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marishachu [46]
3 years ago
13

Chemical equation for heat of formation of c25h52

Chemistry
1 answer:
Alexus [3.1K]3 years ago
8 0
 its exothermic. exothermic means the reaction gives out heat. [exo meaning outside and thermic meaning heat] because its a combustion reaction, fire is there meaning lots of heat. 
endothermic means it takes in heat. a good example is ammonium and water (NH3 +H20) 
<span>most chemical reaction are exothermic</span>
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A solution is saturated in CO2 gas and KNO3 at room temperature. What happens when the solution is warmed to 75°C?
nignag [31]

Answer:

gaseous CO2 bubbles out of the solution

Explanation:

We already know that the dissolution of a gas in water is exothermic. Hence, when the temperature of a solution containing a gas is increased, the solubility of the gas decreases and the gas bubbles out of the solution.

Similarly, the dissolution of KNO3 in water is endothermic. This implies that the solubility of the solid increases with increasing temperature.

Thus the solid becomes more soluble at 75°.

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The element nickel has the following properties. Select the two that are chemical properties of nickel.
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For a fixed amount of gas at a fixed volume, what will happen if the absolute temperature is doubled?
Firlakuza [10]
The answer to this question is: The pressure will be increased two times

The gas law includes an interaction of pressure(p), volume(V) and temperature(T). In this case, the gas has same amount and volume but then the temperature is doubled. The calculation would be:

p1 * V1 / T1= p2*V2 /T2 
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Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
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