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2 years ago

what is the vapor pressure of water if the total vapor pressure is 1.5 atm and the pressure of oxygen is 0.2 atm?

1 answer:

8 0

Since the total pressure of each gas is the sum of the partial pressures of all the individual gases, the vapor pressure of oxygen is 1.3 atm.

From Dalton's law of partial pressure, the total pressure is the sum of the individual partial pressures of the gases;

PT = PA + PB + PC + --------

Given that;

Vapor pressure of oxygen = 0.2 atm

Total pressure = 1.5 atm

Vapor pressure of oxygen = ?

Partial pressure of oxygen = 1.5 atm - 0.2 atm = 1.3 atm

The vapor pressure of oxygen is 1.3 atm.

Learn more: brainly.com/question/2510654

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Which of these accurately describes the products of a reaction?

stiv31 [10]

The one that accurately describes the products of a reaction is : B. new substances that are present at the end of a reaction
For example the process of photosynthesis transform CO2 and other nutrients into O2 and H2O

hope this helps

3 0

2 years ago

Read 2 more answers

a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b

SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .