Answer:
total mass will be = = 1.207g
Explanation:
First what is given
Pressure P= 0.988 atm Room TemperatureT = 23.5°C= 296.5 K
Volume V= 1.042 L
Nitrogen in air is 80 % (moles number) = 0.8
Ideal gas constant R = 0.0821 L atmmol-1K-1
Given mass of N = 14.01 g/mol
Given mass of oxygen O = 16.00 g/mol
Total number of moles = ?
So first we have to find the total number of moles by using formula
Total number of moles
n = PV/ RT
adding the values
moles n= 0.988atm x 1.042L / (0.0821L-atm/mole-K x 296.6K)
= 1.0294 / 24.35
= 0.042 moles (total number of moles)
So by using Nitrogen percentage
Moles of nitrogen = total moles x 80/100
= 0.042moles x 0.8
So moles of O2= Total moles – moles of N2
= 0.042moles - 0.034moles
Moles of O2 = 0.008moles
Now for finding the mass of the N2 and oxygen
Mass of Nitrogen N2 = no of moles x molar mass
= 0.034 x 28
= 0.952 g
Mass of oxygen O2 = no of moles x molar mass
= 0.008 x 32
= 0.256 g
total mass will be = Mass of Nitrogen N2 + Mass of oxygen O2
=0.952 g + 0.256 g
=1.207g
The mass of air in the flask = 1.207g
Given:
Pressure, P= 0.988 atm
Room Temperature, T = 23.5°C= 296.5 K
Volume, V= 1.042 L
Nitrogen in air is 80 % (moles number) = 0.8
Ideal gas constant R = 0.0821 L atm [tex]mol^{-1}K^{-1}[/tex]
Molar mass of N = 14.01 g/mol
Molar mass of oxygen O = 16.00 g/mol
To find:
Total number of moles = ?
Calculation for number of moles:From ideal gas law:
n = PV/ RT
n= 0.988atm * 1.042L / (0.0821 L atm [tex]mol^{-1}K^{-1}[/tex] * 296.6K)
n = 1.0294 / 24.35
n= 0.042 moles
Using mol fraction we will calculate moles for nitrogen and oxygen:
Moles of nitrogen = total moles * 80/100
Moles of nitrogen = 0.042moles * 0.8 = 0.034 moles
So, Moles of O₂ = Total moles – moles of N₂
Moles of O₂ = 0.042 moles - 0.034 moles
Moles of O₂ = 0.008 moles
Calculation for mass:
Mass of Nitrogen N₂ = no of moles x molar mass
= 0.034 x 28
= 0.952 g
Mass of oxygen O₂ = no of moles x molar mass
= 0.008 * 32
= 0.256 g
Total mass will be = Mass of Nitrogen N₂ + Mass of oxygen O₂
=0.952 g + 0.256 g
Total mass = 1.207g
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Consider the dissolution of AB(s):AB(s)⇌A+(aq)+B−(aq)Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.The generic metal hydroxide M(OH)2 has Ksp = 1.05×10−18. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)What is the solubility of M(OH)2 in pure water?
Answer:
S = 6.40 × 10⁻⁷ M
Explanation:
In order to calculate the solubility (S) of M(OH)₂ in pure water we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.
M(OH)₂(s) ⇄ M²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product (Kps) is:
Kps = 1.05 × 10⁻¹⁸ = [M²⁺].[OH⁻]²=S.(2S)²
1.05 × 10⁻¹⁸ = 4S³
S = 6.40 × 10⁻⁷ M
Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge.
Calculate the standard cell potential given the following standard reduction potentials:
Al3++3e-?Al; E?=?1.66 V
Cu2++2e-?Cu; E?=0.340 V
Answer: 2.00 V
Explanation:
The balanced redox reaction is:
[tex]2Al+3Cu^{2+}\rightarrow 2Al^{3+}+3Cu[/tex]
Here Al undergoes oxidation by loss of electrons, thus act as anode. Copper undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]Al^{3+}+3e^-\rightarrow Al[/tex] [tex]E^0_{[Al^{3+}/Al]}=-1.66V[/tex]
[tex]Cu^{2+}+2e^-\rightarrow Cu[/tex] [tex]E^0_{[Cu^{2+}/Cu]}=0.340V[/tex]
[tex]E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Al^{3+}/Al]}[/tex]
[tex]E^0=+0.34- (-1.66V)=2.00V[/tex]
Thus the standard cell potential is 2.00 V
Describe the relationship between predator and prey in a balanced ecosystem.
Predators grow along with their victims. Over time, prey animals are now developing and avoiding themselves to get eaten by their predators. Such tactics and modifications can take many forms that make their work easier, including disguise, mimicry, defense mechanisms, flexibility, distance, habits and even tool use.
Though fact an equilibrium appears to occur within an ecosystem between predators and prey, there are several factors which affect it, including the birth and death rates of predators and preys.
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 665.9 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = + 190.1 k J / m o l what is Δ H for this reaction? X ( s ) + 1 2 O 2 ( g ) + CO 2 ( g ) ⟶ XCO 3 ( s )
Answer:
ΔH of reaction is -856 kJ/mol
Explanation:
The property ΔH is property which can be added to find the net change in enthalpy of reaction.
