Answer:
37064 J
Explanation:
Data Given:
mass of Steam (m) = 16.4 g
heat released (Q) = ?
Solution:
This question is related to the latent heat of condensation.
Latent heat of condensation is the amount of heat released when water vapors condenses to liquid.
Formula used
Q = m x Lc . . . . . (1)
where
Lc = specific latent heat of condensation
Latent heat of vaporization of water is exactly equal to heat of condensation with - charge
So, Latent heat of vaporization of water have a constant value
Latent heat of vaporization of water = 2260 J/g
So
Latent heat of condensation of water will be = - 2260 J/g
Put values in eq. 1
Q = (16.4 g) x (- 2260 J/g)
Q = - 37064 J
So, 37064 J of heat will be released negative sign indicate release of energy
Answer:
Area of rectangle = 108 cm²
Given:
Base = 6 cm
Height = 18 cm
Explanation:
Area of rectangle = Base × Height
= 6 × 18
= 108 cm²
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.