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fomenos
2 years ago
11

If 1.76 g of an ideal gas occupy 1.0 L at standard temperature and pressure (STP), what is the molar mass of the gas?

Chemistry
1 answer:
ycow [4]2 years ago
8 0

Answer:

Explanation:

Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :) 

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Answer:

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Given a block of wood 2 cm long, 4 cm wide, and 3 cm high, determine its volume. You may work on a separate piece of paper. 11 c
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4 years ago
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(b) Why is the heat of sublimation ( ΔH fus ) of a substance greater than its ( ΔHvap ) ?
Ivanshal [37]

This is because solids have less energy than liquids do, hence it takes more energy to excite a solid into its gaseous phase than it does a liquid.

When opposed to merely reducing their separation, from solid to liquid, the energy needed to totally separate the molecules as they move from liquid to gas is substantially higher. The latent heat of vaporization is therefore bigger than the latent heat of fusion for this reason.

<h3>What is heat of sublimation?</h3>

The amount of energy required to change one mole of a substance from its solid to its gaseous state under particular conditions—typically the standard ones—is known as the enthalpy of sublimation or heat of sublimation (STP). A solid's worth is based on its cohesive energy.

<h3>What is heat of vaporization?</h3>

The term "enthalpy of vaporization," which is often referred to as "heat of vaporization" or "heat of evaporation," refers to the amount of energy that must be applied to a liquid substance in order to cause a part of that substance to transform into a gas. Vaporization's enthalpy varies with the pressure at which the transition takes place.

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The correct question is:

Why heat of the sublimation of a substance is greater than the heat of vaporization?

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A solution is made by dissolving 26.42g of (NH4)2so4 in enough h2o to make 50.00ml of solution
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Molarity = moles of ammonium sulfate/volume of solution
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K_c = \frac{[NH_2]^{1}[H_2S]^{1}}{[NH_2HS]^{1}} is the equilibrium constant expression for

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<h3>Definition of equilibrium constant.</h3>

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Equilibrium constant for the NH₂HS(S)   →   NH₂(g) + H₂S(g) will be:

K_c = \frac{[Product]^{coefficient}}{[Reactantt]^{coefficient}}

K_c = \frac{[NH_2]^{1}[H_2S]^{1}}{[NH_2HS]^{1}}

Hence, option A is correct.

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