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sammy [17]
3 years ago
12

The world's total petroleum reserve is estimated at 2.0 x 10^22 joules La joule (J) is the unit of energy where 1J =1kg m^2/s^2)

. At the present rate of consumption, 1.8 x 10^20 joules per year (J/yr), how long would it take to exhaust the supply?
Chemistry
1 answer:
mamaluj [8]3 years ago
3 0

Answer:

It will take 1.1\times 10^{2}years to exhaust the supply

Explanation:

We have to apply unitary method to solve this problem.

Divide total petroleum reserve by petroleum consumption in each year to calculate estimated time.

Presently, 1.8\times 10^{20} joules of petroleum are being consumed per year.

Hence, applying unitary method, 2.0\times 10^{22} joules of petroleum can be consumed in \frac{2.0\times 10^{22}}{1.8\times 10^{20}}years=1.1\times 10^{2}years

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3 years ago
A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘c and 1.00 atm. The molar mass of oxygen is 32
patriot [66]

Answer:

Mass of O₂ present in the room: 185200 g

185.2 kg

Quantity of O₂ = 5787.5

Explanation:

The ideal gases Law can solved, what is ths mass of O₂ present in the room, or the quantity of O₂ (moles), present in the room.

P . V = n . R . T

P = 1 atm

V = volume of the room

n = moles

R = 0.082 L.atm/mol.K

T = T° in K

T° C + 273 → 22°C + 273 = 295K

Let's find out the volume of the room

7 m . 8 m . 2.5 m = 140 m³

Units of R, are in L, so we must convert m³ to dm³

1dm³ = 1 L

140m³ . 1000 = 140000 dm³

1.4x10⁵L

1 atm . 1.4x10⁵L = n . 0.082 L.atm/mol.K . 295K

(1 atm . 1.4x10⁵L) / (0.082 L.atm/mol.K . 295K)  = n

5787.5 moles = n

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3 0
3 years ago
For the reactions below, describe the reactor system and conditions you suggest to maximize the selectivity to make the desired
alexandr1967 [171]

Answer:

hello your question lacks the required reaction pairs below are the missing pairs

Reaction system 1 :

A + B ⇒ D  -r_{1A}  = 10exp[-8000K/T]C_{A}C_{B}

A + B ⇒ U -r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}

Reaction system 2

A + B ⇒ D  -r_{1A} = 10exp( -1000K/T)C_{A}C_{B}

B + D ⇒ U  -r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}

Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

reaction 2 :

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

Explanation:

Reaction system 1 :

A + B ⇒ D  -r_{1A}  = 10exp[-8000K/T]C_{A}C_{B}

A + B ⇒ U -r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}

the selectivity of D is represented  using the relationship below

S_{DU} = \frac{-r1A}{-r2A}

hence SDu = 1/10 * \frac{exp(-800K/T)}{exp(-1000K/T)} * C_{A} ^{0.5} C_{B} ^{-0.5}

description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

Reaction system 2

A + B ⇒ D  -r_{1A} = 10exp( -1000K/T)C_{A}C_{B}

B + D ⇒ U  -r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}

selectivity of D

S_{DU} = \frac{-r1A}{-r2A}

hence Sdu = 1/10^7  *  \frac{exp(-1000K/T)}{exp(-10000K/T)} *\frac{C_{A} }{C_{D} }

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

3 0
4 years ago
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