Answer:
W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.
Explanation:
heat engine:
∴ Th = 485°C
∴ Tc = 42°C
∴ Qh = 9.75 KJ......heat from hot source
first law:
∴ ΔU = Q + W = 0 .........cyclic process
⇒ Q = - W
∴ Q = Qh + Qc = Qh - (- Qc)
∴ Qc: heat from the cold source ( - )
⇒ Qh - ( - Qc) = - W..............(1)
⇒ Qc/Qh = - Tc/Th...........(2)
from (2):
⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)
⇒ Qc = - 0.8443 KJ
replacing in (1):
⇒ - W = 9.75 KJ - ( - 0.8443 KJ)
⇒ - W = 10.5943 KJ
(45 pts) What is the theoretical yield (in g) of iron(III) carbonate that can be produced from 1.72 g of iron(III) nitrate and an excess of sodium carbonate? The molar mass of iron(III) carbonate (Fe2(CO3)3) is 291.73 g∙mol–1 and the molar mass of iron(III) nitrate (Fe(NO3)3) is 241.88 g∙mol–1.
2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)
Answer:
1.04g of iron III carbonate
Explanation:
First, we must put down the equation of reaction because it must guide our work.
2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)
From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.
Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate
Mass of iron III nitrate reacted= 1.72g
Molar mass of iron III nitrate= 241.88 g∙mol–1
Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles
From the equation of the reaction;
2 moles of iron III nitrate yields 1 mole of iron III carbonate
7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate
Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass
Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate
The theoretical yield of Fe₂(CO₃)₃ obtained from the reaction is 1.04 g
We'll begin by calculating the mass of Fe(NO₃)₃ that reacted and the mass of Fe₂(CO₃)₃ produced from the balanced equation.
2Fe(NO₃)₃ + 3Na₂CO₃ —> Fe₂(CO₃)₃ + 6NaNO₃
Molar mass of Fe(NO₃)₃ = 241.88 g/mol
Mass of Fe(NO₃)₃ from the balanced equation = 2 × 241.88 = 483.4 g
Molar mass of Fe₂(CO₃)₃ = 291.73 g/mol
Mass of Fe₂(CO₃)₃ from the balanced equation = 1 × 291.73 = 291.73 g
From the balanced equation above,
483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃
Finally, we shall determine the theoretical yield of Fe₂(CO₃)₃. This can be obtained as follow:From the balanced equation above,
483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃
Therefore,
1.72 g of Fe(NO₃)₃ will react to produce = (1.72 × 291.73) / 483.4 = 1.04 g of Fe₂(CO₃)₃
Thus, the theoretical yield of Fe₂(CO₃)₃ is 1.04 g
Learn more: https://brainly.com/question/25821244
A voltaic cell that uses the reaction PdCl42−(aq)+Cd(s) → Pd(s)+4Cl−(aq)+Cd2+(aq) has a measured standard cell potential of +1.03 V. You may want to reference (Pages 860 - 867) Section 20.4 while completing this problem. Part A Write the half-cell reaction at the cathode.
Answer: The half-cell reaction occurring at cathode is [tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]
Explanation:
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Reduction reaction is defined as the reaction in which a chemical specie accepts electrons. The oxidation state of the chemical specie reduces.
The given balanced chemical equation is:
[tex]PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)[/tex]
The half cell reaction occurring at cathode follows:
[tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]
Hence, the half-cell reaction occurring at cathode is given above.
What mass of potassium bromide (in grams) do you need to make 250.0 mL of a 1.50 M potassium bromide solution?
Answer:
44.63g
Explanation:
First, let us calculate the number of mole of KBr in 1.50M KBr solution.
This is illustrated below:
Data obtained from the question include:
Volume of solution = 250mL = 250/1000 = 0.25L
Molarity of solution = 1.50M
Mole of solute (KBr) =.?
Molarity is simply mole of solute per unit litre of solution
Molarity = mole /Volume
Mole = Molarity x Volume
Mole of solute (KBr) = 1.50 x 0.25
Mole of solute (KBr) = 0.375 mole
Now, we calculate the mass of KBr needed to make the solution as follow:
Molar Mass of KBr = 39 + 80 = 119g/mol
Mole of KBr = 0.375 mole
Mass of KBr =?
Mass = number of mole x molar Mass
Mass of KBr = 0.375 x 119
Mass of KBr = 44.63g
Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution
Final answer:
To make a 1.50 M potassium bromide solution with a volume of 250.0 mL, you would need approximately 44.625 grams of potassium bromide.
Explanation:
To calculate the mass of potassium bromide required, we can use the formula:
mass (g) = concentration (M) x volume (L) x molar mass (g/mol)
The concentration is given as 1.50 M, the volume is 250.0 mL (which can be converted to 0.250 L), and the molar mass of potassium bromide (KBr) is 119.0 g/mol. Plugging in these values, we get:
mass (g) = 1.50 M x 0.250 L x 119.0 g/mol = 44.625 g
Therefore, you would need approximately 44.625 grams of potassium bromide to make 250.0 mL of a 1.50 M potassium bromide solution.
9. How many molecules are contained in 6.44 moles of oxygen gas?
Answer:
3.88 × 10²⁴ molecules
Explanation:
In order to solve this question, we need to consider the Avogadro's number. know that 1 mole of particles contains 6.02 × 10²³ particles. This applies to different kinds of particles: atoms, molecules, electrons.
In this case, 1 mole of molecules of oxygen gas contains 6.02 × 10²³ molecules of oxygen gas. We will use this relation to find the number of molecules of oxygen gas in 6.44 moles of oxygen gas.
[tex]6.44mol \times \frac{6.02 \times 10^{23}molecules }{1mol} = 3.88 \times 10^{24}molecules[/tex]
What is the volume, in liters, occupied by 2.79 moles of N2 gas at 439 mmHg pressure and a
temperature of 64°C?
Answer:
132.93 L
Explanation:
Step 1:
Data obtained from the question:
Volume (V) =?
Number of mole (n) = 2.79 moles
Pressure (P) = 439 mmHg
Temperature (T) = 64°C
Step 2:
Conversion to appropriate unit
For pressure:
760mmHg = 1atm
Therefore, 439 mmHg = 439/760 = 0.58 atm
For temperature:
Temperature (Kelvin) = temperature (celsius) + 273
temperature (celsius) = 64°C
Temperature (Kelvin) = 64°C + 273 = 337 K
Step 3:
Determination of the volume.
The volume occupied by N2 can be obtained by using the ideal gas equation as follow:
Volume (V) =?
Number of mole (n) = 2.79 moles
Pressure (P) = 0.58 atm
Temperature (T) = 337 K
Gas constant (R) = 0.082atm.L/Kmol
PV = nRT
0.58 x V = 2.79 x 0.082 x 337
Divide both side by 0.58
V = (2.79 x 0.082 x 337)/0.58
V = 132.93 L
Therefore, the volume occupied by N2 is 132.93 L
The pH of a 1.1M solution of hexanoic acid HC6H11O2 is measured to be 2.40. Calculate the acid dissociation constant Ka of hexanoic acid. Be sure your answer has the correct number of significant digits.
Answer:
Ka = 1.4 x 10⁻⁴
Explanation:
HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻
At equilibrium 1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M
Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵
Answer:
Ka = 1.4 x 10⁻⁴
Explanation:
HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻
At equilibrium
1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M
Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵
Given the thermochemical equations below, What is the standard heat of formation of CuO(s)? 2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ
The standard heat of formation of [tex]\( \text{CuO(s)} \) is \( -149.6 \text{ kJ/mol} \).[/tex]
To find the standard heat of formation of [tex]\( \text{CuO(s)} \)[/tex], we can use the given thermochemical equations and apply Hess's Law.
Given Equations:
1. [tex]\( 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \)[/tex]
2. [tex]\( \text{Cu}_2\text{O(s)} \rightarrow \text{CuO(s)} + \text{Cu(s)} \quad \Delta H^\circ = 11.3 \text{ kJ} \)[/tex]
Goal:
Find the standard heat of formation of [tex]\( \text{CuO(s)} \), \( \Delta H_f^\circ \text{(CuO(s))} \).[/tex]
Steps:
1. Write the formation reaction for [tex]\( \text{CuO(s)} \)[/tex]
[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]
2. Manipulate the given equations to match the formation reaction:
a. Reverse Equation 2 to get [tex]\( \text{CuO(s)} \) and \( \text{Cu(s)} \)[/tex] on the reactant side:
[tex]\[ \text{CuO(s)} + \text{Cu(s)} \rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \][/tex]
b. Add this to Equation 1 (which needs no change) to eliminate[tex]\( \text{Cu}_2\text{O(s)} \): \[ \begin{align*} \text{CuO(s)} + \text{Cu(s)} &\rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \\ 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} &\rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \\ \end{align*} \][/tex]
Adding these equations:
[tex]\[ \text{CuO(s)} + \text{Cu(s)} + 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)} + 4 \text{CuO(s)} \][/tex]
Simplify:
[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]
3. Combine the enthalpy changes:
[tex]\[ \Delta H^\circ = -287.9 \text{ kJ} + (-11.3 \text{ kJ}) = -299.2 \text{ kJ} \][/tex]
Since the above reaction is for 2 moles of [tex]\( \text{CuO(s)} \)[/tex], the enthalpy change for the formation of 1 mole of [tex]\( \text{CuO(s)} \)[/tex] is:
[tex]\[ \Delta H_f^\circ \text{(CuO(s))} = \frac{-299.2 \text{ kJ}}{2} = -149.6 \text{ kJ/mol} \][/tex]
How much energy is evolved during the formation of 98.7 g of Fe, according to the reaction below? Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ΔH°rxn = -852 kJ How much energy is evolved during the formation of 98.7 g of Fe, according to the reaction below? Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ΔH°rxn = -852 kJ 753 kJ 482 kJ 1.51 x 103 kJ 4.20 x 103 kJ 241 kJ
The formation of 98.7 grams of Fe from the reaction Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with an enthalpy change of -852 kJ per 112 grams of Fe formed, would result in approximately -753 kJ of energy being released. The reaction is exothermic, meaning it releases energy as heat.
Explanation:The question asks about the energy evolved during the formation of 98.7 grams of iron (Fe) from the reaction given: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with a enthalpy change (ΔH°rxn) of -852 kJ. The reaction is exothermic, meaning it releases energy as heat. The ΔH for the reaction is -852 kJ per 112 grams of Fe formed (the molar mass of Fe is 56 g/mol, so 2 mol of Fe is 112 g).
To find out how much energy is released in forming 98.7 g of Fe, you can use simple proportion: (98.7 g Fe / 112 g Fe) x -852 kJ = approx. -753 kJ. So, the formation of 98.7 grams of Fe would result in approximately -753 kJ of energy being released, pointing to answer 753 kJ among the options given.
Learn more about Thermochemistry here:https://brainly.com/question/34429178
#SPJ12
The energy evolved during the formation of 98.7 g of Fe is 753 kJ.
Explanation:The given reaction is:
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)
The enthalpy change for the reaction is -852 kJ.
To calculate the energy evolved during the formation of 98.7 g of Fe, we need to use the molar mass of Fe.
The molar mass of Fe is 55.845 g/mol.
By converting the given mass of Fe to moles and using the stoichiometry of the balanced equation, we can calculate the energy evolved.
98.7 g Fe * (1 mol Fe / 55.845 g Fe) * (-852 kJ / 2 mol Fe) = -753 kJ
Therefore, the energy evolved during the formation of 98.7 g of Fe is 753 kJ.
Learn more about Enthalpy change in reaction here:https://brainly.com/question/29556033
#SPJ11
A 25.0 mL sample of a 0.100 M solution of acetic acid is titrated with a 0.125 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The Ka for acetic acid is 1.76×10–5.
Explanation:
10mil 76 the experllon is the most numbers
Answer:
Ater adding 10.0 mL the pH = 4.74
After adding 20.0 mL the pH = 8.75
After adding 30.0 mL the pH = 12.36
Explanation:
Step 1: Data given
Volume of a 0.100 M acetic acid solution = 25.0 mL
Molarity of NaOH solution = 0.125 M
A) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH
Calculate moles
Equation: NaOH + HAc --> NaAc + H20
Moles HAc = 0.025 L* 0.100 mol /L = 0.00250 moles
Moles NaOH = 0.01 L * 0.125mol/L = 0.00125 moles
Initial moles
NaOH = 0.00125 moles
HAc = 0.00250 moles
NaAC = 0 moles
Moles at the equilibrium
NaOH = 0.00125 - 0.00125 = 0 moles
HAc = 0.00250 - 0.00125 = 0.00125 moles
NaAC = 0.00125 moles
Calculate molarity
[HAc] = [Ac-] = 0.00125 moles / 0.035 L = 0.0357M
Concentration at equilibrium
HAc <--> Ac- + H+ Ka = 1.76 * 10^-5
[HAc] = 0.0357 - x M
[Ac-] = 0.0357 + X M
[H+] = XM
Ka=1.76*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]
Ka=1.76*10^-5 = [0.00357][x]/[0.0357] usually good approx.
Ka = 1.76* 10-5 = x = [H+]
pH = -log [H+] = -log [1.76*10^-5] = 4.74
B) 25.0mL HAc and 20.0mL NaOH
Calculate moles
Equation: NaOH + HAc --> NaAc + H20
Moles HAc = 0.025 L* 0.100 mol /L = 0.00250 moles
Moles NaOH = 0.02 L * 0.125mol/L = 0.00250 moles
Initial moles
NaOH = 0.00250 moles
HAc = 0.00250 moles
NaAC = 0 moles
Moles at the equilibrium
NaOH = 0.00250 - 0.00250 = 0 moles
HAc = 0.00250 - 0.00250 = 0 moles
NaAC = 0.00250 moles
Calculate molarity
[NaAC] = 0.00250 moles / 0.045 L = 0.0556M
Concentration at equilibrium
Ac- + H20 <--> HAc + OH- Ka = 1.76 * 10^-5
[Ac] = 0.0556 - x M
[OH-] = XM
[HAc] = XM
**Kw = 1*10^-14 at STP I'll assume this for this problem
**Kw = Ka*Kb ⇒ Kb = Kw/Ka
**Kb = (1*10^-14)/(1.76*10^-5)= 5.68*10^-10
Kb = ([HAc][OH-])/[Ac-] = ([x][x])/[0.0556]
Kb = x² /0.0556
5.68*10^-10 = x² /0.0556
so x =5.62*10^-6
pOH = -log [OH-] = 5.250
pH = 14 - pOH
pH 14 - 5.25 = 8.75
C)25.0mL HAc and 30.0mL NaOH
NaOH + HAc --> NaAc + H20
Initial numbers of moles
Moles NaOH = 0.125M * 0.03 L = 0.00375 moles
Moles HAc = 0.00250 moles
moles NaAC = 0 moles
Moles at the equilibrium
Moles NaOH = 0.00375 moles - 0.00250 = 0.00125 moles
Moles HAc = 0.00250 moles - 0.00250 = 0 moles
moles NaAC = 0.00250 moles
After the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.
