Moles = mass/molar mass, so n(C2H6O)= 16.0 / (12+12+(1x6)+16)
=0.348 (to correct sig figs)
Answer:
A. Carbon double bonded to oxygen and a hydroxyl group (OH).
This should help :)
Example 1: A 36.0 g sample of water is initially at 10.0 °C.
How much energy is required to turn it into steam at 200.0 °C? (This
example starts with a temperature change, then a phase change followed
by another temperature change.)
Solution:
<span>q = (36.0 g) (90.0 °C) (4.184 J g¯1 °C¯1) = 13,556 J = 13.556 kJ
q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ
q = (36.0 g) (100.0 °C) (2.02 J g¯1 °C¯1) = 7272 J = 7.272 kJ
q = 102 kJ (rounded to the appropriate number of significant figures)
</span>
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>
<em />
As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
<em>Where X is reaction coefficient.</em>
<em />
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X²
]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
<h3>[HI] = 0.264M</h3>
