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3 years ago

How many moles are present in 63.80L of oxygen gas, O2?

Chemistry

1 answer:

matrenka [14]3 years ago

4 0

Answer:

2.85moles of oxygen gas

Explanation:

Given parameters:

Volume of oxygen gas = 63.8L

Unknown:

Number of moles = ?

Solution:

We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:

1 mole of a gas at STP occupies a volume of 22.4L

So;

63.8L of oxygen gas will take up a volume of

$\frac{63.8}{22.4}$ = 2.85moles of oxygen gas

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What happens when carbon dioxide gas is cooled to -78 degree Celsius

katrin2010 [14]

Answer:

At temperatures below −78 °C, carbon dioxide changes directly from a gas to a white solid called dry ice through a process called deposition.

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2 years ago

Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.

Sever21 [200]

Answer:

The spectator ions is: $K^+$ and $SO^{2-}_4$

Explanation:

The equation of reaction between H₂ SO₄ and KOH is:

$H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}$

Rewriting this equation as ionic;

$[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+ SO_4^{2-} + 2H_2O ]$

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

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3 years ago

What is the volume of 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?

steposvetlana [31]

STP condition mean we have P=1 atm. T=273K. R=ideal gas constant, but make sure you use the one that has the same units of pressure, temperature that you are using. In this case R=0.0821 L*atm K^-1mol^1. You are provided with n=2.1 moles.
V=nRTP
Input your values and solve.

3 0

3 years ago

The period of a ripple on a lake is half a second. what is the frequency of the wave

Nataly_w [17]

Answer:

2 Hertz

Explanation:

<em>The frequency would be 2 Hertz.</em>

<u>The frequency of a wave is defined as the rate at which the particles of a medium vibrates when the wave is passed through it while the period of a wave is the time it takes the particles to make a complete cycle of vibration.</u>

The frequency of a wave is inversely related to its period and is defined by the following equation:

f = 1/t, where f is the frequency (in hertz) and t is the period (in seconds).

Hence, if the period of a ripple is 1/2 or 0.5 seconds, the frequency becomes;

f = 1/0.5 = 2 Hertz

7 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago