The combustion of 1 mole of CO according to the reaction CO(g) + ½O2(g) → CO2(g) + 67.6 kcal gives off how much heat?

1 answer:

5 0

When the enthalpy value is given, we can calculate how much heat is use or produces in a given equation.

67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)

so 1 mole of CO gives us 67.6 kCal of heat.

calculation:

1 mol CO $\frac{67.5 kcal}{1 mol CO} = 67.5 kcal$

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An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

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3 years ago

A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung

BlackZzzverrR [31]

Answer:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol