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3 years ago

If a gas occupies 79.5 mL at -1.4°C, what temperature, in Kelvin, would it

Chemistry

1 answer:

Anna35 [415]3 years ago

5 0

Answer:

121 K

Explanation:

Step 1: Given data

Initial volume (V₁): 79.5 mL

Initial temperature (T₁): -1.4°C

Final volume (V₂): 35.3 mL

Step 2: Convert "-1.4°C" to Kelvin

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

Step 3: Calculate the final temperature of the gas (T₂)

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

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An electron can be added to halogen atom to force a halide ion with 8 valence electrons

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2 years ago

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Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

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The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

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3 years ago

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3 years ago

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A piece of iron is heated to 95.0 celsius and then placed in an insulated vessel containing 250. grams of water at 25.0 celsius.

Nana76 [90]

Answer:

c) 387g

Explanation:

Water;

Mass = 250g

Specific heat = 4.184

Initial Temp, T1 = 25 + 273 = 298K

Final Temp, T2 = 35 + 273 = 308K

Heat = ?

H = mc(T2 - T1)

H = 250 * 4.184 (308 - 298)

H = 10460 J

Iron;

Initial Temp, T2 = 95 + 273 = 368K (Upon converting to kelvin temperature)

Mass = ?

Final Temp, T1 = 35 + 273 = 308

Heat = 10460 (Heat lost by iron is qual to heat gained by water)

Specific heat = 0.45

H = mc(T2-T1)

M = 10460 / [0.45 (308 - 368)]

M = 10460 / 27

M = 387g

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3 years ago