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tekilochka [14]
2 years ago
14

If you had done the experiment hands-on and the environmental air pressure changed, how would your results have been affected?

Chemistry
1 answer:
lukranit [14]2 years ago
8 0

Boyle's law states the relationship between pressure and volume of a gas.

From Boyle's law, we know that volume and pressure of a given mass of gas are inversely proportional to each other. Hence, PV = k.

Now, in carrying out this experiment, the atmospheric pressure is assumed to be constant. If the atmospheric pressure changes during the experiment, here would have been a change in the measurements of balloon volume because air pressure changes would have interfered with the results.

Learn more about Boyle's law:  brainly.com/question/1437490

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Explanation:

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Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The
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Answer:

           \large\boxed{\large\boxed{24.6kJ}}

Explanation:

<u>1. Energy to heat the liquid water from 55ºC to 100ºC</u>

     Q=m\times C\times \Delta T

  • m = 10.1g
  • C = 4.18g/JºC
  • ΔT = 100ºC - 55ºC = 45ºC

     Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J

<u>2. Energy to change the liquid to steam at 100ºC</u>

      L=\lambda \times n

  • λ = 40.6kJ/mol
  • n = 10.1g / 18.015g/mol = 0.5606mol

      L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J

<u>3. Total energy</u>

       1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ

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Ce cantitate de O se gaseste in 5,8 g hidroxid de magneziu
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Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess
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Answer:

The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

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PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The relation of K_c\& K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p= Equilibrium constant in terms of partial pressure.=98.1

K_c= Equilibrium constant in terms of concentration  =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

\Delta n_g = Difference between gaseous moles on product side and reactant side=n_{g,p}-n_{g.r}=1-2=-1

98.1=K_c(RT)^{-1}

98.1 =\frac{K_c}{RT}

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The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

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