How much heat is absorbed in the formation of 1.00mole of HI(g) from H2 and I2?

How has the pressure of the gas changed?

9966 [12]

Cold you please repost the question
THIS IS A QUESTION REGARDING
ideal gas equation
which states that
PV = nRT
where;
P : Pressure (atm)
V : Volume (litres)
n : Number of moles of that gas(mol)
R : Real gas constant = 0.0821
T : Temperature of that gas(Kelvin)
hope this helps

6 0

3 years ago

How many moles would there be in 1.5x10^24 molecules of water?

Nookie1986 [14]

Answer:

5.85 moles

Explanation:

Moles in (3.52 X 10^24) molecules of water. 3.52 x 10^24 molecules x `1mol/6.02 x 10^23 molecules = 5.85 moles of H2O.

3 0

3 years ago

Wolves prey heavily on pavers in a temperature forest ecosystem the diet of the beavers consist of leaves and branches and bark

Virty [35]

Answer:

<u>B. Primary Consumer</u>

Explanation:

It is the top of the food chain in this example.

5 0

3 years ago

Which reaction leads to the formation of a precipitate?

Brrunno [24]

2Ca(OH)2(aq) + 2FeCl3(aq) on the dead locs

6 0

3 years ago

Read 2 more answers

If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite

Elodia [21]

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of $Al_2O_3$ = 13.2 kg = 13200 g

Mass of $NaOH$ = 50.4 kg = 50400 g

Mass of $HF$ = 50.4 kg = 50400 g

Molar mass of $Al_2O_3$ = 101.9 g/mole

Molar mass of $NaOH$ = 40 g/mole

Molar mass of $HF$ = 20 g/mole

Molar mass of $Na_3AlF_6$ = 209.9 g/mole

First we have to calculate the moles of $Al_2O_3,NaOH$ and $HF$ .

$\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=\frac{13200g}{101.9g/mole}=129.54moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{50400g}{40g/mole}=1260moles$

$\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=\frac{50400g}{20g/mole}=2520moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Al_2O_3+6NaOH+12HF\rightarrow 2Na_3AlF_6+9H_2O$

From the balanced reaction we conclude that

The mole ratio of $Al_2O_3,NaOH$ and $HF$ is, 1 : 6 : 12

And, the ratio of given moles of $Al_2O_3,NaOH$ and $HF$ is, 129.54 : 1260 : 2520

From this we conclude that, $NaOH,HF$ is an excess reagent because the given moles are greater than the required moles and $Al_2O_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Na_3AlF_6$

From the reaction, we conclude that

As, 1 mole of $Al_2O_3$ react to give 2 mole of $Na_3AlF_6$

So, 129.54 moles of $Al_2O_3$ react to give $129.54\times 2=259.08$ moles of $Na_3AlF_6$

Now we have to calculate the mass of $Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=\text{ Moles of }Na_3AlF_6\times \text{ Molar mass of }Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=(259.08moles)\times (209.9g/mole)=54380.892g=54.38kg$

Thus, the mass of cryolite produced will be, 54.38 kg

5 0

3 years ago