Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant [tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
The equilibrium constant [tex]K_c= \dfrac{(0.17 )(0.17)}{0.74}[/tex]
The equilibrium constant [tex]K_c= 0.03905[/tex]
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :
[tex]H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17[/tex]
Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
[tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
[tex]0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}[/tex]
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095 x -x²
x² - 0.13095 x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
1L of bleach has a mass of 1,100 grams, 7.25% of the mass of bleach is NaClO, 1 mol of NaClO has a mass of 74.44 grams. What is the molarity (mol/L) of NaClO in the bleach? A.0.097 B.0.93 C. 1.07 D.79.75
Answer:
C. 1.07 M.
Explanation:
Hello,
In this case, we can define the molarity of the bleach as shown below:
[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]
In such a way, given the mass of bleach in a 1-L solution, we can compute the density:
[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]
In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:
[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]
Afterwards, using the by-mass percent of bleach we compute the mass:
[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]
And the moles:
[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]
Therefore, the molarity turns out:
[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]
Thus, answer is C. 1.07 M.
Regards.
What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of the ions in solution were [Cr3+] = 0.75 M and [Fe2+] = 0.25 M at 298K?
Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
E°/V
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
E°/V
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
3Fe ⇌ 3Fe²⁺ + 6e-; -0.41
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation
[tex]E = E^{\circ} - \dfrac{RT}{zF}\ln Q[/tex]
(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
[tex]Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}[/tex]
A chemistry student weighs out of chloroacetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
Answer:
11.6mL of the 0.1400M NaOH solution
Explanation:
0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.
The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:
ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O
Where 1 mole of the acid reacts per mole of the base.
That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.
You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:
0.154g ₓ (1mol / 94.5g) = 1.63x10⁻³ moles of ClCH₂COOH
To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:
1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =
11.6mL of the 0.1400M NaOH solutionWhich of the following reagents should be used to convert to Question 26 options: A) H2, Lindlar's catalyst B) Na, liquid NH3 C) H2 / Nickel D) H2SO4, H2O
Answer: The question is is not complete...here is the complete question.
Which of the following reagents should be used to convert 3-Hexyne to E-3-hexene
Option B.
Na, liquid NH3.
Explanation:
3-Hexyne to E-3-hexene can be converted with by using the reagent of Na, NH3 (birch reduction) and this can be done by hdrogenation of H2.
The reagents NaNH3 convert 3-Hexyne to E-3-hexene because it is a reducing agents that convert or has the ability to reduce alkynes to trans alkenes.
3 Hexyne is an alkynes and it is converted to E- 3 hexene Na and NH3.
If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?
The Average atomic mass of phosphorus is 29.9.
What is Average atomic mass ?The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.
Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.
An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.
The versions of an element with different neutrons have different masses and are called isotopes.
The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,
Average atomic mass of P = ∑(Isotope mass) (its abundance)
∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)
Abundance of isotope = % of the isotope / 100.
∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.
Hence , The Average atomic mass of phosphorus is 29.9.
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bonding is similar to ionic bonding, except there are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate. Group of answer choices
The question is incomplete, the complete question is;
["covalent", "Van der Waals", "ionic", "hydrogen", "metallic"] bonding is similar to ionic bonding, except their are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate.
Answer:
metallic
Explanation:
The metallic bond bears a striking similarity to the ionic bonds only that there are no electronegative elements present to accept electrons in a metallic bond as in an ionic bond.
Most metals usually have a few valence electrons which are loosely bound to the outermost shell of the metal atom. Metallic bonds are usually comprised of metal ions bonded together by a sea of mobile electrons
These mobile electrons exert an attractive force on the positive ions and hold them together in the metallic crystal lattice. This force of attraction that holds the metal atoms together in the metallic crystal lattice is known as the metallic bond.
The volume of ammonia gas at 1.14 atm of pressure is gradually decreased from 90.3 mL to 43.4 mL. What is the final pressure of ammonia if there is no change in temperature
Answer:
THE FINAL PRESSURE OF AMMONIA IF THERE IS NO CHANGE IN TEMPERATURE AND A DECREASE IN VOLUME FROM 90.3 mL TO 43.4 mL IS 2.91 atm.
Explanation:
At constant temperature, the pressure of a given mass of gas is inversely proportional to the volume. This question follows Boyle's law of gas laws.
Mathematically written as:
P1V1 = P2V2
Re-arranging the formula by making P2 the subject of the formula;
P2 = P1V1 / T2
P1 = 1.4 atm
V1 = 90.3 mL
V2 = 43.4 mL
P2 = unknown
So therefore, we have:
P2 = 1.4 * 90.3 / 43.4
P2 = 2.91 atm
The final pressure of ammonia is therefore 2.91 atm.
Answer:
2.37 atm
Explanation:
Step 1: Given data
Initial pressure of ammonia (P₁): 1.14 atmInitial volume of ammonia (V₁): 90.3 mLFinal pressure of ammonia (P₂): ?Final volume of ammonia (V₂): 43.4 mLConstant temperatureStep 2: Calculate the final pressure of ammonia
Since the temperature is kept constant, we can calculate the final pressure of ammonia using Boyle's law.
[tex]P_1 \times V_1 = P_2 \times V_2\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{1.14atm \times 90.3mL}{43.4mL} = 2.37 atm[/tex]
Choose the compound that exhibits hydrogen bonding as its strongest intermolecular force.
A. C2H6
B. CH3OH
C. CH2Br2
D. SBr2
E. None of the above compounds exhibit hydrogen bonding.
Answer:
B
Explanation:
To form hydrogen bondings between the molecules, the compound needs a highly electronegative atom (usually N, O, or F) bonded with a hydrogen atom;
and that the highly electronegative atom has lone pair outermost shell electrons.
In the 5 options, only B (CH3OH) has an N, O, or F atom that has lone pair outermost shell electrons (2 lone pairs on each O atom), so it can form hydrogen bonds within its molecules.
Hydrogen bonds are stronger than the van der Waals' forces between its molecules (that exist regardless of whether there are hydrogen bonds).
The compound that exhibits hydrogen bonding as its strongest intermolecular force is CH₃OH as electronegative oxygen atom is bonded to hydrogen atom.
What is compound?Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:
1)Molecular compounds where in atoms are joined by covalent bonds.
2) ionic compounds where atoms are joined by ionic bond.
3)Inter-metallic compounds where atoms are held by metallic bonds
4) co-ordination complexes where atoms are held by co-ordinate bonds.
They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.
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A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution
Answer:
0.0913 M
Explanation:
We'll begin by writing the balanced equation for the reaction.
This is given below:
H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O
From the balanced equation above, we obtained the following:
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 2
Data obtained from the question include:
Molarity of base, NaOH (Mb) = 0.0990 M
Volume of base, NaOH (Vb) = 20.52 mL
Volume of acid, H2C3H2O4 (Va) = 11.13 mL
Molarity of acid, H2C3H2O4 (Ma) =..?
The molarity of the acid, H2C3H2O4 can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 11.13 / 0.0990 x 20.52 = 1/2
Cross multiply
Ma x 11.13 x 2 = 0.0990 x 20.52 x 1
Divide both side by 11.13 x 2
Ma = (0.0990 x 20.52)/ (11.13 x 2)
Ma = 0.0913 M
Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M
Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 
Answer:
ΔH = [tex]q_{p}[/tex]
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
[tex]q_{com}[/tex] = [tex]q_{p}[/tex]
where
[tex]q_{p}[/tex] = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
[tex]q_{com}[/tex] = the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is [tex]q_{p}[/tex]
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = [tex]q_{p}[/tex] + 0
==> ΔH = [tex]q_{p}[/tex]
What is it’s molecular formula for C5H4 if it’s molar mass is 128.17g/mol
✔ C5H4 has a molecular molar mass of :
M(C5H4) = 5 x M(C) + 4 x M(H)
M(C5H4) = 5 x 12 + 4 x 1 M(C5H4) = 60 + 4 M(C5H4) = 64 g/mol✔ The molecular mass of C5H4 is therefore 64 g/mol.
But, 128/64 = 2 This is double the molar mass of C5H4, this molecule has the formula 2C5H4.Answer:
C10H8
Explanation:
I clicked on that answer, and it is correct.
Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-
Answer:
See explanation
Explanation:
A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.
Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.
The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.
what is the pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4
Answer:
2
Explanation:
First, find the hydronium ion concentration of the solution with a pH of 4.
[H₃O⁺] = 10^-pH
[H₃O⁺] = 10⁻⁴
[H₃O⁺] = 1 × 10⁻⁴
Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.
[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01
Lastly, find the pH.
pH = -log [H₃O⁺]
pH = -log (0.01)
pH = 2
The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.
Hope this helps.
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 15.51 mL . How many moles of K I O 3 were titrated
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
some students believe that teachers are full of hot air. If I inhale 3.5 liters of gas at a temperature of 19 degrees Celsius and it heats to a temperature of 58 degrees celsius in my lungs. what is the new volume of the gas?
Answer:
3.97 L
Explanation:
Data obtained from the question include the following:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 19 °C
Final temperature (T2) = 58 °C
Final volume (V2) =..?
Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:
Temperature (K) = temperature (°C) + 273
T (K) = T (°C) + 273
Initial temperature (T1) = 19 °C
Initial temperature (T1) = 19 °C + 273 = 292 K
Final temperature (T2) = 58 °C
Final temperature (T2) = 58 °C + 273 = 331 K
Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 292 K
Final temperature (T2) = 331 K
Final volume (V2) =..?
V1 /T1 = V2 /T2
3.5 /292 = V2 /331
Cross multiply
292 x V2 = 3.5 x 331
Divide both side by 292
V2 = (3.5 x 331) / 292
V2 = 3.97 L
Therefore, the new volume of the gas is 3.97 L.
Draw the major product(s) obtained when the following compounds are treated with bromine in the presence of iron tribromide.
a. Bromobenzene
b. ortho-Xylene
c. Benzene sulfonic acid
d. Benzaldehyde
e. meta-Nitrotoluene
f, para-Dibromobenzene
g. Nitrobenzene tert-Butylbenzene
h. Benzoic acid
i. Dibromobenzene
Answer:
The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.
Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.
HBr will always be the side product of the bromine reactions along with the major compound.
Explanation:
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Answer:
the enthalpy of the second intermediate equation is halved and has its sign changed.
Explanation:
Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Hence the overall equation is now;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;
Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
Answer:
A.
Explanation:
Did it on Edge.
A 8.22-g sample of solid calcium reacted in excess fluorine gas to give a 16-g sample of pure solid CaF2. The heat given off in this reaction was 251 kJ at constant pressure. Given this information, what is the enthalpy of formation of CaF2(s)
Answer:
The enthalpy of formation of CaF₂ is -1224.4 kJ.
Explanation:
The enthalpy of formation of CaF₂ can be calculated as follows:
[tex] \Delta H_{f} = \frac{q}{n_{CaF_{2}}} [/tex]
Where:
q: is the heat liberated in the reaction = -251 kJ
The number of moles of CaF₂ is:
[tex] n_{CaF_{2}} = \frac{m}{M} [/tex]
Where:
m: is the mass of CaF₂ = 16 g
M: is the molar mass of CaF₂ = 78.07 g/mol
[tex] n_{CaF_{2}} = \frac{m}{M} = \frac{16 g}{78.07 g/mol} = 0.205 moles [/tex]
Now, the enthalpy of formation of CaF₂ is:
[tex]\Delta H_{f} = \frac{q}{n_{CaF_{2}}} = \frac{-251 \cdot 10^{3} J}{0.205 moles} = -1224.4 kJ/mol[/tex]
Therefore, the enthalpy of formation of CaF₂ is -1224.4 kJ.
I hope it helps you!
Use your periodic table and calculator as needed for the following question.
How much stock solution is needed to make 250 mL of a 6.0M solution. The molarity of the stock solution is 18M.
Selections may be rounded so choose the best answer.
