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BaLLatris [955]
3 years ago
14

Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KClO3(s)2KCl(s) + 3O2(g) The prod

uct gas, O2, is collected over water at a temperature of 25 °C and a pressure of 749 mm Hg. If the wet O2 gas formed occupies a volume of 5.76 L, the number of moles of KClO3 reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

The number of moles of KClO₃ reacted was 0,15 mol

Explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)

V is volume (<em>5,76L</em>)

R is gas constant (<em>0,082 atmL/molK</em>)

And T is temperature (25°C ≡ <em>298,15K</em>)

Replacing, number of moles of O₂ are <em>0,2248 moles</em>

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂×\frac{2 mol KClO_{3}}{3 mol O_{2}} = <em>0,15 mol of KClO₃</em>

I hope it helps!

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3 years ago
A 110 g copper bowl contains 240 g of water, both at 21.0°C. A very hot 410 g copper cylinder is dropped into the water, causing
vlada-n [284]

Answer:

There is 98.76 kJ energy transfered to the water as heat.

Explanation:

<u>Step 1:</u> Data given

Mass of copper bowl = 110 grams

Mass of water = 240 grams

Temperature of water and copper = 21.0 °C

Mass of the hot copper cylinder = 410 grams

8.6 grams being converted to steam

Final temperature = 100 °C

<u>Step 2:</u> Calculate the energy gained by the water:

Q = m(water)*C(water)*ΔT + m(vapor)*Lw

⇒with mass of water = 0.240 kg

⇒ with C(water) = the heat capacity of water = 4184 J/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 100 °C - 21.0 = 79°C

⇒ with mass of vapor = 8.60 grams = 0.0086 kg

⇒ with Lw = The latent heat of vaporization (water to steam) = 22.6 *10^5 J/kg

Q = 0.24kg * 4184 J/kg°C *79°C + 0.0086 kg*22.6*10^5 J/kg

Q = 79328.64 + 19436 = 98764.64 J = 98.76 kJ

There is 98.76 kJ energy transfered to the water as heat.

4 0
3 years ago
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a_sh-v [17]

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5 0
2 years ago
Calculate the volume of 2.0116 x 10^23 molecules of nh3 gas trapped in a container at a pressure of 2280.0 mmhg and a temperatur
OLEGan [10]
The problem in this question can be solved using the ideal gas formula. Ideal gas formula show interaction between the pressure, volume, temperature and the number of molecules.

n= number of molecule/ avogadro number
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T= celcius + 273.15
T= (41+273.15) K=314.15

P=2280 mmHg / (760mmHg/atm)= 3 atm

PV=nRT
V= nrT/ P
V=  1/3 moles * (<span>0.08205 L atm / mol·K) </span>  * 314.15 K / 3 atm
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3 years ago
Ammonia, NH3, acts as an Arrhenius base because it Ammonia, N H 3, acts as an Arrhenius base because it blank the concentration
sergejj [24]

Answer:

Ammonia acts as an Arrhenius base because it increases the concentration of OH⁻ in aqueous solution.

Explanation:

The acid-base theory of Arrhenius explains that in aqueous solutions both acid and base dissociate, releasing ions in the solution. The acid release the ion H⁺ and some anion, and the base release the ion OH⁻ and some cation.

In water, the reaction of ammonia is:

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Because of that, ammonia is an Arrhenius base.

3 0
3 years ago
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