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Greeley [361]
2 years ago
8

A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t

emperature is changed to 336 K and the volume is decreased to 450. milliliters?
(1) 0.177 atm (3) 5.64 atm
(2) 4.43 atm (4) 14.1 atm
Chemistry
1 answer:
Sergio039 [100]2 years ago
5 0

<span>The combined Boyle's and Charles' gas law says P1V1/T1 = P2V2/T2 </span>

<span>In this case we have (2.5 atm) (900 ml) / 298 K = (P2) (450 ml) / 336 K -----> </span>
<span>7.55 = 1.34P2 -----> P2 = 7.55 / 1.34 = 5.63 atm</span>
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Answer:

FAS concentration = 1.61*10^-4M

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Beer Lambert's law relates the absorbance (A) of a substance to its concentration (c) as:

A = \epsilon lc----(1)

where ε = molar absorption coefficient

l = path length

A plot of 'A' vs 'c' gives a straight line with slope = εl

In addition absorbance (A) is related to % Transmittance (%T) as:

A = 2-log%T----(2)

For the FAS solution, the corresponding calibration fit is given as:

y = 3678(x) + 0.056

This implies that the slope = εl = 3678

It is given that %T = 25.6%

A = 2-log(25.6)=0.592

Based on equation(1):

c = \frac{A}{\epsilon l}=\frac{0.592}{3678}=1.61*10^{-4}M

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What mass, in grams, of CO2 and H2O<br> is formed from 2.55 mol of propane?
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Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
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  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}

mass of H₂O= 183.6 grams

<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>

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