ΔH of first reaction is -665.9 kJ/mol
ΔH of second reaction is 190.1 kJ/mol
carefully looking at the third equation,
it is first reaction - second reaction.
thus, by Hess's law,
Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
ΔH of third reaction is = [tex]-665.9 - (190.1)[/tex]
ΔH = -856 kJ/mol
Which of the following best describe the particles present in a 2.0 M aqueous solution of Al(NO3)3?2 M Al3+(aq) and 18 M NO3-(aq)2 M Al(NO3)3(aq)2 M Al3+(aq) and 6 M NO3-(aq)2 M Al3+(aq) and 3 M NO3-(aq)
Answer:
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
Explanation:
Step 1: Data given
2.0 M Al(NO3)3
Step 2:
Al(NO3)3 in water will dissociate as following:
Al(NO3)3 → Al^3+ + 3NO3^-
For 1 mol of Al(NO3)3 we will have 1 mol of Al^3+ and 3 moles of NO3^-
We know that the molarity of Al(NO3)3 = 2.0 M, this means 2.0 mol/ L
The mol ratio Al(NO3)3 and Al^3+ is 1:1 so the molarity of Al^3+ is 2.0 M
The mol ratio Al(NO3)3 and NO3^- is 1:3 so the molarity of NO3^- is 6.0M
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
Consider a sample of helium gas in a container fitted with a piston as pictured below. The piston is frictionless, but has a mass of 10.0 kg. How many of the following processes will cause the piston to move away from the base and decrease the pressure of the gas? Assume ideal behavior.
I. Heating the helium.
II. Removing some of the helium from the container.
III. Turning the container on its side.
IV. Decreasing the pressure outside the container.
a) 0
b) 1
c) 2
d) 3
e) 4
Answer:
I. heating the helium
Explanation:
Final answer:
(I) Heating the helium and (IV) decreasing the pressure outside the container are the two processes that will cause the piston to move away from the base and decrease the pressure of the gas. Hence, (c) is the correct option.
Explanation:
Assuming ideal gas behavior, the question asks which processes will lead the piston to travel away from the base and lower the helium gas's pressure. Let's examine each of the scenarios that are presented:
I. Heating the helium: Heating increases the internal energy of the gas molecules, making them move faster. This increases the force they exert against the piston, pushing it outward and increasing the volume, thus reducing the pressure according to Boyle's Law (P₁V₁ = P₂V₂).
II. Removing some of the helium from the container: If you remove some of the helium, there will be fewer molecules to exert force on the piston, leading to a decrease in pressure. However, the piston will not necessarily move since the external pressure remains the same. Without a corresponding change in external pressure, there is no force to move the piston outward.
III. Turning the container on its side: Turning the container on its side will have no effect on the pressure or volume of the gas if the external conditions remain the same. The position of the piston is influenced by forces and pressures, not by its orientation in space.
IV. Decreasing the pressure outside the container: Decreasing the external pressure acting on the piston will allow the internal pressure of the gas to push the piston outward, increasing the volume and therefore decreasing the gas pressure, as explained by the ideal gas law (PV = nRT).
Therefore, the processes that will cause the piston to move away from the base and decrease the pressure of the gas are heating the helium and decreasing the pressure outside the container. So the correct answer to the question is (c) 2.
The normal freezing point of a certain liquid
X
is
0.4°C
, but when
5.90g
of urea
NH22CO
are dissolved in
450.g
of
X
, it is found that the solution freezes at
−0.5°C
instead. Use this information to calculate the molal freezing point depression constant
Kf
of
X
the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].
To calculate the molal freezing point depression constant (\(K_f\)) of liquid X, we can use the formula:
[tex]\[ \Delta T_f = K_f \times m \][/tex]
Where:
- [tex]\( \Delta T_f \)[/tex] is the freezing point depression (given as [tex]\(0.4^\circ \text{C} - (-0.5^\circ \text{C}) = 0.9^\circ \text{C}\)[/tex]),
- [tex]\( m \)[/tex] is the molality of the solution,
- [tex]\( K_f \)[/tex] is the molal freezing point depression constant.
First, we need to calculate the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.
Given:
- Mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] = 5.90 g
- Mass of liquid X = 450.0 g
We need to find the moles of urea first:
[tex]\[ \text{moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} \][/tex]
The molar mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] is the sum of the molar masses of nitrogen, hydrogen, carbon, and oxygen:
[tex]\[ \text{molar mass of urea} = 14.01 + 2(1.01) + 12.01 + 16.00 = 60.03 \, \text{g/mol} \][/tex]
[tex]\[ \text{moles of urea} = \frac{5.90 \, \text{g}}{60.03 \, \text{g/mol}} = 0.0983 \, \text{mol} \][/tex]
Now, we can calculate the molality of the solution:
[tex]\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \][/tex]
[tex]\[ \text{molality} = \frac{0.0983 \, \text{mol}}{0.450 \, \text{kg}} = 0.218 \, \text{mol/kg} \][/tex]
Now, we can rearrange the formula for [tex]\(K_f\)[/tex] and solve for it:
[tex]\[ K_f = \frac{\Delta T_f}{m} \][/tex]
[tex]\[ K_f = \frac{0.9^\circ \text{C}}{0.218 \, \text{mol/kg}} \][/tex]
[tex]\[ K_f \approx 4.13 \, \text{°C/molal} \][/tex]
Therefore, the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].
The molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).
To calculate the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X, we can use the formula for freezing point depression:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
where:
[tex]\( \Delta T_f \)[/tex] is the freezing point depression, which is the difference between the normal freezing point of the solvent and the freezing point of the solution.
i is the van 't Hoff factor, which is the number of moles of solute particles per mole of solute dissolved. For urea, i is typically 1 because urea does not dissociate in solution.
