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3 years ago

Compute Reaction Rates for All Seven Trials

Chemistry

2 answers:

jeka943 years ago

7 0

Answer : The reaction rate at $3^oC,24^oC,40^oC\text{ and }65^oC$ are $36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec$

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

The given formula will be,

$\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}$

Now we have to calculate the reaction rate at different temperatures and reaction time.

$\text{Reaction rate at }3^oC=\frac{1000mg/0.200L}{138.5sec}=\frac{5000mg/L}{138.5sec}=36mg/L/sec$

$\text{Reaction rate at }24^oC=\frac{1000mg/0.200L}{34.2sec}=\frac{5000mg/L}{34.2sec}=142mg/L/sec$

$\text{Reaction rate at }40^oC=\frac{1000mg/0.200L}{26.3sec}=\frac{5000mg/L}{26.3sec}=190mg/L/sec$

$\text{Reaction rate at }65^oC=\frac{1000mg/0.200L}{14.2sec}=\frac{5000mg/L}{14.2sec}=352mg/L/sec$

Therefore, the reaction rate at $3^oC,24^oC,40^oC\text{ and }65^oC$ are $36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec$

Mila [183]3 years ago

5 0

Answer:

36, 146, 190, 352

Explanation:

That is the correct answer on edg

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what pressure is generated when 5 mol of ethane is stored in a volume of 8 dm 3 at 25°C? Base calculations on each of the follow

Goshia [24]

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b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

PV = nRT

where:

P = pressure [atm]
V = volume [L]
n = number of mole of gas [n]
R= gas constant = 0,08205 [atm.L/mol.°K]
T=absolute temperature [°K]

Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.

The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the intermolecular forces and the space occupied by the gas

$\frac{Pv}{RT} = 1 + \frac{B}{v}$

where:

v is the molar volume [L/mol]
B is the second virial coefficient [L/mol]
P the pressure [atm]
R the gas constant = 0,08205 [atm.L/mol.°K]

a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

$\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}$

and v = 3 L/5 moles = 0,6 L/mol

$P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm$

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3 years ago