56 mL
83 mL
2.3 mL
4.7 ml
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. What is the theoretical yield of liquid iron, in grams? Just enter a numerical value. Do not enter units.
Answer: 313.6
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450g}{160g/mol}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260g}{28g/mol}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron is 313.6 g
Bayer Villiger Provide a balanced chemical equation of the reaction performed in this experiment. Use structures and compound names to show ALL reactants and products involved. Baeyer-Villiger Reaction of Acetophenone Data Results
• Moles of acetophenone used: (Show calculations) 0.020 moles (2.40g/120.151 g mol-1 =0.0199 moles)
• Moles of mCPBA used: (Show calculations) 0.036 moles_(6.25 grams/ 172.56 g.mol-1)
• Expected mass of the product: (Show calculation. Clearly show the limiting and excess reactants)
Answer:
See the explanations
Explanation:
In the Baeyer-Villiger reaction, we will produce an ester from a ketone (see the first reaction). In our case, the ketone is Acetophenone therefore phenyl acetate would be produced.
Now, for the mass calculation, we have to keep in mind that we have a reaction with a 1:1 ratio. So, if we have 0.02 moles of acetophenone and 0.036 moles of m-CPBA the limiting reagent would be the smallest value in this case acetophenone.
Additionally, if we have a 1:1 ratio and the limiting reagent is 0.02 moles of acetophenone we will have as product 0.02 of phenyl acetate, if we take into account the molar mass of phenyl acetate (136.05 g/mol), we can do the final calculation:
[tex]0.02~mol~acetophenone\frac{1~mol~phenyl acetate}{1~mol~acetophenone}\frac{136.05~g~phenyl acetate}{1~mol~phenyl acetate}=2.72~g~phenyl acetate[/tex]
I hope it helps!
Which Group is in the second column of the periodic table?
A. Noble gases
B. Halogens
C. Alkali metals
D. Alkaline earth metals
Answer:
Hey there!
That would be the alkaline earth metals.
Hope this helps :)
Answer: alkaline earth metals
Explanation:
A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.
Answer:
Cu₂S
Explanation:
From the question,
Cu S
Mass: 40.80 g 51.09-40.80 = 10.29 g
Mole ratio: 40.80/63.5 10.29/32.1
0.64 : 0.32
Divide by the smallest,
0.64/0.32 : 0.32/0.32
2 : 1
Therefore,
Empirical formula = Cu₂S.
A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and measures the initial reaction rates (The data from the three experiments is in the table). 1. Write the rate law 2. Solve for k.
Answer:
1. [tex]Rate =k [NO]^{2}[Cl_{2}][/tex]
2. [tex]k= 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]
Explanation:
[tex]Rate =k [NO]^{m}[Cl_{2}]^{n}[/tex]
[tex]Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2[/tex]
[tex]Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1[/tex]
[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}[/tex]
[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]
Find the percentage composition of each element in the compound having 9.8 grams of nitrogen,0.7 grams of hydrogen and 33.6 grams of oxygen
Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.
Explanation:
Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.
Percentage composition=[tex]\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%[/tex]
a) [tex]{\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%[/tex]
b) [tex]{\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%[/tex]
c) [tex]{\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%[/tex]
The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.
if your acetic acid buret was still wet inside with deionized water when you filled it with acetic acid?
Answer:
The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.
Describe the process of scientific inquiry ?
Answer:
It usually consists of six steps: question, observation or investigation, hypothesis, experiment, analysis of data (reviewing what happened during the experiment), and conclusion. Scientific inquiry, on the other hand, is non-linear, which means it does not follow a consistent step-by-step process.
Explanation:
Hope it helps
Which of the following is true regarding a voltaic (or galvanic) cell? a) It can only be used with hydrogen. b) It produces electrical current spontaneously. c) It consumes electrical current to drive a nonspontaneous chemical reaction.
Answer:
b) It produces electrical current spontaneously.
Explanation:
Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.
The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.
Which of the following metals has a low melting point?
2 A. Rubidium
B. Potassium
C. Calcium
D. Sodium
Answer:
Rubidium
Explanation: