 9.475

Explanation:

Hello there!

In this case, according to the basic ionization of the hydroxylamine:

$HONH_2+H_2O\rightarrow HONH_3^++OH^-$

The resulting equilibrium expression would be:

$Kb=\frac{[HONH_3^+][OH^-]}{[HONH_2]} =1.1x10^{-8}$

Thus, we first need to compute the initial concentration of this base by considering its molar mass (33.03 g/mol):

$[HONH_2]_0=\frac{1.34g/(33.03g/mol)}{0.500L} =0.0811M$

Now, we introduce $x$ as the reaction extent which provides the concentration of the hydroxyl ions to subsequently compute the pOH:

$1.1x10^{-8}=\frac{x^2}{0.0811-x}$

However, since Kb<<<<1, it is possible to solve for $x$ by easily neglecting it on the bottom to obtain:

$x=[OH^-]=\sqrt{1.1x10^{-8}*0.0811}= 2.99x10^{-5}$

Thus, the pOH is:

$pOH=-log(2.99x10^{-5})=4.525$

And the pH:

$pH=14-4.525\\\\pH=9.475$

Regards!

5 0

3 years ago

What is the total number of grams of h2o produced when 22 grams of propane undergoes complete combustion?

SSSSS [86.1K]

The combustion reaction of propane would be expressed as:

C3H8 + 5O2 = 3CO2 + 4H2O

To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:

moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O

Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.

5 0

3 years ago