If a candle is burning in an open room, which reactant is in excess?

Suppose you carry out a titration involving 3.00 molar HCl and an unknown concentration of KOH. To bring the reaction to its end

snow_tiger [21]

Answer: The concentration of the KOH solution is 1.01 M

Explanation:

According to neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of $HCl$ = 1

$n_2$ = acidity of KOH = 1

$M_1$ = concentration of HCl = 3.00 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of HCl = 35.3 ml

$V_2$ = volume of KOH = 105.0 ml

Putting the values we get:

$1\times 3.00\times 35.3=1\times M_2\times 105.0$

$M_2=1.01$

Thus the concentration of the KOH solution is 1.01 M

3 0

4 years ago

Predict the sign of ΔS for each process:(c) Cl₂(g) (100°C and 1 atm) → Cl₂(g) (10°C and 1 atm)

dexar [7]

The entropy for the process:

Cu(s) (350°C and 2.5 atm) → Cu(s) (450°C and 2.5 atm) decreases. The sign of entropy is negative.

<h3>What is Entropy? </h3>

Entropy is defined as the measure of the thermal energy of a system per unit temperature which is unavailable for doing useful work. As the work is obtained from ordered molecular motion, therefore the amount of entropy is also gives a measure of the molecular disorder, or randomness, of a system.

<h3>Effect of Temperature</h3>

There is increase in the Entropy as temperature increases. An increase in the temperature results that particles of substance have greater kinetic energy. The fast moving particles have more disorder than particles which are moving slowly at the given lower temperature.

<h3>Effect of Pressure</h3>

The entropy increases with increases in the pressure on the substance.

Since the temperature of given reaction decreases. So, entropy also decreases.

Thus, we concluded that the entropy for the process:

Cu(s) (350°C and 2.5 atm) → Cu(s) (450°C and 2.5 atm) decreases. So, the sign of entropy is negative.

learn more about Entropy:

brainly.com/question/14131507

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7 0

2 years ago

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank

marysya [2.9K]

The rate of Formation of Carbocation mainly depends on two factors'

1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

R-I > R-Cl > R-F

B > C > A