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Natali [406]
3 years ago
9

What mass, in grams, of CO2 and H20 N is formed from 2.55 mol of propane?

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer: The mass of CO_2 and H_2O produced are 336.6 g and 183.6 g respectively.

Explanation:

The combustion reaction between propane and oxygen leads to formation of carbon dioxide and water.

Law of Conservation of mass states that the mass will remain constant for a balanced equation. This is carried out when the total number of atoms on reactant side is same as the total number of atoms on the product side. Thus the equation must be balanced.

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

a) 1 mol of propane produces = 3 moles of CO_2

Thus 2.55 mol of propane produces = \frac{3}{1}\times 2.55=7.65 moles of [tex]CO_2

mass of CO_2=moles\times {\text {molar mass}}=7.65mol\times 44g/mol=336.6g

b) 1 mol of propane produces = 4 moles of H_2O

Thus 2.55 mol of propane produces = \frac{4}{1}\times 2.55=10.2 moles of [tex]H_2O

mass of H_2O=moles\times {\text {molar mass}}=10.2mol\times 18g/mol=183.6g

The mass of CO_2 and H_2O produced are 336.6 g and 183.6 g respectively.

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How much heat must your body transfer to 500.0g of water to heat it from 25.0°C to body temperature, 37.0°C?
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Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

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Q=m\times c\times \Delta T

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Q=m\times c\times (T_2-T_1)

where,

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c = specific heat of water = 4.18J/g^oC

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T_2 = final temperature  = 37.0^oC

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

Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K

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3 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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