All categories

4 years ago

How co you find anhydrides?

1 answer:

3 0

took two ethanoic acid molecules and removed a molecule of water between them you would get the acid anhydride, ethanoic anhydride. The old name: acetic anhydride.

You might be interested in

What volume of stock solution and water must you add to prepare 36.25ml of a 1.25M solution

Serjik [45]

Given :

Number of moles , n = 36.25 mol .

Molarity , M = 1.25 M .

To Find :

The volume of water required .

Solution :

Moarity is given by :

$M=\dfrac{n}{V}$

So , $V=\dfrac{n}{M}$

Here , n is number of moles and M is molarity .

Putting all values in above equation , we get :

$V=\dfrac{36.25}{1.25}\\\\V=29\ L$

Therefore , volume of water required is 29 L .

5 0

3 years ago

1. Which is not a characteristic of background radiation?

siniylev [52]

Answer: C.) It burns and causes lung cancer

5 0

4 years ago

Read 2 more answers

What is the process by which molecules of water vapor in the air become liquid water?

mote1985 [20]

That is called condensation

5 0

4 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$