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sweet [91]
2 years ago
14

2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to

relieve his heartburn. Unfortunately, he felt reaction happened in his stomach and the product which is CO2 gas was produced. Assume the pressure in Matt’s stomach was 780 torr and the amount of the gas was 400mL. What was the volume of the gas after it was released outside of Matt’s body? Assume human body temperature is 40°C and room temperature is 25°C. The pressure of the room is 1 atm. 760 torr = 1atm
Chemistry
1 answer:
evablogger [386]2 years ago
3 0

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

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Some bacteria cells make people sick by releasing large molecules called toxins that affect your bodies. What
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3 years ago
Read 2 more answers
The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are
Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>

For the zero-order reactions, concentration-time equation can be written as follows:

                                          [A] = - Kt + [Ao]

where:

  • [A]: concentration of the reactant A at the <em>t </em>time,
  • [A]o: initial concentration of the reactant A,
  • K: rate constant,
  • t: elapsed time of the reaction

<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>

Data:

K = 4.2 ×10−3atm/s,  

[A]o=[HBr]o= 2 atm,  

[A]=[HBr]=0 atm (all HBr(g) is gone)

<em>We clear the incognita :</em>

[A] = - Kt + [Ao]............. Kt =  [Ao] - [A]

                                        t  = ([Ao] - [A])/K

<em>We replace the numerical values:</em>

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0
3 years ago
PLEASE HELP! Class is almost ending, I don't want to repeat.
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ANSWERS:

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3. 2 C_{10}  H_{22} +  31O_{2} ⇒ 20C O_{2} + 22 H_{2} O

4. 4 H_{2} + Fe_{3} O_{4} ⇒ 3Fe +4 H_{2} O

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I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)

6 0
3 years ago
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