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expeople1 [14]
2 years ago
8

How many moles of gas occupy 98L at a pressure of 2.8atm and a temperature of 292k? and what law would you use?

Chemistry
1 answer:
algol132 years ago
8 0
Let's assume that the gas has ideal gas behavior.

Then we can use ideal gas equation,
PV = nRT

Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)

The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?

By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
                                       n = 11.45 mol

Hence,moles of gas is 11.45 mol.
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Since the membrane of the bag is semipermeable, then the fact that the bag in the beaker decreased in size, lost volume, and became flaccid indicates that the solution in the bag is of lower solute concentration than the solution in the beaker hence the movement of water molecules into the beaker by osmosis.

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morpeh [17]

T eg(1) : 97.5°C, T eg(2): 98.5°C, T eg(3): 99.2°C

∆T water(1): -2.5°C, ∆T water(2): -1.5°C, ∆T water(3): -0.8°C

∆T metal(1): 77.5°C, ∆T metal(2): 80.5°C, ∆T metal(3): 80.2°C.

ft= (m1 cp1 t1 + m2 cp2 t2 + .... + mn cpn tn) / (m1 cp1 + m2 cp2 + .... + mn cpn) (1)

where,

1000g = 1kg

ft(t eg)= final mixed temperature (°C)

m = mass of substance (kg)

cp = specific heat of substance (J/kg°C)

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The half-life of a reaction, t1/2, is the time required for one-half of a reactant to be consumed. It is the time during which t
ycow [4]

<u>Answer:</u> The half life of the reaction is 593.8 seconds

<u>Explanation:</u>

We are given:

Rate constant = 0.0016mol/L.s

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\text{Unit}=\frac{(Concentration)^{1-n}}{Time}

where, n = order of reaction

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The unit of time is, second or 's'

Evaluating the value of 'n' from above equation:

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The equation used to calculate half life for zero order kinetics:

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