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2 years ago

A scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

Chemistry

1 answer:

pav-90 [236]2 years ago

3 0

Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Amount of moles of helium within the balloon</h3>

In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?

$amount of moles of helium=\frac{75 Lx1 mole}{22.4 L}$

<u><em>amount of moles of helium= 3.35 moles</em></u>

Finally, 3.35 moles of helium are contained within the balloon.

Learn more about STP conditions:

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An aqueous solution is 40.0 % by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. The mole fraction of hydrochloric

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The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

<em>Molar concentration = mass of HCl/ molar mass of HCl</em>

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We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

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3 years ago

The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be

Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature $T_1 =$ 1001 K

Final temperature $T_2 =$ 298 K

Applying the equation of Arrhenius theory.

$In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})$

where ;

R gas constant = 8.314 J/K/mol

$In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})$

$In \dfrac{k_2}{0.380 }= -70.8655$

$\dfrac{k_2}{0.380 }= e^{-70.8655}$

$\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}$

${k_2}= 1.67303256 \times 10^{-31} \times {0.380 }$

${k_2}= 6.3575 \times 10^{-32}$ /M .sec

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At 1001 K.

$t_{1/2} = \dfrac{In_2}{k_1}$

$t_{1/2} = \dfrac{0.693}{0.38}$

$t_{1/2} =$ 1.82368 secc

At 298 K:

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$t_{1/2} =1.09 \times 10^{31} \ sec$

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