Calculate NaOH molarity
(0.00125moles/0.055 L= 0.0227M NaOH
Strong Bases dissociate 100% so [OH-] = 0.0227M
pOH= -log[0.0227] = 1.644
pH = 14-pOH = 12.36
What is the maximum mass in grams of NH3 that can be produced by the reaction of of 2.5 g N2 with 2.5 g of H2 via the equation below?
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Answer: The mass of [tex]NH_3[/tex] produced is, 3.03 grams.
Explanation : Given,
Mass of [tex]N_2[/tex] = 2.5 g
Mass of [tex]H_2[/tex] = 2.5 g
Molar mass of [tex]N_2[/tex] = 28 g/mol
Molar mass of [tex]H_2[/tex] = 2 g/mol
First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].
[tex]\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=\frac{2.5g}{28g/mol}=0.089mol[/tex]
and,
[tex]\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=\frac{2.5g}{2g/mol}=1.25mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]
So, 0.089 moles of [tex]N_2[/tex] react with [tex]0.089\times 3=0.267[/tex] moles of [tex]H_2[/tex]
From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NH_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]
So, 0.089 mole of [tex]N_2[/tex] react to give [tex]0.089\times 2=0.178[/tex] mole of [tex]NH_3[/tex]
Now we have to calculate the mass of [tex]NH_3[/tex]
[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]
Molar mass of [tex]NH_3[/tex] = 17 g/mole
[tex]\text{ Mass of }NH_3=(0.178moles)\times (17g/mole)=3.03g[/tex]
Therefore, the mass of [tex]NH_3[/tex] produced is, 3.03 grams.
Sulfurous Acid, H2SO3, is a diprotic acid, with the following acid dissociation constants: Ka1 = 1.4 × 10-2 Ka2 = 6.7 × 10-8 What is the Kb for SO32- at 25ºC? Group of answer choices 9.4 × 10-10 1.5 × 10-7 6.7 × 10-8 7.2 × 10-13
Answer:
Kb = 7.1 x 10⁻¹³
Explanation:
Ka x Kb = Kw => Kb = 1 x 10⁻¹⁴/1.4 x 10⁻² = 7.1 x 10⁻¹³
The Kb for SO32- at 25ºC is approximately 1.5 x 10^-7.
therefore correct option (b).
Explanation:To find the Kb for SO32-, we can use the relationship between Ka and Kb. Since the diprotic acid, H2SO3, forms SO32- in the second dissociation step, we can use the formula Kb = Kw/Ka. Kb is the equilibrium constant for the reaction of a base with water to form the hydroxide ion. Kw is the ion product of water, which is 1.0 x 10^-14 at 25ºC. So, using the given Ka2 value of 6.7 x 10^-8, we can calculate the Kb for SO32-:
Kb = Kw/Ka = (1.0 x 10^-14) / (6.7 x 10^-8)
Kb ≈ 1.5 x 10^-7
Learn more about Kb for SO32- here:https://brainly.com/question/34152049
#SPJ6
what is the ratio of the atoms in molecular formulae (NCl3) nitrogen trichloride
Answer:
3:1 M ratio at pH 3-4
Explanation:
A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?
Answer:
=16.49 L
Explanation:
Using the equation
P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273
P1V1/T1= P2V2/T2
0.6×30/298= 1×V2/273
V2=16.49L
Name the following compound:
(ignore the dots, I tried to space them out but it wouldn't let me)
........... O
.............ll
CH3 - C - CH2 - CH2 - CH2 - CH3
Methyl butanone
2-hexanol
2-hexanone
Methyl butyl ether
Answer:
2-hexanone
Explanation:
First, we'll begin by:
1. Locating the longest continuous chain i.e haxane
2. Determine the functional group in the compound. The functional group in ketone (C=O). This changes the name from hexane to hexanone i.e replacing the -e in at the end in hexane with -one to make it hexanone.
3. Give the functional group the lowest low count. In doing this, we'll start counting from the left. The functional group is at carbon 2.
Note: no substitute group is attached.
Now, in naming the compound, we must indicate the position of functional group as illustrated below:
The name of the compound is:
2-hexanone
The compound is called 2-hexanone. This is because there is a six carbon chain with a ketone functional group attached to the second carbon.
Explanation:You are dealing with an organic compound formula. Let's identify the compound based on its structure. The compound you have given is CH3 - C - CH2 - CH2 - CH2 - CH3. This is a 6 carbon chain and there is double bond between Oxygen and the second carbon. Now, when naming compounds in chemistry we firstly count the length of the chain, it is hexane. Then, we look for any functional group and where it is attached in the chain. A ketone functional group is CO, so this would make it a hexanone. As the CO is attached at the second carbon, it is 2-hexanone. So the correct answer for the name of the compound would be 2-hexanone.
Learn more about Chemical Nomenclature here:https://brainly.com/question/36634852
#SPJ6
Almost all energy on Earth came from the blank
Answer:
sun
Explanation:
Answer:
The sun is where almost all energy on Earth came from.
If 25.21 ml of NaOH solution is required to react completely with 0.550 g KHP, what is the molarity of the NaOH solution? Answer MUST be in Mol NaOH/L solution. Do not put in scientific notation.
Answer:
[NaOH] = 0.107M
Explanation:
25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O
The reaction is 1 to 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)
moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole
moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...
Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.
Answer:
[NaOH] = 0.107M
Explanation:
25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O
The reaction is 1 to ratio 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)
moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole
moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...
Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.
A chemist prepares hydrogen fluoride by means of the following reaction:
CaF2 + H2SO4 --> CaSO4 + 2HF
The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.
(b) Calculate the percent yield of HF.
Answer:
39.3%
Explanation:
Our guide in solving the problem must be the reaction equation hence it is pertinent to put down first:
CaF2 + H2SO4 --> CaSO4 + 2HF
We have a very important information in the question, sulphuric acid is present in excess. This implies that calcium fluoride is the limiting reactant.
Number of moles of calcium fluoride reacted= mass of calcium fluoride reacted/ molar mass of calcium fluoride
Molar mass of calcium fluoride= 78.07 g/mol
Number of moles of calcium fluoride= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of 0.28 moles of hydrogen fluoride = number of moles× molar mass
Molar mass of hydrogen flouride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g this is the theoretical yield of HF
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
Answer:
A. The theoretical yield of HF is 5.64g
B. The percentage yield of HF is 39%
Explanation:
Step 1:
The balanced equation for the reaction:
CaF2 + H2SO4 --> CaSO4 + 2HF
Step 2:
Determination of the mass of CaF2 that reacted and the mass of HF produced from the balanced equation. This is illustrated below:
Molar Mass of CaF2 = 40 + (19x2) = 40 + 38 = 78g/mol
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF from the balanced equation = 2 x 20 = 40g.
From the balanced equation above,
78g of CaF2 reacted and 40g of HF were produced.
A. Determination of the theoretical yield of HF.
This is illustrated below:
From the balanced equation above,
78g of CaF2 reacted to produce 40g of HF.
Therefore, 11g of CaF2 will react to produce = (11 x 40)/78 = 5.64g of HF.
The theoretical yield of HF is 5.64g
B. Determination of the percentage yield.
The percentage yield of HF can be obtained as follow:
Actual yield = 2.2g
Theoretical yield = 5.64g
Percentage yield =?
Percentage yield = Actual yield/Theoretical yield x100
Percentage yield = 2.2/5.64 x 100
Percentage yield = 39%
The percentage yield of HF is 39%
A reaction was performed in which 3.4 g of benzoic acid was reacted with excess methanol to make 1.2 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer : The theoretical yield and percent yield for this reaction is, 3.78 grams and 31.7 % respectively.
Explanation : Given,
Mass of [tex]C_6H_5COOH[/tex] = 3.4 g
Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mol
First we have to calculate the moles of [tex]C_6H_5COOH[/tex]
[tex]\text{Moles of }C_6H_5COOH=\frac{\text{Given mass }C_6H_5COOH}{\text{Molar mass }C_6H_5COOH}[/tex]
[tex]\text{Moles of }C_6H_5COOH=\frac{3.4g}{122.12g/mol}=0.0278mol[/tex]
Now we have to calculate the moles of [tex]C_6H_5COOCH_3[/tex]
The balanced chemical equation is:
[tex]C_6H_5COOH+CH_3OH\rightarrow C_6H_5COOCH_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]C_6H_5COOH[/tex] react to give 1 mole of [tex]C_6H_5COOCH_3[/tex]
So, 0.0278 mole of [tex]C_6H_5COOH[/tex] react to give 0.0278 mole of [tex]C_6H_5COOCH_3[/tex]
Now we have to calculate the mass of [tex]C_6H_5COOCH_3[/tex]
[tex]\text{ Mass of }C_6H_5COOCH_3=\text{ Moles of }C_6H_5COOCH_3\times \text{ Molar mass of }C_6H_5COOCH_3[/tex]
Molar mass of = 136.14 g/mole
[tex]\text{ Mass of }C_6H_5COOCH_3=(0.0278moles)\times (136.14g/mole)=3.78g[/tex]
The theoretical yield of [tex]C_6H_5COOCH_3[/tex] produced is, 3.78 grams.
Now we have to calculate the percent yield of the reaction.
Theoretical yield of the reaction = 3.78 g
Experimental yield of the reaction = 1.2 g
The formula used for the percent yield will be :
[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Percent yield}=\frac{1.2g}{3.78g}\times 100=31.7\%[/tex]
The percent yield of the reaction is, 31.7 %
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final volumes added, which are recorded below. The student then followed the directions in the manual to make the stock solution. Volumes Used to Create Stock Solution Volume 0.200 M Fe(NO3)3 (mL) Volume 0.00200 M KSCN (mL) Stock Solution 10.63 mL 1.42 mL Using the stock solution above, the student made additional dilutions, with the final volumes below. Calculate the [FeSCN2 ] in Standard 2. Report your answer in mM. Volumes Used to Create Standard Solutions Solution Volume Stock solution (mL) Volume Water (mL) Standard 1 7.43 2.53 Standard 2 4.63 5.17 Standard 3 2.66 6.58 Standard 4 0.57 9.36 Note: Reporting your answer in mM is for grading purposes only. It is not necessary during lab. Report your answer to three places after the decimal.
Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
The value of the (F-e-S-C-N⁺²) = 0.11 m-M when the student made additional dilutions.
Calculation of the value of the (F-e-S-C-N⁺²):Since
M (F-e(N-O₃)₃ = 0.200 M
V (F-e(N-O₃)₃ = 10.63 mL
n (F-e(N-O₃)₃ = 0.200*10.63
= 2.126 mmol
M (K-S-C-N) = 0.00200 M
V (K-S-C-N) = 1.42 mL
And,
n (KS-C-N) = 0.00200 * 1.42 = 0.00284 mmol
Now
Total volume = V (F-e(N-O₃)₃ + V (K-S-C-N)
= 10.63 + 1.42
= 12.05 mL
Also, Limiting reactant = K-S-C-N
So,
F-e-S-C-N⁺² = 0.00284 mmol
M (F-e-S-C-N⁺²) = 0.00284/12.05
= 0.000236 M
Now
Excess reactant = (F-e(N-O₃)₃
n(F-e(N-O₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
Now For standard 2:
n (F-e-S-C-N⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (F-e-S-C-N⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (F-e-S-C-N⁺²) = 0.11 mM
Learn more about volume here: https://brainly.com/question/22695394
The combustion of 1.771 g of propanol ( C 3 H 7 OH ) increases the temperature of a bomb calorimeter from 298.00 K to 302.34 K . The heat capacity of the bomb calorimeter is 13.70 kJ/K . Determine Δ H for the combustion of propanol to carbon dioxide gas and liquid water.
Answer: ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=C\times \Delta T[/tex]
Q = Heat absorbed by calorimeter =?
C = heat capacity of calorimeter = 13.70 kJ/K
Initial temperature of the calorimeter = [tex]T_i[/tex] = 298.00 K
Final temperature of the calorimeter = [tex]T_f[/tex] = 302.34 K
Change in temperature ,[tex]\Delta T=T_f-T_i=(302.34-298.00)K=4.34K[/tex]
Putting in the values, we get:
[tex]Q=13.70kJ/K\times 4.34K=59.4kJ[/tex]
As heat absorbed by calorimeter is equal to heat released by combustion of propanol
[tex]Q=q[/tex]
[tex]\text{Moles of propanol}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{1.771g}{60g/mol}=0.030mol[/tex]
Heat released by 0.030 moles of propanol = 59.4 kJ
Heat released by 1 mole of propanol = [tex]\frac{59.4}{0.030}\times 1=1980kJ[/tex]
ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ/mol
A person can burn 550 Calories by running at a moderate pace for one hour. Calculate the number of Calories that would be burned if the person only ran for 22 minutes. (1 hr = 60 min) Write only the number (answer) with 3 sig. fig.
Answer:
calories burned = 202 calories
Explanation:
22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.
Answer:
202 calories is the estimate of calories burned
Explanation:
22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.
At a certain temperature, the K p for the decomposition of H 2 S is 0.739 . H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g ) Initially, only H 2 S is present at a pressure of 0.215 atm in a closed container. What is the total pressure in the container at equilibrium?
Answer:
The total pressure in the container is 0.389 atm
Explanation:
Step 1: Data given
Kp = 0.739
The initial pressure of H2S = 0.215 atm
Step 2: The balanced equation
H2S(g) ⇆ H2(g) + S(g)
Step 3: The initial pressures
pH2S = 0.215 atm
pH2 = 0 atm
pS = 0 atm
Step 4: The pressures at the equilibrium
pH2S = 0.215 - X atm
pH2 = X atm
pS = X atm
Step 5:
Kp = 0.739 = (pS)*(pH2) / (pH2S)
0.739 = X*X / (0.215 - X)
0.739 = X² / (0.215 - X)
X² = 0.739*(0.215-X)
X² = 0.1589 - 0.739X
X² +0.739X - 0.1589 = 0
X = 0.174
pH2S = 0.215 - 0.174 atm = 0.041 atm
pH2 = 0.174 atm
pS = 0.174 atm
Step 6: Calculate the total pressure in the container
Total pressure = 0.041 atm + 0.174 atm + 0.174 atm
Total pressure = 0.389 atm
The total pressure in the container is 0.389 atm
Consider two equal-volume balloons under the same conditions of temperature and pressure. One contains helium, and the other contains carbon dioxide.
Which of the following are the SAME for these two gas samples. Select all choices that apply.
Number of gas particles (atoms or molecules)
Density
Number of moles of gas
Average kinetic energy
Average particle velocity
Mass
Answer:
1. Number of gas particles (atoms or molecules)
2. Number of moles of gas
3. Average kinetic energy
Explanation:
Since the two gas has the same volume and are under the same conditions of temperature and pressure,
Then:
1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole
2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.
3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.
The answer choices which are the SAME for these two gas samples are
1. Number of gas particles (atoms or molecules)2. Number of moles of gas3. Average kinetic energyWhat is Kinetic Energy?This refers to the type of energy which has to do with the motion of an object.
Hence, we can see that since the two gas has the same volume and are under the same conditions of temperature and pressure,
We would infer that:
They have the same number of moles They have the same number of molecules based on Avogadro's number.They have the same average kinetic energy because of the similar conditions of temperature and pressureRead more about kinetic energy here:
https://brainly.com/question/25959744
For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate. barium chloride (aq) potassium sulfate (aq) barium sulfate (s) potassium chloride (aq)
For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.
