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siniylev [52]
2 years ago
12

A sample 0.100 moles of a gas is collected at at stp. what is the volume of the gas in liters? group of answer choices 2.44 l

Chemistry
1 answer:
bearhunter [10]2 years ago
8 0

n = 0.100 moles

At STP (standard temperature pressure), where gas is collected, the temperature is 0°C and the pressure is 1 atm.

\\$\therefore$ Temperature $(T)=0^{\circ} \mathrm{C}=0^{\circ}+273

                              =273K

          Pressure $(P)=1$ atm.

The ideal gas equation indicates that

\begin{aligned}& P V=n R T \Rightarrow V=\frac{n R T}{P} \\\therefore & R=\text { gas constant }=0.0821 \mathrm{~L} \cdot \mathrm{cetm} / \mathrm{mo} / \cdot K\end{aligned}

 \begin{aligned}&V=\frac{0.100 \mathrm{~mol} \cdot \times 0.0821 \mathrm{~L} \cdot \mathrm{atm} / \mathrm{mol} . \mathrm{K} \times 273 \mathrm{~K}}{1 \mathrm{~atm}} \\&V=\frac{2.24 L}{1}\end{aligned}\\

V=2.24L

Group of answer choices 2.44L .The volume occupied by  mole of a given gas at a given temperature and pressure is expressed as the gas's molar volume.

The most typical illustration is the molar volume of a gas at STP, which is equal to 2.44L for 1 mole of any ideal gas at a temperature of 273.15 Kand a pressure of 1 atm.

Hence, the volume of the gas is 2.24L.

<h3 /><h3>What is the STP?</h3>

A unit is stated to have a temperature of absolute zero (273 Kelvins) and an atmospheric pressure of one atmosphere, or 1 atm, at standard temperature and pressure. Additionally, at STP, a mole of any gas takes up 22.414 L of space. Keep in mind that this idea only applies to gases.

For experimental measurements to be established under standard conditions that allow for comparisons between various sets of data, standard temperature and pressure must be met.

To learn more about STP, Visit:

brainly.com/question/1626157

#SPJ4

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Solve the ideal gas law equation for pressure.
posledela

Answer:

p=\frac{nRT}{V}

Explanation:

The ideal gas law equation is an equation that relates some of the quantities that describe a gas: pressure, volume and temperature.

The equation is:

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R is the gas constant

T is the absolute temperature of the gas (must be expressed in Kelvin)

Here we want to solve the equation isolating p, the pressure of the gas.

We can do that simply by dividing both terms by the volume, V. We find:

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants
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Explanation:

Reaction equation is as follows.

      Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)

Here, 1 mole of Na_{2}SO_{3} produces 2 moles of cations.

[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58

                                  = 1.16 M

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As,     K_{b} = \frac{K_{w}}{K_{a_{2}}}

                       = \frac{10^{-14}}{6.3 \times 10^{-8}}

                       = 1.59 \times 10^{-7}

According to the ICE table for the given reaction,

          SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

        K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}

 1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}

        x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)

                x = 0.0003 M

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The reaction will be as follows.

              HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

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        K_{b} = \frac{K_{w}}{K_{a_{1}}}

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As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  x^{2} = 7.14 \times 10^{-13} \times 0.0003

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                = 3.33 \times 10^{-11} M

Thus, we can conclude that the concentration of spectator ion is 3.33 \times 10^{-11} M.

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