[tex]\( K_f \)[/tex] is the molal freezing point depression constant for the solvent.
m is the molality of the solution, which is the number of moles of solute per kilogram of solvent.
First, we calculate the freezing point depression [tex]\( \Delta T_f \)[/tex]:
[tex]\[ \Delta T_f = T_f^0 - T_f = 0.4C - (-0.5C) = 0.9C \][/tex]
Next, we need to calculate the molality m of the solution. The molar mass of urea [tex]\( NH_2CONH_2 \)[/tex] is 60.06 g/mol. The number of moles of urea is:
[tex]\[ n_{urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{5.90 \text{ g}}{60.06 \text{ g/mol}} \] \[ n_{urea} = 0.0982 \text{ mol} \][/tex]
The molality m is:
[tex]\[ m = \frac{n_{urea}}{\text{mass of solvent in kg}} = \frac{0.0982 \text{ mol}}{0.450 \text{ kg}} \] \[ m = 0.2182 \text{ mol/kg} \][/tex]
Now we can rearrange the freezing point depression formula to solve for [tex]\( K_f \)[/tex]:
[tex]\[ K_f = \frac{\Delta T_f}{i \cdot m} \] Since \( i = 1 \) for urea, we have: \[ K_f = \frac{0.9C}{1 \cdot 0.2182 \text{ mol/kg}} \] \[ K_f = 4.124 \text{ C/(mol/kg)} \][/tex]
Therefore, the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).
For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
The entropy change ΔSo for a reaction can be obtained by considering the sum of decompositions and formations of reactants and products. In this scenario, while we were only able to compute the enthalpy change ΔHxn of -136.80 kJ, the entropy change, ΔSo, cannot be calculated without additional data.
Explanation:To find the entropy change ΔSo for the reaction 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g), we start by writing this as the sum of decompositions of 3NO₂(g) and 1H₂O(1) into their constituent elements. Similarly, we identify the formation of 2HNO3(aq) and 1NO(g) from their constituent elements. By summing the enthalpy changes obtained from standard enthalpy changes of formation (ΔHf) for these compounds, we find the result for ΔHxn, which in this case equates to -136.80 kJ. However, the original question asked for ΔSo, not ΔHxn. Without knowing the ΔSo for the individual substances in this reaction, we cannot calculate ΔSo for the entire reaction.
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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calorimeter at 20.4 degrees Celsius. At equilibrium the temperature of the water and metal was 39.6 degrees Celsius. Calculate the specific heat of the metal. The specific heat of the water is 4.184 J/g/C
Answer : The specific heat of metal is [tex]0.481J/g^oC[/tex].
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of metal = ?
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]m_1[/tex] = mass of metal = 129.00 g
[tex]m_2[/tex] = mass of water = 45.00 g
[tex]T_f[/tex] = final temperature = [tex]39.6^oC[/tex]
[tex]T_1[/tex] = initial temperature of metal = [tex]97.8^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]20.4^oC[/tex]
Now put all the given values in the above formula, we get
[tex]129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC[/tex]
[tex]c_1=0.481J/g^oC[/tex]
Therefore, the specific heat of metal is [tex]0.481J/g^oC[/tex].
To calculate the specific heat of the metal, the heat transfer equation q=mcΔT is used. By setting the heat lost by the metal equal to the heat gained by the water, and substituting the known values into the equation, we can solve for the specific heat of the metal.
Explanation:The specific heat of a metal can be calculated by using the concept of heat transfer, where heat lost by the metal is equal to the heat gained by the water in a calorimetry experiment. The equation is q = mcΔT, where q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For the water, the heat gained can be calculated as qwater = mwatercwaterΔTwater. For the metal, the heat lost is qmetal = mmetalcmetalΔTmetal. Given that the heat lost by the metal equals the heat gained by the water, the equation can be set up to solve for the specific heat of the metal, cmetal = (qwater / (mmetalΔTmetal)). We know the following: mwater = 45.00 g, cwater = 4.184 J/g/C, ΔTwater = final temperature - initial temperature of water, mmetal = 129.00 g, ΔTmetal = initial temperature of metal - final temperature. By substituting these values into the equation, we can find the specific heat of the metal.
What change occurs during the reaction:MnO4- --->Mn2+?Five electrons are lost.Three electrons are lost.Five electrons are gained.Three electrons are gained.
Answer:
Five electrons are gained.
Explanation:
Oxidation number or oxidation state of an atom in a chemical compound is the number of electrons lost or gained. It is also defined as the degree of oxidation of the atom in the compound.
This is a theoretical number which can help to decipher the oxidation and reduction in a redox reaction.
Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.
The given reaction is shown below as:
[tex]MnO_4^-\rightarrow Mn^{2+}[/tex]
Manganese in [tex]MnO_4^-[/tex] has oxidation state of +7
Manganese in [tex]Mn^{2+}[/tex] has an oxidation state of +2
It reduces from +7 to +2 . It means that 5 electrons are gained.
Cytochromes are critical participants in the electron transport chains used in photosynthesis and cellular respiration. How do cytochromes donate and accept electrons?
1) Every cytochrome's iron‑containing heme group accepts electrons from oxygen and donates the electrons to the next cytochrome in the chain.
2) The cytochromes donate electrons excited by photons to other cytochromes that accept electrons as replacements for lost electrons.
3) Cytochromes donate electrons directly to the energy‑carrier molecules and and accept electrons from less electronegative substances.
4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.
