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3 years ago

How is enthalpy used to predict whether a reaction is endothermic or exothermic?

Chemistry

2 answers:

lana [24]3 years ago

5 0

The correct option is this: WHEN THE ENTHALPY CHANGE OF THE REACTION IS POSITIVE, THE REACTION IS ENDOTHERMIC.
All chemical reactions are divided into two classes, these are exothermic and endothermic reaction. An exothermic reaction is a type of reaction in which energy is lost to the environment, thus the enthalpy change of the reaction is negative. An endothermic reaction is a type of reaction in which energy is absorbed from the environment, thus the enthaply change of the reaction is positive.

UkoKoshka [18]3 years ago

5 0

Answer: When the enthalpy change of the reaction is positive, the reaction is endothermic.

Explanation: Exothermic reactions are defined as the reactions in which energy of the product is less than the energy of the reactants. The total energy is released in the form of heat and $\Delta H$ for the reaction comes out to be negative.

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 $\Delta H$ for the reaction comes out to be positive.

$\Delta H=H_{products}-H_{reactants}$

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how many liters of a 60% antifreeze solution must be added to 8L of a 10% antifreeze solution to produce a 20% antifreeze soluti

Tasya [4]

Answer: 2Liters

Explanation:

The expression used will be :

$M_1V_1+M_2V_2=M_3V_3$

where,

$M_1$ = concentration of first antifreeze= 60%

$M_2$ = concentration of second antifreeze= 10%

$V_1$ = volume of first antifreeze = x L

$V_2$ = volume of second antifreeze = 8 L

$M_3$ = concentration of final antifreeze solution= 20%

$V_3$ = volume of final antifreeze = (x+8) L

Now put all the given values in the above law, we get the volume of antifreeze added

$60\times x+10\times 8=20\times (x+8)$

$x=2L$

Therefore, the volume of 60% antifreeze solution that must be added is 2L

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3 years ago

Stirring will A. Increase a reaction B. Decrease a reaction C. Cause no change D. Create a catalyst

lbvjy [14]

A. Increase reaction

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2 years ago

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Answer:

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2 years ago

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What occurs along a convergent plate boundary

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3 years ago

What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.100 M NaOH? Assume that the volu

klemol [59]

Answer:

The pH of the solution is 12.78.

Explanation:

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution. The pOH is defined as the negative logarithm of the activity of the hydroxide ions. That is, the concentration of OH- ions:

pOH= - log [OH-]

On the other side, Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume. Molarity is calculated as:

$Molarity= \frac{number of moles}{volume}$

Molarity is expressed in units: $\frac{moles}{liter}$

In this case, the solution is prepared by mixing 100 ml (equal to 0.1 L, where 1000 mL = 1 L) of Ca(OH)₂ 0.020 M with 50 ml (equal to 0.05 L) of 0.100 M NaOH. Then, Ca(OH)₂ and NaOH are strong bases, so they dissociate completely. In the case of the first hydroxide, for each mole of Ca(OH)₂,

form two moles of OH-. In the case of sodium hydroxide, for each mole of hydroxide, one mole of OH- is formed. So, taking into account the definition of molarity, the number of moles of OH- that each hydroxide contributes to the solution is calculated as:

From Ca(OH)₂: 0.1 L* 0.02 M*2 = 0.004 moles

From NaOH: 0.05 L* 0.1 M= 0.005 moles

So, the amount of total moles of OH- is the sum that each hydroxide contributes to the solution: 0.004 moles + 0.005 moles= 0.009 moles

On the other hand, volumes are additive. Then: 0.1 L + 0.05 L= 0.15 L

Replacing in the definition of molarity the number of moles and the volume:

$[OH-]=\frac{0.009 moles}{0.15 L}$

Solving:

[OH-]= 0.06 $\frac{moles}{liter}$

Replacing in the definition of pOH:

pOH= - log 0.06

pOH= 1.22

The following relationship can be established between pH and pOH:

pH + pOH= 14

Being pOH= 1.22 and replacing:

pH + 1.22= 14

pH= 14 - 1.22

pH= 12.78

<u><em>The pH of the solution is 12.78.</em></u>

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2 years ago