In this solution we will need two steps, 1) calculate the heat required to get water to the temp of 100 C and 2) find the latent heat of vapourization required.
First, change the 11.8g to kg: = 0.0118kg
Equations: Q = mcT, Q=mL
c for water = 4186 J/(kg C)
L for water = 22.6 x 10^5 J/kg
1) Q = mcT
Q = 0.0118 kg x 4186 J/(kg C) x (100 C - 65.5 C)
Q = 1704.12 J
2) Q = mL
Q = 0.0118 kg x 22.6 x 10^5
Q = 26668 J
Qf = 1704.12J + 26668J
Qf = 28372.12J
With sig figs: <u>2.84 x 10^4 J</u>
Answer:
Explanation:
The compound 
is .
The reaction included are:
(because bromine is less reactive than chlorine)
But , because of hexane the solution get dilute and its color changes to orange.
Now, NaI is added and we know Br is more reactive than I.
Therefore it replace it.
Reaction: 
Purple color develop due to formation of Iodine.
Answer: 780 kcal are required to produce 12.00 mol of 
Explanation:
Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and written along with reactants.
The balanced chemical reaction is:
According to stoichiometry :
2 moles of are produced by absorption of energy = 130 kcal
Thus 12.00 moles of are produced by absorption of energy = 
Thus 780 kcal are required to produce 12.00 mol of
Not always ammonium salts of weak acids form neutral solutions.
When formic acid reacts with ammonia, ammonium formate is produced:
HCO2H + NH3 ----> NH4HCO2
You already know that the weak conjugate bases of NH3 and HCO2H are NH4+ and HCO2, respectively.
How can the pH of the solution be calculated if the salt's anion causes the pH to rise and the salt's cation causes it to fall? The relative intensities of the basic anion and the acidic cation hold the key to the solution.
As was already established, formate is a weak base and will create hydroxide ions in water, whereas ammonium is a weak acid and will make hydronium ions in water.
NH4⁺ + H2O -----> NH3 + H3O⁺
HCO2⁻ + H2O -----> HCO2H + OH⁻
Since the acid ionization of NH4+ is more favored than the base ionization of HCO2-, the solution will be acidic.
To learn more about ammonium salts:
brainly.com/question/10874844
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Answer : The pH of a 0.1 M phosphate buffer is, 6.86
Explanation : Given,
Concentration of acid = 0.1 M
Concentration of conjugate base (salt) = 0.1 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of a 0.1 M phosphate buffer is, 6.86