Answer:
4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.
Explanation:
The cytochromes are proteins that contain heme prosthetic groups. Cytochromes undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome:
[tex]Cytochrome-Fe²⁺ ⇄ cytochrome-Fe³⁺-e⁻[/tex]
The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes b and c₁, and a Fe-S center. This second complex reduces cytochrome c, a water-soluble membrane peripheral protein. Cytochrome c, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes a, a₃ and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.
Each transporter "downstream" is more electronegative than its neighbor "upstream"; oxygen is located in the inferior part of the chain. Thus, the electrons fall in an energetic gradient in the electron chain transport to a more stable localization in the electronegative oxygen atom.
Why is a mechanism needed to remove carbon dioxide gas from your cells?
Answer:
Respiration is responsible for removing carbon dioxide and providing oxygen to our body cells. Carbondioxide is eliminated from the body as a byproduct.
Explanation:
In calculating the concentration of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, the stepwise formation constants are as follows: K1=1.90×104 K2=3.90×103 K3=1.00×103 K4=1.50×102 Calculate the overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+.
Answer:
Kf = 1.11x10¹³
Explanation:
The value of Kf for a multistep process that involves an equilibrium at each step, is the multiplication of the constant of the equilibrium of each step.
Kf = K1xK2xK3xK4
Kf = 1.90x10⁴ x 3.90x10³ x 1.00x10³ x 1.50x10²
Kf = 1.11x10¹³
Final answer:
The overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ is 2.1 x 10^13.
Explanation:
The overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ can be calculated by multiplying the stepwise formation constants (K1, K2, K3, K4). In this case, the value of Kf is calculated as follows:
Kf = K1 * K2 * K3 * K4 = 1.90×10^4 * 3.90×10^3 * 1.00×10^3 * 1.50×10^2 = 2.1 x 10^13
Therefore, the overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ is 2.1 x 10^13.
A solenoid 2 cm long consists of 5653 loops of wire. If the magnetic field inside the solenoid is 4 T, what is the magnitude of the current that flows through it? A solenoid 2 cm long consists of 5653 loops of wire. If the magnetic field inside the solenoid is 4 T, what is the magnitude of the current that flows through it? 140 A 9.1×10−2 A 11 A 2.0 A
Answer:
11 A
Explanation:
The magnetic field inside a solenoid can be calculated by the equation:
B = μ*(N/L)*i
Where B is the magnetic field, μ is the magnetic permeability (which is 1.256x10⁻⁶ T/m.A at vacuum), N is the number os loops, L is the length of the solenoid (2 cm = 0.02 m), and i the current.
4 = 1.256x10⁻⁶ *(5653/0.02)*i
0.355i = 4
i ≅ 11 A
Complexes containing metals with d10 electron configurations are typically colorless because:
a) there are no d electrons to form bonds to ligands.
b) here is no d electron that can be promoted via the absorption of visible light.
c) d electrons must be emitted by the complex in order for it to appear colored.
d) a complex must be charged to be colored.
Answer:
d
Explanation:
The complexes that involve metal having d10 electrons arrangement are usually colorless because:
A). There is no d electron that can be promoted via the absorption of visible light.
The reason for which the complexes that have been made with metal possessing electron configuration of d10 being colorless is that no d electron could be encouraged through absorption of viewable light.The absence of d electron will result in leaving the arrangement to remain colorless.The key cause of d electrons' absence is that no d electron could be inspired for absorbing light that is viewable.Thus, option A is the correct answer.
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Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol, and arrange the correct mechanism for its formation. Be sure to answer all parts.
Answer:
2-methoxy-2-methylpropane
Explanation:
The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as [tex]H^+[/tex]. (See figure)
In the water molecule, the H-O-H bond angle is 105º. Which distribution of electrons around the central atom provides the best explanation for this bond angle?
4 shared pairs
2 shared pairs, 2 lone pairs
1 shared pair, 3 lone pairs
3 shared pairs, 1 lone pair
The water molecule has an H-O-H bond angle of 104.5° which is due to the central oxygen atom having 2 shared pairs and 2 lone pairs of electrons, resulting in a bent molecular geometry due to the repulsion of lone pairs and sp³ hybridization.
Explanation:The bond angle in a water molecule is closer to 104.5° rather than the stated 105°, and the distribution of electrons around the central oxygen atom that best explains this bond angle is 2 shared pairs, 2 lone pairs.
In the water molecule, the central oxygen atom is sp³ hybridized, with its four hybrid orbitals occupied by two lone pairs of electrons and two bonding pairs that form covalent bonds with hydrogen atoms.
Lone pairs require more space than bonding pairs, which leads to a repulsion that pushes the hydrogen atoms closer together, resulting in the H-O-H bond angle being slightly less than the ideal 109.5° angle of a tetrahedron.
For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H2PO4-?
A) H2PO4-(aq) + H3O+(aq)--> H3PO4(aq) + H2O(l)
B) H2PO4-(aq) + H2O(l)--> H3PO4(aq) + OH-(aq)
C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42(aq)
D) H3PO4(aq) + H2O(l)--> H3O+(aq) + H2PO4-(aq)
E) HPO42-(aq) + H2O(l)--> H2PO4-(aq) + OH-(aq)
Answer:
The correct answer is C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42-(aq)
Explanation:
The acid dissociation equilibrium involves the loss of a proton of the acid to give the conjugated acid. In this case, the acid is H₂PO₄⁻ and it losses a proton (H⁺) to give the conjugated acid HPO₄²⁻ (without a proton and with 1 more negative charge). In the aqueous equilibrium, the proton is taken by H₂O molecule to give the hydronium ion H₃O⁺.
