Chemistry, 06.12.2019 04:31 Frankia

Suppose that a fully charged lead-acid battery contains 1.50 l of 5.00 m h2so4. what will be the concentration of h2so4 in the battery after 2.20 a of current is drawn from the battery for 9.50 hours ? express your answer with the appropriate units.

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Answer from: leximae7720

The molar concentration of the solution is 4.46 M.

Explanation:

From the given,

Molarity of sulfuric acid = 5.00 M

Volume of sulfuric aci = 1.50L

Amperes of battery = 2.20 A

Time = 9.50 h

$Charge= Current \times time(s)$

$=2.30 \times 9.50 \times 3600 s = 78660 \, C$

$Moles\,of\,electrons = \frac{charge}{Faraday\,constant}$

$= \frac{78660}{96485}= 0.8153 mol$

$Mole\,of\,H_{2}SO_{4}\,consumed = moles\,of\,electrons = 0.8153\,mol$

$Initial\,moles\,of\,H_{2}SO_{4}\,=Volume \,\times Initial\,concentration$

$1.50 \times 5.00 = 7.5\,mol$

$Final\,moles\,of\,H_{2}SO_{4}=7.5-0.8513=6.6847\,mol$

$Final\,concentration\,of\,H_{2}SO_{4}= \frac{final\,moles}{Volume}$

$= \frac{6.6847}{1.50}=4.46M$

Therefore, The molar concentration of the solution is 4.46 M.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

1.from liquid state to gaseous state is evaporation 3.solid to liquid is melting 5. solid to gaseous state is sublimation 6. gas to solid state is decomposition 7. a change of state is known as the states of matters

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Suppose that a fully charged lead-acid battery contains 1.50 l of 5.00 m h2so4. what will be the con...

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