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Aleksandr [31]
3 years ago
15

Qué volumen ocupan 5 moles de un gas que se encuentran a 28°C y 3 atm de presión?

Chemistry
1 answer:
lys-0071 [83]3 years ago
5 0

<em>What volume do 5 moles of a gas occupy at 28 ° C and 3 atm of pressure?</em>

<em />

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

n= 5 moles

T=28 +273=301 K

P=3 atm

The volume of the gas :

\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{5\times 0.082\times 301}{3}\\\\V=\boxed{\bold{41.136~L}}

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5 0
3 years ago
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

4 0
3 years ago
What is the correct formula for xenon hexafluoride? XeF XeF6 Xe6F6 Xe6F
Vinil7 [7]
Hexa- means six
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Answer:
xenon hexafluoride → XeFl_{6}

Any questions?
3 0
3 years ago
Read 2 more answers
What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0
3 years ago
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eimsori [14]
2 boxes of A
Because C = A + B
2 of A = 20 grams
at the other hand we have 2 of B = 10
So 20 + 10 = 30 grams
3 0
3 years ago
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