H₂PO₄⁻(aq) + H₂O(l)--> H₃O⁺(aq) + HPO₄²⁻(aq)
Answer:
C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)
Explanation:
For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H₂PO₄⁻?
A) H₂PO₄⁻(aq) + H₃O⁺(aq) → H₃PO₄(aq) + H₂O(l)
NO. This is the inverse of the acid dissociation of H₃PO₄.
B) H₂PO₄⁻(aq) + H₂O(l) → H₃PO₄(aq) + OH⁻(aq)
NO. This is the basic dissociation of H₂PO₄⁻.
C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)
YES. This is the acid dissociation of H₂PO₄⁻. The acid-dissociation constant is:
[tex]Ka=\frac{[H_{3}O^{+}].[HPO_{4}^{2-} ]}{[H_{2}PO_{4}^{-} ]}[/tex]
D) H₃PO₄(aq) + H₂O(l) → H₃O⁺(aq) + H₂PO₄⁻(aq)
No. This is the acid dissociation of H₃PO₄.
E) HPO₄²⁻(aq) + H₂O(l) → H₂PO₄⁻(aq) + OH⁻(aq)
NO. This is the basic dissociation of HPO₄²⁻.
An atom of 105In has a mass of 104.914558 amu. Calculate the binding energy in MeV per NUCLEON. Enter your answer with 3 significant figures and no units. Use the masses:
mass of 1H atom = 1.007825 amu
mass of a neutron = 1.008665 amu
1 amu = 931.5 MeV
Answer:
8.46
Explanation:
Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.
Also, atomic number of I = 549
Thus, the number of protons = 49
Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.
Mass number = Number of protons + Number of neutrons
105 = 49 + Number of neutrons
Number of neutrons = 56
Mass of neutron = 1.008665 amu
Mass of proton = 1.007825 amu
Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron
Thus,
Calculated mass = (49*1.007825 + 56*1.008665) amu = 105.868665 amu
Mass defect = Δm = |105.868665 - 104.914558| amu = 0.954107 amu
The conversion of amu to MeV is shown below as:-
1 amu = 931.5 MeV
So, Energy = 0.954107*931.5 MeV/atom = 888.750671 MeV/atom
Also, 1 atom has 105 nucleons (Protons+neutrons)
So, Energy = 888.750671 MeV/105nucleons = 8.46 MeV/nucleon
Answer:- 8.46
The binding energy per nucleon for an atom of 105In is calculated by determining the mass defect, converting it to energy using Einstein's mass-energy equivalence, and dividing by the number of nucleons.
Explanation:To calculate the binding energy per nucleon in MeV for an atom of 105In with a mass of 104.914558 amu, we must first identify the number of protons and neutrons in the nucleus. Indium-105 has 49 protons (since In is the element with atomic number 49), and subtracting this from the mass number 105 gives us 56 neutons. The mass of the protons is 49 x 1.007825 amu, and the mass of the neutrons is 56 x 1.008665 amu.
Next, we calculate the mass defect by subtracting the atomic mass of the 105In from the combined mass of the protons and neutrons. To convert the mass defect into energy, we multiply by 931.5 MeV/amu, according to Einstein’s mass-energy equivalence principle (E=mc2). Lastly, we divide this energy by the total number of nucleons (protons + neutrons) to find the binding energy per nucleon.
Standard Reduction Potentials in Aqueous Solution at 25 °C
Acidic Solution (V) 1.77 1.68 1.50 1.08 0.96 0.799 0.789 0.337 0.27
–0.14 –0.25 –0.28 –0.41 –0.44 –0.763 –0.91 –1.66 –2.37
Which metal ion from the selection is the weakest oxidizing agent? Which metal ion from the selection is the strongest oxidizing agent? Which metal from the selection is the strongest reducing agent? Which metal from the selection is the weakest reducing agent? Will reduce to ? yes no Will reduce to ? yes no Which metal from the selection can be oxidized by ?
Answer:
Hope this helps:)
Explanation:
The values for the table entries are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.
Elemental fluorine, for example, is the strongest common oxidizing agent.
Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution. The standard reduction potentials can be interpreted as a ranking of substances according to their oxidizing and reducing power
You are running a rather large scale reaction where you prepare the grignard reagent phenylmagnesium bromide by reacting 210.14 grams of magnesium with 772 ml of bromobenzene. How many moles of grignard reagent would you expect to form? (the density of bromobenzene is 1.495 g/ml, Mg = 24.3 g/mol, bromobenzene=157.01 g/mol)
Answer:
We would expect to form 7.35 moles of grignard reagent.