The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.
Answer: a) The reaction consumes 0.365 moles of barium chloride.
b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles[/tex]
[tex]\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles[/tex]
[tex]BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]
According to stoichiometry :
1 mole of [tex]BaCl_2[/tex] require 1 mole of [tex]K_2SO_4[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.365=0.365moles[/tex] of [tex]K_2SO_4[/tex]
Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]K_2SO_4[/tex] is the excess reagent.
As 1 moles of [tex]BaCl_2[/tex] give = 1 moles of [tex]BaSO_4[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{1}{1}\times 0.365=0.365moles[/tex] of [tex]BaSO_4[/tex]
As 1 moles of [tex]BaCl_2[/tex] give = 2 moles of [tex]KCl[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{2}{1}\times 0.365=0.730moles[/tex] of [tex]KCl[/tex]
Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.
How many grams of lithium sulfate are required to make 2500 mL of a 2.67 M solution of lithium sulfate? *
A. 248 g
B. 734 g
C. 527 g
D. 812 g
Answer:
The correct answer is B. 734 g
Explanation:
The chemical formulae of lithium sulfate is Li₂SO₄. With this, we can calculate the molecular weight (MM) of lithium sulfate as follows:
MM(Li₂SO₄) = (2 x molar mass Li) + molar mass S + (4 x molar mass O)
= (2 x 6.9 g/mol) + 32 g/mol + (4 x 16 g/mol)
= 109.9 g/mol
We need to prepare a solution with a molarity of 2.67 M. That means that the solution has to have 2.67 moles of Li₂SO₄ per liter of solution. We can convert from mol to grams with the calculated molecular weight and then we have to multiply by the volume of solution (2500 ml= 2.5 L), as follows:
Mass = 2.67 mol/L x 109.9 g/mol x 2.5 L ≅ 734 g
A technician plates a faucet with 2.68 g of Cr metal by electrolysis of aqueous Cr2(SO4)3. If 15.2 min is allowed for the plating, what current is needed? Use 96500 C/mol e- for the Faraday constant. Enter a number to 2 decimal places.
Answer:
16.36A
Explanation:
We'll begin by writing a balanced dissociation equation of aqueous Cr2(SO4)3. This is illustrated below:
Cr2(SO4)3 —> 2Cr^3+ 3(SO4)^2-
From the above, we can see that Cr is trivalent.
Next, let us determine the number of faraday needed to deposit metallic Cr. This is illustrated below:
Cr^3+ 3e- —> Cr
From the above equation, 3 faradays are needed to deposit metallic Cr
1 faraday = 96500C
Therefore, 3 faraday = 3 x 96500C = 289500C.
Molar Mass of Cr = 52g/mol
Now let us determine the quantity of electricity needed for 2.68g of Cr metal
This is shown below:
52g of Cr required 289500C.
Therefore, 2.68g of Cr will require = (2.68 x 289500)/52 = 14920.38C
Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:
Q (quantity of electricity) = 14920.38C
t (time) = 15.2mins = 15.2 x 60 = 912secs
I (current) =?
Apply the equation Q = It
Q = It
14920.38 = I x 912
Divide both side by 912
I = 14920.38/912
I = 16.36A
Therefore, a current of 16.36A is needed for the process.
The current needed is 16.36 A. The quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process.
Balanced dissociation equation:[tex]Cr_2(SO_4)_3 ---- > 2Cr^{3+}+ 3SO_4^{2-}[/tex]
The number of faradays needed to deposit metallic Cr. This is illustrated below:
[tex]Cr^{3+}+ 3e^- ---- > Cr[/tex]
Given:
3 faradays are needed to deposit metallic Cr
1 faraday = 96500C
Therefore, 3 faraday = 3 * 96500C = 289500C.
Molar Mass of Cr = 52g/mol
52g of Cr required 289500C.
Therefore, 2.68g of Cr will require = (2.68 * 289500)/52 = 14920.38C
Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:
Q (quantity of electricity) = 14920.38C
t (time) = 15.2mins = 15.2 x 60 = 912secs
To find:
I (current) =?
Apply the equation,
Q = It
14920.38 = I * 912
I = 14920.38/912
I = 16.36A
Therefore, a current of 16.36A is needed for the process.
Find more information Dissociation equation here:
brainly.com/question/25854432
At 25 oC the solubility of lead(II) chloride is 1.59 x 10-2 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]
Answer:
The Ksp at 25°C is 1.6 * 10^-5
Explanation:
Step 1: Data given
Temperature = 25°C
Solubility of lead(II) chloride = 1.59 * 10^-2 mol/L
Step 2: The balanced equation
PbCl2(s) <===> Pb2+(aq) + 2Cl-(aq)
Step 3: Calculate the Ksp
Ksp = [Pb2+][Cl-]²
Ksp = [Pb2+][Cl-]²
[Pb2+] = 1.59 *10-2 = 0.0159 M
[Br-] = 2 x 1.59*10-2 = 3.18 *10-2 M
Ksp = (1.59*10-2)(3.18*10-2)²
Ksp =1.6 * 10^-5
The Ksp at 25°C is 1.6 * 10^-5
The value of solubility constant (Ksp) at 25°C in scientific notation is 1.6 × 10-⁵.
