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love history [14]
3 years ago
10

A gas fills a balloon at a temperature of 27 C and 101.325 kPa of pressure. What will the pressure of the balloon be if the gas

is heated to 127 C?
Chemistry
1 answer:
JulijaS [17]3 years ago
3 0

135.1‬kPa

Explanation:

Given parameters:

T1 = 27°C

P1  = 101.325 kPa

T2 = 127°C

Unknown:

P2 = ?

Solution:

Using a derivative of the combined gas law where we assume that the gas has a constant volume, we can solve for the unknown.

 At constant volume:

           \frac{P1}{T1}  = \frac{P2}{T2}

P1 is the initial pressure

T1 is the initial temperature

P2 is the final pressure

T2 is the final temperature

  Take the given temperature to K

T1 = 27 + 273 = 300K

T2 = 127 + 273  = 400K

 Input the variables:

    \frac{101.325}{300}  = \frac{P2}{400}

  P2 = 135.1‬kPa

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

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Given a solution with a hydroxide ion concentration, oh=3.73×10^-5,what is the ph
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Answer:

  • <em><u>pH </u></em><u>= 9.73</u>

<u></u>

Explanation:

<em>pOH</em> is the negative logarithm of the molar <em>concentration</em> of<em> hydroxide ions </em>and pH is the negative logarithm of the concentration of hydronium ions.

The mathematical expression to determine pOH is:

         pOH=-\log [OH^-]

Where [OH⁻] means the molar concentration of hydroxide ions.

Substituting:

        pOH=-\log {(3.73\times 10^{-5})

       pOH=4.43

<em>pH </em>and <em>pOH</em> are related by the equation:

           pH+pOH=14

From which:

           pH=14-pOH\\\\pH=14 - 4.43\\\\pH=9.57\leftarrow answer

8 0
3 years ago
Please help due soon.
Vikentia [17]

Answer:

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Explanation:

Hopes it helps

3 0
3 years ago
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vova2212 [387]

STEM is an acronym meaning

Science

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One of the main focuses of STEM courses is on science, so you will be learning about various types of science

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8 0
3 years ago
For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:
Leviafan [203]
1) Chemical reaction

HCl        +       NaOH      --->      NaCl + H2O

25.0 ml            
0.150 M            0.250M

2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution

0.001875 mol HCl => 0.001875 mol H(+)

Volume = Volume of HCl solution + Volumen of NaOH solution added

Volume of HCl solution = 0.0250 l

Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l

Total volume = 0.0250 l + 0.0075 l = 0.0325 l

[H+] = 0.001875 mol / 0.0325 l = 0.05769 M

pH = - log [H+] = - log (0.05769) = 1.23

Answer: 1.23

3) Equivalence point

0.02500 l * 0.150 M = 0.250M * V

=> V = 0.02500 * 0.150 / 0.250 = 0.015 l

4) 1.00 ml NaOH added beyond the equivalence point

1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess

0.00025 mol NaOH = 0.00025 mol OH-

Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l

[OH-] = 0.00025 mol / 0.041 l = 0.00610 M

pOH = - log (0.00610) = 2.21

pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76

Answer: 11.76
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