Explanation:
Step 1: Data given
Mass of magnesium = 210.14 grams
Volume bromobenzene = 772 mL
Density of bromobenzene = 1.495 g/mL
Molar mass of Mg = 24.3 g/mol
Molar mass of bromobenzene = 157.01 g/mol
Step 2: The balanced equation
C6H5Br + Mg ⇒ C6H5MgBr
Step 3: Calculate mass of bromobenzene
Mass bromobenzene = density bromobenzene * volume
Mass bromobenzene = 1.495 g/mL * 772 mL
Mass bromobenzene = 1154.14 grams
Step 4: Calculate number of moles bromobenzene
Moles bromobenzene = mass bromobenzene / molar mass bromobenzene
Moles bromobenzene = 1154.14g / 157.01 g/mol
Moles bromobenzene = 7.35 moles
Step 5: Calculate moles of Mg
Moles Mg = 210.14 grams /24.3 g/mol
Moles Mg = 8.65 moles
Step 6: The limiting reactant
The mole ratio is 1:1 So the bromobenzene has the smallest amount of moles, so it's the limiting reactant. It will be completely consumed ( 7.35 moles). Magnesium is in excess, There will react 7.35 moles. There will remain 8.65 - 7.35 = 1.30 moles
Step 7: Calculate moles of phenylmagnesium bromide
For 1 mole of bromobenzene, we need 1 mole of Mg to produce 1 mole of phenylmagnesium bromide
For 7.35 moles bromobenzene, we have 7.35 moles phenylmagnesium bromide
We would expect to form 7.35 moles of grignard reagent.
For which process is ΔS negative? Group of answer choices A. grinding a large crystal of KCl to powder B. raising the temperature of 100 g Cu from 275 K to 295 K C. compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 D. mixing 5 mL ethanol with 25 mL water E. evaporation of 1 mol of CCl4(l)
Answer:
C.
Explanation:
The entropy (S) is the measure of the randomness of a system, and ΔS = Sfinal - Sinitial. As higher is the disorder of the system, as higher is the entropy.
A. When KCl is fractionated in power, there'll be more portions of it, so, the disorder must be higher, then ΔS is positive.
B. As higher is the temperature, higher is the kinetic energy of the system, and because of that, the disorder is also higher, so ΔS is positive.
C. The decrease in the volume (compression) decreases the distance between the molecules, so the system will be more organized, then ΔS is negative.
D. The volume before the mixing will be higher, and the ethanol will dissociate, so it will be more particles, the disorder will increase, and ΔS is positive.
E. Sgas > Sliquid > Ssolid because of the disorder of the molecules, then ΔS is positive.
Final answer:
The process with a negative change in entropy (ΔS) is compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L, as it leads to a decrease in volume and increases the orderliness of the system.
Explanation:
The question asks for which process the change in entropy (ΔS) is negative. Entropy generally refers to the measure of disorder or randomness in a system. A negative ΔS indicates a decrease in entropy, meaning the system becomes more ordered. Considering the options:
A. grinding a large crystal of KCl to powder - Increases disorder by breaking down the crystal structure, so ΔS is positive.
B. raising the temperature of 100 g Cu from 275 K to 295 K - Increases thermal motion and disorder, so ΔS is positive.
C. compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L - Decreases the volume and increases order, so ΔS is negative.
D. mixing 5 mL ethanol with 25 mL water - Mixing increases disorder, so ΔS is positive.
E. evaporation of 1 mol of CCl4(l) - Changing from liquid to gas increases disorder, so ΔS is positive.
Therefore, the process with a negative ΔS is C. compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L.
Carboxylic acids boil at considerably higher temperatures than do alcohols, ketones, or aldehydes of similar molecular weights. This is because they: A. have a greater oxygen content. B. are more acidic. C. form stable hydrogen-bonded dimers. D. are hydrophobic. E. none of the above
Answer:
The correct option is: C. form stable hydrogen-bonded dimers
Explanation:
Boiling point is the temperature at which a particular substance changes from liquid state to vapor state.
The boiling point of a chemical substance depends upon the intermolecular forces present between the molecules.
Carboxylic acids are the organic molecules containing carboxyl functional group (COOH). They tend to have greater boiling point than alcohols, ketones, or aldehydes.
This is because only carboxylic acids are capable of forming dimers that are stabilized by hydrogen bonding.
At a particular temperature, 12.6 mol of SO3 is placed into a 4.0-L rigid container, and the SO3 dissociates by the reaction given below. 2 SO3(g) equilibrium reaction arrow 2 SO2(g) + O2(g) At equilibrium, 3.4 mol of SO2 is present. Calculate K for this reaction.
Answer:
At equilibrium, Kc = 0.058
Explanation:
Step 1: Data given
Mol SO3 = 12.6
Volume = 4.0 L
At equilibrium we have:
3.4 mol of SO2
Step 2: The balanced equation
2SO3(g) + ⇆ 2SO2(g) + O2(g)
Step 3: ICE-chart
The initial number of moles are:
SO3: 12.6 moles
SO2 : 0 mol
O2: 0 mol
There will react:
SO3: -2x
SO2: +2x
O2: +x
The number of moles at the equilibrium are:
SO3: 12.6 - 2x
SO2: 2x = 3.4 mol
O2: x
Since at the equilibrium, we have 2x = 3.4 mol. x = 1.7 mol
This means at the equilibrium we have 1.7 mol of O2 and 12.6 -3.4 = 9.2 mol of SO3
Step 4: Calculate the equilibrium constant Kc
Kc = [3.4/4]² *[1.7/4] / [9.2/4]²
Kc = 0.058
At equilibrium, Kc = 0.058
When solid Fe metal is put into an aqueous solution of Sn(NO3)2, solid Sn metal and a solution of Fe(NO3)2 result. Write the net ionic equation for the reaction. (Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer:
Fe(s) + Sn²⁺(aq) ⇒ Sn(s) + Fe²⁺(aq)
Explanation:
When solid Fe metal is put into an aqueous solution of Sn(NO₃)₂, solid Sn metal and a solution of Fe(NO₃)₂ result. The resulting molecular equation is:
Fe(s) + Sn(NO₃)₂(aq) ⇒ Sn(s) + Fe(NO₃)₂
The full ionic equation includes all the ions and the species that do not dissociate in water.
