Answer:
2
Explanation:
In two reactions energy is released.
1) C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂ + heat
It is cellular respiration reaction.It involves the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.
Glucose + oxygen → carbon dioxide + water + 38ATP
2) 2H₂ + O₂ → 2H₂O ΔH = -486 kj/mol
The given reaction is formation of water. In this reaction oxygen and hydrogen react to form water and 486 kj/mol is also released.
The reaction in which heat is released is called exothermic reaction.
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
24.6 ℃
<h3>Explanation</h3>
Hydrochloric acid and sodium hydroxide reacts by the following equation:
which is equivalent to
The question states that the second equation has an enthalpy, or "heat", of neutralization of 
. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 
or 
of energy.
500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 
of energy.
Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.
The question has given the heat capacity of the calorimeter directly.
The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.
The calorimeter contains 1.00 liters or 
of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 
.
The solution has a specific heat of 
. The solution thus have a heat capacity of 
. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.
The calorimeter-solution system thus has a heat capacity of 
, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 
are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.
Answer:
kf = 1.16 x 10¹⁸
Explanation:
Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹
Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹
Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹
________________________________________________________
Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r
ΔG°r = ΔG°1 + ΔG°2 + ΔG°3
ΔG°r = -42.9 - 35.8 - 24.3
ΔG°r = -103.0 kJmol⁻¹
ΔG°r = -RTlnKf
-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf
kf = e ^(-103,000/-8.31x298)
kf = e ^41.59
kf = 1.16 x 10¹⁸
The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
What is meant by concentration?
Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.
Concentration of hydroxide ions can be calculated by,
M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.
where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.
Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
To learn more about concentration click on the given link brainly.com/question/17206790
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