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1 year ago

What pressure does 3.54 moles of chlorine gas at 376 k exert on the walls of it 51.2 l container? I'm Lazy sooooo

Chemistry

1 answer:

Lilit [14]1 year ago

6 0

Answer:

The answer to your question is P = 2.13 atm

Explanation:

Data

Pressure = ?

number of moles = 3.54

Temperature = 376 °K

Volume = 51.2 L

R = 0.08205 atm L/mol°K

Formula

PV = nRT

- Solve for P

P = nRT / V

- Substitution

P = (3.54)(0.08205)(376) / 51.2

- Simplification

P = 109.21 / 51.2

Result

P = 2.13 atm

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Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago

Would an alkali metal make a good replacement for tin in a tin can? Explain.

Igoryamba

Answer:

Explanation:

If tin is heated, it can react with alkalis' with the release of hydrogen.

6 0

1 year ago

6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy ( $\Delta G$ ) should be negative.

We know, $\Delta G=\Delta H-T\Delta S$ , where T is temperature in Kelvin scale.

Reaction A: $\Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{10.5\times 10^{3}}{30}$

or, $T> 350$

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: $\Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{1.8\times 10^{3}}{-113}$

or, $T> -16$

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0

2 years ago

What is the mass of a sample of iron if that sample lost 2300J of heat energy when it cooled from 80 oC to 30°C? The specific he

irakobra [83]

102 grams.
Equation:
Quantify of heat = mass x specific heat x difference in temperature
We have: quantity of heat : 2300J
specific heat: .449 J/g
difference in t: 80 - 30 = 50
Solve for mass: 2300 = mass x 0.449 x 50
mass = 102.449
2 sig-figs --> 102 grams

7 0

2 years ago

ANSWER ASAP! WILL GIVE BRAINLIEST! A pan containing 20.0 grams of water was allowed to cook from a temperature of 95.0 degrees C

vodka [1.7K]

Answer:

C. 81 degrees Celsius

Explanation:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67 °C ≅ 81.0 °C.

So, the right choice is: C. 81 degrees Celsius

3 0

2 years ago

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