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ohaa [14]
3 years ago
14

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem

perature (in K) is needed to reduce the volume to 2.50 L? Hint: Use Charles' Law V1/T1 = V2/T2 where V1= initial volume; T1 = initial temperature; V2 = final volume and T2 = final temperature
Chemistry
1 answer:
Ludmilka [50]3 years ago
8 0

Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L

⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K

⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

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1) Molar mass C8H9NO2

Element    Atomic mass    # of atoms   mass
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C               12                         8               12*8 = 96
H                 1                         9                1*9 =    9
N               14                         1               14*1 = 14
O               16                        2                16*2 = 32

                              molar mass =   96 + 9 + 14 + 32 = 151 g/mol

2) Number of mols in a tablet

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3 years ago
When 2.00 g of methane are burned in a bomb calorimeter, the change in temperature is 3.08°C. The heat capacity of the calorimet
melisa1 [442]

Answer:

The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

q=c\times \Delta T

where,

q = Heat gained = ?

c = Specific heat = 2.68 kJ/^oC

ΔT =  The change in temperature = 3.08°C

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

q=2.68 kJ/^oC\times 3.08^oC

q=8.2544 kJ

Now we have to calculate molar enthalpy of combustion of this substance :

\Delta H_{comb}=-\frac{q}{n}

where,

\Delta H_{comb} = enthalpy change = ?

q = heat gained = 8.2544kJ

n = number of moles methane = \frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole

\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole

Therefore,  the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

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3 years ago
When an organic molecule loses hydrogen atoms it is said to be:_______.
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C. Oxidized and reduced are the same.
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What mass of carbon dioxide gas occupies a volume of 81.3L at 204 kPa and a temperature of 95.0°C?
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Answer:

236.9g

Explanation:

Given parameters:

Volume of gas  = 81.3L

Pressure of gas = 204kPa

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Unknown:

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Solution:

To solve this problem, the ideal gas law will be well suited. The ideal gas law is a fusion of Boyle's law, Charles's law and Avogadro's law.

Mathematically, it is expressed as;

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Since the unknown is mass;

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