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exis [7]
4 years ago
15

The volume of oxygen, collected over water, is 185 mL at 25 degrees Celsius and 600 torr. calculate the dry volume of the oxygen

at STP
(for the question, you have to convert from mL to L, Celcius to Kelvin, and torr to atmospheres)​
Chemistry
1 answer:
ivolga24 [154]4 years ago
4 0

Answer:

0.1593 L.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>P₁V₁T₂ = P₂V₂T₁</em>

<em></em>

P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.

P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.

<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>

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3 0
3 years ago
Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expres
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Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

Ksp=[Ag^+][Cl^-]

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

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I          -                   0             0

C        -                   +x           +x

E        -                    x             x

Which leads to the following modified equilibrium expression:

Ksp=x^2

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

6 0
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Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is
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<span>294400 cal The heating of the water will have 3 phases 1. Melting of the ice, the temperature will remain constant at 0 degrees C 2. Heating of water to boiling, the temperature will rise 3. Boiling of water, temperature will remain constant at 100 degrees C So, let's see how many cal are needed for each phase. We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion. 80 cal/g * 320 g = 25600 cal Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C 420 * 100 = 42000 cal Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away. 420 g * 540 cal/g = 226800 cal So the total number of cal used is 25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
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