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Eva8 [605]
4 years ago
11

At the same conditions of pressure and temperature ammonia gas is less dense than air why is this true

Chemistry
1 answer:
Paha777 [63]4 years ago
8 0
According to Marriotte's law, we can use a formula that's a derivation of the original law. That formula says that pressure multiplied by the molar mass, and then divided by R (a constant value) multiplied by the temperature, equal the specific mass of a gas, for example. If we are considering the same temperature and pressure for the air and the ammonia gas then the only difference is really on the molar mass of the gas. The molar mas for air is 0.029 kg/mole and the molar mass for the ammonia gas is <span>0.017 kg/mol. Just from here we see that the ammonia gas will be less dense then the air.
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Calculate the volume of a sample of mercury with a density of 14.6 g/mL and a mass of 1.00 g.
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Answer:

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4 0
3 years ago
The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th
Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically  Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their  standard states.

  Compound                 ΔfHº (kJmol⁻¹)

        SO₂                             -296.8

         O₂                                    0

         SO₃                            -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) =  1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction  will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

                         = -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº +  CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

                     = -99,787.8 J x 1 kJ/1000 J  = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

3 0
3 years ago
What are the stable isotopes of nitrogen?
klio [65]

nitrogen-14, which makes up the vast majority of naturally occurring nitrogen, and nitrogen-15, which is less common.

6 0
4 years ago
PLEASE ANSWER
romanna [79]

Answer:    D

Explanation:

4 0
3 years ago
What are the values for % Cl and % F, respectively, for Sample II?
horsena [70]

Answer:

Explanation:

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6 0
4 years ago
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