Answer:
The total heat required is 4088.6 J
Explanation:
We have three processes which involve heat absorption. We have to calculate the heat of each process and then to calculate the total heat.
1- liquid ethanol is heated from 25ºC (298 K) to the boiling point 78.5ºc (351.5 K). We use specifi heat of liquid ethanol to calculate the heat absorbed in this part:
H1= m x Sh x ΔT
H1= m x Sh x (Tfinal - Tinitial)
H1= 3.95 g x 2.45 J/g.K x (351.5 K -298 K)
H1= 517.7 JExplanation:
The molarity of a 27%(v/v) aqueous ethanol solution is 4.63 M
calculation
convert 27%(v/v) to fraction = 27ml/100 ml
use density to convert 27ml to grams = 27 ml x0.79 g/ml = 21.33 g
find the number of moles of C2H6O
moles = mass/molar mass of C2H6O(46.07 g/mol)
moles is therefore = 21.33 g /46.07 g/mol =0.463 moles
find the molarity = moles /volume in liters
volume in liter = 100 ml/1000 = 0.1 L
molarity is therefore = 0.463 mole/0.1 L = 4.63 M
By atomic radius, the bottom left
Answer:
Which is the ph of a solution in which the concentration of hydroxide ion is greater than the concentration of hydrogen ion?
An acidic solution is one in which the hydrogen ion concentration is greater than the hydroxide ion concentration; in other words, the hydrogen ion concentration is greater than 1 X 10-7 M, and the hydroxide ion concentration is less than 1 X 10-7 M. In terms of pH, an acidic solution has a pH less than 7
Explanation:
Answer:
Kc for this equilibrium is 2.30*10⁻⁶
Explanation:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.
Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is defined as:
![Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%2A%5BD%5D%5E%7Bd%7D%20%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.
In this case, being:
2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)
the equilibrium constant Kc is:
![Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BN_%7B2%7D%20%5D%2A%5BH_%7B2%7D%20%5D%5E%7B3%7D%20%20%7D%7B%5BNH_%7B3%7D%20%5D%5E%7B2%7D%20%7D)
Being:
- [N₂]= 0.0551 M
- [H₂]= 0.0183 M
- [NH₃]= 0.383 M
and replacing:
you get:
Kc= 2.30*10⁻⁶
<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>
