How many formula units (f.u.'s) of GaF3 are present in 145 g of GaF3?

State two applications of simple distillation of separating mixture

Salsk061 [2.6K]

Answer:

Purify drinking water of unwanted

chemicals and

minerals such as salt.

Hope this helps! :D

3 0

3 years ago

What are the two most common elements found in the universe?

cluponka [151]

Answer:

That would be hydrogen and helium! :)

3 0

2 years ago

Read 2 more answers

Information about tectonic plate boundaries has been primarily found by scientists studying data from

taurus [48]

Studying data from earthquakes!

8 0

3 years ago

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What is the value for AG at 5000 K if AH = -220 kJ/mol and S= -0.05 kJ/(mol-K)?

Serhud [2]

Answer:

C. 30 kJ

Explanation:

Hello there!

In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:

$\Delta G= \Delta H-T\Delta S$

It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:

$\Delta G=-220kJ/mol-5000K*-0.05kJ/mol*K\\\\\Delta G=30kJ/mol$

Therefore, the answer is C. 30 kJ .

Best regards!

4 0

2 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:

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

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago