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3 years ago

An aqueous solution contains 0.23 M potassium hypochlorite.

Chemistry

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of 0.22 mol HClO will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, 0.11mol HBr will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of 0.25mol NH₄Cl will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of 0.12 mol HCl will produce NH₄⁺. 0.25mol HCl consume all NH₃.

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when fully inflated a hot air balloon has a volume of 1.6 *10^5 L ( liter) an average temperature of 373k and 0 967 atm . 1) Ass

Sidana [21]

1. First, you have to find the number of moles 1.6z10^5L of gas is at 373K and 0.967atm using PV=nRT solving for n. (n=PV/RT). Everything is in the correct units and we know R is going to be 0.08206atmL/molK since it is a constant.
n=(0.967atmx160000L)/(0.08206atmL/molKx373K)
n=5054.8mol gas
Then you have to find the the number grams which can be found using the molar mass given as 29g/mol. multiply 29g/mol by the number of moles of gas we found in the previous step.
5054.8molx29g/mol=146589.9g of gas
Lastly, to find the density of the gas you need to divide the mass of the gas by its volume.
146589.9g/160000L=0.916g/L

2. The dinsity of the gas at STP should be higher than the density of gas with the given conditions. This is due to the fact that the given conditions involves a higher temperature than that of at STP which will cause the gas to expand therefore increasing the volume with out increasing the mass. The reason why the pressure is not building up even though the pressure is higher is that the balloon is not sealed meaning the gas can maintain about atmospheric pressure while expanding since the excess are just leaves the balloon.
the answer to part 2 can be proven by the fallowing:
To find the density of the gas at STP you first multiply the molar volume of gas at STP by the number of moles of gas from part 1 to get the volume of the gas at STP.
5054.8molx22.4L/mol=113228L
Then you divide the mass form part by the new volume to get the new density.
146589.9g/113228L=1.30g/L

I hope this helps. Let me know in the comments if any of it is unclear.

4 0

3 years ago

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denis23 [38]

Answer:

Personal Question

Explanation:

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3 years ago

How many moles are there in 13g of KMnO4?

Hoochie [10]

Answer:

0.08 moles

Explanation:

Relative atomic mass of KMnO4 = K + Mn + O4 = 39 + 55 + 16 x 4 = 158

=> Moles in 13g of KMnO4 = $\frac{Mass}{Relative-atomic-mass}$ = $\frac{13}{158}$ = 0.082278.. = 0.08 moles

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3 years ago

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An object has a density of 16.3 g/mL and a volume of 0.46 L. Calculate the

deff fn [24]

Answer:

16.53 pounds

Explanation:

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3 years ago

Equal moles of H2, N2, O2, and He are placed into separate containers at the same temperature. Assuming each gas behaves ideally

lbvjy [14]

Answer:

They would all exhibit the same pressure.

Explanation:

Since the same number of mole of each gas is placed in different containers, it means the gas will occupy the same volume.

Now, the gases were observed at the same temperature. This means they will all have the same pressure as their volume is the same.

Now we can further understand this by doing a simple calculation as follow:

Assumptions:

For H2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, H2 has a pressure of 1 atm.

For N2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, N2 has a pressure of 1 atm

For O2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, O2 has a pressure of 1 atm

For He:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, He has a pressure of 1 atm.

From the above illustrations we can see that the gases have the same pressure since they have the same number of mole, volume and were observed at the same temperature.

4 0

3 years ago