How to calculate solubility constant?According to this question, the following parameters are given:
Temperature = 25°CSolubility of lead(II) chloride = 1.59 × 10-² mol/LThe balanced equation is as follows:
PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)
The solubility constant can be calculated as follows:
Ksp = [Pb2+][Cl-]²
Ksp = [Pb2+][Cl-]²
[Pb2+] = 1.59 × 10-² = 0.0159 M
[Br-] = 2 x 1.59*10-² = 3.18 × 10-² M
Ksp = (1.59 × 10-²)(3.18 × 10-²)²
Ksp =1.6 × 10-⁵
Therefore, the solubility constant (Ksp) at 25°C is 1.6 × 10-⁵.
Learn more about solubility constant at: https://brainly.com/question/7176737
Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [H2]eq = 0.14 M, [Cl2]eq = 0.39 M, [HCl]eq = 1.6 M. H2(g) + Cl2(g) ⇌ 2 HCl(g) Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [H2]eq = 0.14 M, [Cl2]eq = 0.39 M, [HCl]eq = 1.6 M. H2(g) + Cl2(g) ⇌ 2 HCl(g) 12 29 2.1 × 10-2 8.7 × 10-2 47
Question:
A) 12
B) 29
C) 2.1 × 10⁻²
D) 8.7 × 10⁻²
E) 47
Answer:
The correct option is;
E) 47
Explanation:
Kc, which is the equilibrium constant of a chemical reaction is derived by finding the ratio between the product of the equilibrium concentration of the product raised to their respective coefficients to the product of the equilibrium concentration of the reactants also raised to their respective coefficients.
Here we have;
[H₂] = 0.14 M
[Cl₂] = 0.39 M
[HCl] = 1.6
The reaction is given as follows;
H₂ (g) + Cl₂ (g) ⇄ 2HCl (g)
The formula for Kc is given as follows;
[tex]Kc = \frac{[HCl]^2}{[H_2][Cl_2]} = \frac{1.6^2}{0.14 \times 0.39} = 46.886[/tex]
Therefore, the Kc for the reaction is approximately equal to 47.
According to the experimental procedure of Experiment F1, 135 microliters of acetophenone (120.15 g/mol, 1.03 g/mL) was reacted with 127 mg of 4-nitrobenzaldehyde (151.12 g/mol). What is the theoretical yield, in milligrams (mg), of trans-4-nitrochalcone (253.25 g/mol)? Enter your answer as digits only (no units), using the proper number of significant figures.
Answer: The theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2[/tex]
Explanation : Given,
Volume of acetophenone = 135 microliters = 135 × 10⁻⁶ L = 0.135 mL
conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)
Density of acetophenone = 1.03 g/mL
Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g
Mass of 4-nitrobenzaldehyde = 127 mg = 0.127 g
Conversion used : (1 mg = 0.001 g)
First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde
[tex]\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}[/tex]
[tex]\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol[/tex]
and,
[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}[/tex]
[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone
So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone
From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of 4-nitrochalcone
From the reaction, we conclude that
As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone
So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone
Now we have to calculate the mass of 4-nitrochalcone
[tex]\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}[/tex]
Molar mass of 4-nitrochalcone = 253.25 g/mole
[tex]\text{ Mass of 4-nitrochalcone}=(0.000840moles)\times (253.25g/mole)=0.21273g=212.73mg=2.13\times 10^2mg[/tex]
(1 g = 1000 g)
Therefore, the theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2mg[/tex]
The reaction you actually perform in this experiment has some complexity in the mechanism (Solomons p904), but you can analyze a more simple two step version. Draw the arrow-pushing mechanism of the first step of this mechanism to form an imine. b) Draw the arrow-pushing mechanism of the second step of this mechanism to convert the imine to an amine using sodium borohydride.
Answer:
Explanation:
Attach is the solution
Final answer:
The formation of an imine involves nucleophilic addition, carbinolamine formation, and subsequent deprotonations leading to the imine, which is then reduced to an amine by sodium borohydride through nucleophilic attack.
Explanation:
Arrow-pushing Mechanism for Imine and Amine Formation
The question involves describing the arrow-pushing mechanisms for the formation of an imine from an aldehyde and ammonia, followed by the conversion of the imine to an amine using sodium borohydride. Let's address this step by step.
Step 1: Formation of Imine
The formation of an imine involves several key steps:
Nucleophilic addition of the amine to the carbonyl carbon of the aldehyde.Formation of carbinolamine via proton transfer.Protonation of carbinolamine oxygen turns it into a better leaving group, facilitating its elimination as water and resulting in the formation of an iminium ion.Finally, deprotonation of the nitrogen atom yields the imine.
Step 2: Conversion of Imine to Amine
In the next step, sodium borohydride (NaBH4) is used as a reducing agent to convert the imine into an amine. This involves the nucleophilic attack of hydride (H:-) from NaBH4 on the carbon atom of the iminium ion, leading to the formation of an amine.