Fe(s) + Sn²⁺(aq) + 2 NO₃⁻(aq) ⇒ Sn(s) + Fe²⁺(aq) + 2 NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and species that do not dissociate in water.
Fe(s) + Sn²⁺(aq) ⇒ Sn(s) + Fe²⁺(aq)
Final answer:
The net ionic equation for the reaction of solid Fe metal with an aqueous solution of Sn(NO3)2 is Fe(s) + Sn2+(aq)
ightarrow Fe2+(aq) + Sn(s), a single displacement redox reaction.
Explanation:
When solid Fe metal is introduced to an aqueous solution of Sn(NO3)2, a single displacement reaction occurs where Fe displaces Sn from the stannous nitrate. The net ionic equation for this chemical reaction, considering that nitrates are soluble and metallic tin will precipitate out of the solution as a solid, can be written as follows:
Fe(s) + Sn2+(aq)
ightarrow Fe2+(aq) + Sn(s)
In this equation, solid iron (Fe) reacts with stannous ions (Sn2+) in solution to form ferrous ions (Fe2+) and solid tin (Sn). This type of reaction is also referred to as a redox reaction.
The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following:
Sn2+(aq)+2e−→Sn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.140 V E∘red=+0.337 V
What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Express your answer to three decimal places and include the appropriate units.
Answer:
0.477 V
Explanation:
When a substance is gaining electrons, it's reducing, and when the substance loses electrons, it's oxidizing. In a galvanic cell, one substance oxides giving electrons for the other, which reduces. Then, the substance with higher reduction potential must reduce and the other must oxide.
E°cell = E°red(red) - E°red(oxid)
Where, E°red(red) is the reduction potential of the substance that reduces, and E°red(oxid) is the reduction potential of the substance that oxides. For the value given, Cu⁺² reduces, so:
E°cell = +0.337 - (-0.140)
E°cell = 0.477 V
The standard cell potential for this galvanic cell is [tex]0.477 \ V[/tex].
To calculate the standard cell potential [tex](\( E^\circ_\text{cell} \))[/tex] for the galvanic cell using the given reduction potentials, we use the formula:
[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]
Given reduction potentials:
[tex]\( E^\circ_\text{red}(\text{Sn}^{2+} \rightarrow \text{Sn}) = -0.140 \, \text{V} \)[/tex]
[tex]\( E^\circ_\text{red}(\text{Cu}^{2+} \rightarrow \text{Cu}) = +0.337 \, \text{V} \)[/tex]
Since the reduction potential for [tex]\( \text{Cu}^{2+} \)[/tex] is more positive, it acts as the cathode:
[tex]\[ E^\circ_\text{cathode} = +0.337 \, \text{V} \][/tex]
And the reduction potential for [tex]\( \text{Sn}^{2+} \)[/tex] is less positive (more negative), so it acts as the anode:
[tex]\[ E^\circ_\text{anode} = -0.140 \, \text{V} \][/tex]
Now, calculate the standard cell potential:
[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]
[tex]\[ E^\circ_\text{cell} = (+0.337 \, \text{V}) - (-0.140 \, \text{V}) \][/tex]
[tex]\[ E^\circ_\text{cell} = +0.337 \, \text{V} + 0.140 \, \text{V} \][/tex]
[tex]\[ E^\circ_\text{cell} = +0.477 \, \text{V} \][/tex]
What problem would arise if the excess acetyl‑CoA were not converted to ketone bodies?
A. Although acetyl‑CoA is not toxic, it inhibits D‑β‑hydroxyacyl‑CoA dehydrogenase.
B. Although acetyl‑CoA is not toxic, it prevents transfer of fatty acids into the mitochondria by forming acetyl carnitine.
C. Fatty acid oxidation would stop when all of the CoA is bound as acetyl‑CoA.
D. Formation of acetyl‑CoA requires uptake of a proton, and the increased pH inhibits further oxidation.
Answer:
The correct answer is C fatty acid oxidation would stop when all of the CoA is bound as acetyl CoA.
Explanation:
Acetyl CoA is the principle end product of beta oxidation of even chain fatty acid such as palmitic acid.
When the cellelar label of actyl CoA increases at that time the excess acetyl CoA is converted to ketone bodies by the process called ketogenesis.
According to the question if the excess acetyl CoA is not converted to ketone bodies then it will interfere with the oxidation of fatty acid because fatty acid molecules will not get any CoA SH molecule to activate themselves to initiate a new round of beta oxidation.
As a result fatty acid oxidation will stop.
Answer:
C. Fatty acid oxidation would stop when all of the CoA is bound as acetyl‑CoA.
Explanation:
Hello,
In this case, due to the fact that the mitochondrial pool of the CoA is short, thus, such cofactor must be recycled from acetyl-CoA through the production of ketone-like bodies. Therefore, the operation of the beta-oxidation pathway is performed, as it is necessary for energy production, in such a way one concludes that fatty acid oxidation would stop when all of the CoA is bound as acetyl‑CoA.
Best regards.
Define A, Z, and X in the following notation used to specify a nuclide: AZX. Check all that apply.
(A) Z is the mass number (number of protons - neutrons).
(B) A is the chemical symbol of the element.
(C) Z is the chemical symbol of the element.
(D) A is the atomic number (number of protons).
(F) X is the chemical symbol of the element.
(G) X is the mass number (number of protons neutrons).
(H) Z is the atomic number (number of protons).
(I) A is the mass number (number of protons neutrons).
(J) X is the atomic number (number of protons).
Answer:
X is the chemical symbol of the element.
A is the mass number (number of protons neutrons).
Z is the atomic number (number of protons).
Explanation:
In stating the chemical representation of an element, the AZX symbol is used.
The symbol of the element may either come from its Latin or English name. For instance, the symbol of the element sodium, comes from its Latin name natrium (Na).
Its atomic number is the number of protons in the nucleus of the atom. For sodium, the atomic number is 11.
The mass number refers to the sum of the number of protons and neutrons in the atom. For sodium the mass number is 23.
Hence the AZX symbol for sodium is
23_11Na.
The correct option as regarding the definition of A, Z and X in ᴬ₂X are:
A is the mass number (option I)Z is the atomic number (option H)X is the chemical symbol of the element (option F)How to determine what A, Z and X represents?Nuclide of elements are generally represented according to the following notation:
ᴬ₂X
Where
A is defined as the mass number of the element and it the sum of the protons and neutrons in the nucleus of the atom.Z is the atomic number element which is equally the proton number of the elementX is the symbol of the element which is used to identity the elementWith the above information, we can determine the options that is correct from the question.
Thus, the correct options are:
Option I A is the mass number (number of protons + neutrons)Option H Z is the atomic number (number of protons)Option F X is the chemical symbol of the element.Learn more about nuclide notation:
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The vapor pressure of the liquid SO2 is measure at different temperatures. the following vapor pressure data are obtained.Temp (K) Pressure mmHg241.3 271.2259.3 641.8Calculate the enthalpy of vaporization(delta H vap) in KJ/mol for this liquid.
Answer:
24.895 kJ/mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
[tex]\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c [/tex]
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314 J /mol K)
c is the constant.
For two situations and phases, the equation becomes:
[tex]\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)[/tex]
Given:
[tex]P_1[/tex] = 271.2 mmHg
[tex]P_2[/tex] = 641.8 mmHg
[tex]T_1[/tex] = 241.3 K
[tex]T_2[/tex] = 259.3 K
So,
[tex]\ln \:\left(\:\frac{271.2}{641.8}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{259.3}-\:\frac{1}{241.3}\:\right)[/tex]
[tex]\Delta \:H_{vap}=\ln \left(\frac{271.2}{641.8}\right)\frac{8.314}{\left(\frac{1}{259.3}-\:\frac{1}{241.3}\right)}\ J/mol[/tex]
[tex]\Delta \:H_{vap}=\left(-\frac{520199.41426}{18}\right)\left(\ln \left(271.2\right)-\ln \left(641.8\right)\right)\ J/mol[/tex]
ΔHvap = 24895.015 J/mol = 24.895 kJ/mol ( 1 J = 0.001 kJ )
Final answer:
This detailed answer explains how to calculate the enthalpy of vaporization for liquid SO2 using the Clausius-Clapeyron equation and vapor pressure data at different temperatures.
Explanation:
The enthalpy of vaporization (ΔHvap) can be calculated using the Clausius-Clapeyron equation:
In this case, the enthalpy of vaporization for SO2 can be calculated using the given vapor pressure data at different temperatures:
Convert temperatures to Kelvin.
Use the Clausius-Clapeyron equation to find ΔHvap.
By applying the equation with the provided data, the calculated enthalpy of vaporization for liquid SO2 is 24.895 kJ/mol.
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?
Answer:
There is 1.3 kJ heat produced(released)
Explanation:
Step 1: Data given
Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L
Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L
Initial temperature = 21.9 °C
Final temperature = 23.5 °C
Solid AgCl will be formed
Step 2: The balanced equation:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)
Step 3: Define the formula
Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.
Q=m*c*ΔT
⇒ Q = the heat transfer (in joule)
⇒ m =the mass (in grams)
⇒ c= the heat capacity (J/g°C)
⇒ ΔT = Change in temperature = T2- T1
Step 4: Calculate heat
Let's vonsider the density the same as the density of water (1g/mL)
Mass = volume * density
Mass = 200 mL * 1g/mL
Mass = 200 grams
Q= m*c*ΔT
⇒ m = 200 grams
⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C
⇒ ΔT = 23.5 -21.9 = 1.6°C
Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ
There is 1.3 kJ heat produced(released)
Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)
The increased temperature after the precipitation reaction of NaCl and AgNO3 in a coffee cup calorimeter indicates that heat is released. Calculation of the heat released depends on the specific heat capacity of the solution and its mass. Some assumptions made include the solution having the same heat capacity and density as water, and perfect insulation of the calorimeter.
Explanation:The heat released by the precipitation reaction of NaCl and AgNO3 to form AgCl can be deduced from the increase of temperature observed in the coffee cup calorimeter. The process happens when Ag+ from AgNO3(aq) and Cl- ion from NaCl(aq) react to form the solid AgCl. This reaction is exothermic as it leads to an increase in the temperature of surroundings, that is, the solution in the calorimeter.
To calculate the heat produced, we need the specific heat capacity of the solution, which is assumed to be approximately equal to that of water (4.18 J/g°C), and the total mass of the solution which is calculated by adding the volumes of the two solutions since the density of the solution is assumed to be approximately 1 g/mL (same as water).
The approximations made in this calculation include treating the solution as having the same specific heat capacity and density as pure water, and assuming that the calorimeter perfectly insulates the solution so that no heat is lost to the environment.
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