Answer:
First the aqueous solution of iodine is heated mildly and then collection of the iodine crystals is done from its vapors.
Explanation:
Iodine is one of the elements that can get recovered easily from a given solution by going through the process of mild heating. For doing this, first, the aqueous solution is heated mildly over a low flame with a dish placed over the flame. As the process of mild heating continues, the fumes of the iodine start to originate that slowly get condense around the dish's cooler parts.
With condensation, the formation of pure iodine crystals takes place. These iodine crystals can now be extracted easily by a physical method.
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 mL of pure cyclohexene product. Calculate the student’s percent yield for this reaction. Show your work.
Answer:
[tex]49.45~%[/tex]
Explanation:
In this case, we have to start with the chemical reaction:
[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]
So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:
[tex](6*12)+(10*1)=82~g/mol[/tex]
With this value we can calculate the grams:
[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]
Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:
[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]
Finally, with these values we can calculate the yield:
[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
Based on the data given, the percentage yield of the student's work is 49.45 %.
What is the equation of the reaction?The equation of the clay-catalyzed dehydration of cyclohexanol is given below:
C₆H₁₂O ----> C₆H₁₀ + H₂OFrom the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.
Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.
What is the moles of cyclohexene obtained?First we determine the mass of cyclohexene obtained.
Mass = density * volume
volume of cyclohexene = 500 mL
density = 0.811 g/mL
mass of cyclohexene = 500 * 0.811
mass of cyclohexene = 405.5 g
Number of moles of cyclohexene = mass/molar mass
molar mass of cyclohexene = 82 g
moles of cyclohexene = 405.5 g/82
moles of cyclohexene = 4.945 moles
What is the percentage yield?Percentage yield = actual yield /expected yield * 100%Percentage yield = 4.945/10 * 100%
Percentage yield = 49.45%
Therefore, the percentage yield of the student's work is 49.45 %.
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The mass of an empty flask plus stopper is 44.232 g. When the flask is completely filled with water the new mass is 153.617 g. The flask is emptied and dried, and a piece of metal is added. The mass of the flask, stopper and metal is 143.557 g. Next, water is added to the flask containing the metal and the mass is found to be 226.196 g. What is the density of the metal (in g/cm3) to the proper number of significant figures
Answer:
3.7136g/cm³
Explanation:
Density is defined as the ratio between mass of a substance and its volume.
First, we will find the mass of the piece of the metal that is the difference between mass of metal + flask and mass of empty flask. That is:
Mass metal:
143.557g - 44.232g = 99.325g of the metal
Now, to find its volume you must know first the volume of the flask that can be obtained from the mass of water in the filled flask, that is:
153.617g - 44.232g = 109.385g of water = cm³ of water
In the second experiment, the mass of water = its volume is:
226.196g - 143.557g = 82.639g = 82.639cm³ of water
That means the volume the piece of metal is occupying is:
109.385cm³ - 82.639cm³ = 26.746cm³ of piece of metal
And its density is:
99.325g / 26.746cm³ =
3.7136g/cm³
1.60 mL of a suspension of 320.0 mg/5.00 mL aluminum hydroxide is
added to 2.80 mL of hydrochloric acid. What is the molarity of the
hydrochloric acid?
Answer:
1.41 M
Explanation:
First we must use the information provided to determine the concentration of the aluminum hydroxide.
Mass of aluminum hydroxide= 320mg = 0.32 g
Molar mass of aluminum hydroxide= 78 g/mol
Volume of the solution= 5.00 ml
From;
m/M= CV
Where;
m= mass of aluminum hydroxide= 0.32 g
M= molar mass of aluminum hydroxide = 78 g/mol
C= concentration of aluminum hydroxide solution = the unknown
V= volume of aluminum hydroxide solution = 5.0 ml
0.32 g/78 g/mol = C × 5/1000
C = 4.1×10^-3/5×10^-3
C= 0.82 M
Reaction equation;
Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)
Concentration of base CB= 0.82 M
Volume of base VB= 1.60 ml
Concentration of acid CA= the unknown
Volume of acid VA= 2.80 ml
Number of moles of acid NA = 3
Number of moles of base NB= 1
Using;
CA VA/CB VB = NA/NB
CAVANB = CBVBNA
CA= CB VB NA/VA NB
CA= 0.82 × 1.60 × 3/ 2.80 ×1
CA= 1.41 M
Therefore the concentration of HCl is 1.41 M
4. 1.00 mole of Ca(NO3)2 contains the same mass of N as 1.00 mole of NaNO3. True or False
5. The theoretical yield of a reaction is generally less than the actual yield. True or False
6. Which of the following statements is/are TRUE concerning the "mole"? (a) One mole of a substance contains as many particles as exactly 12 amu of carbon-12. (b) One mole of a substance contains 6.022 x 1023 particles of that substance. (c) There are 6.022 x 1023 carbon atoms in 12.00 g of carbon-12. (d) Because it is heavier, a mole of iodine atoms contains more particles than a mole of bromine atoms.
7. The reaction of sulfuric acid with hematite (Fe2O3) , a common mineral, produces iron (III) sulfate and water, as follows: .
Fe2O3 + H2SO4 _________> Fe2(SO4)3 + H2O
When this equation is balanced, it will contain the following term(s) with the appropriate coefficient: (a) 3 H2SO4 (b) 2 Fe2O3 (c) Fe2(SO4)3 (d) 4 H2O
8. In a recent year, electric power plants in the US consumed 1.14 x 1011 kg of natural gas. Assume that natural gas is entirely methane, CH4, and calculate the number of hydrogen atoms in this amount. (a) 4.56 x 1011 atoms H (b) 1.71 x 1037 atoms H (c) 2.84 x 1013 atoms H (d) 4.28 x 1036 atoms H If you could also explain how you get the answer that would be really helpful
Answer:
4. False
5. False.
6. (b) One mole of a substance contains 6.022 x 10²³ particles of that substance
(c) There are 6.022 x 10²³ carbon atoms in 12.00 g of carbon-12.
7. The correct option is (A) 3H₂SO₄
8. (b) 1.71 * 10³⁷ atoms of hydrogen
Explanation:
4. 1.00 mole of Ca(NO3)2 contains the same mass of N as 1.00 mole of NaNO3 is False because 1 mole of Ca(NO3)2 contains 2 moles (28 g) of N atoms whereas 1 mole of NaNO3 contains 1 mole (14 g) of N atoms
5. The theoretical yield of a reaction is generally less than the actual yield is False because it is the actual yield of a reaction that is always less than the theoretical yield due to some incomplete reactions.
The actual yield is obtained from carrying out the actual reaction while the theoretical yield is calculated from the equation of the reaction.
6. A mole of a substance is defined as the amount of that substance which contains as many elementary particles as there are in 12 g of carbon-12.
From experiments, it was discovered that 12 g of carbon -12 contains 6.02 * 10²³ atoms, therefore, a mole of a substance can also be defined as the amount of that substance which contains 6.02 * 10²³ particles of that substance.
(a) One mole of a substance contains as many particles as exactly 12 amu of carbon-12 is false because 12 amu is the mass of 1 atom of carbon-12 and not a mole of carbon-12.
(b) One mole of a substance contains 6.022 x 1023 particles of that substance is true from the definitions above
(c) There are 6.022 x 1023 carbon atoms in 12.00 g of carbon-12 is true from the definitions above.
(d) Because it is heavier, a mole of iodine atoms contains more particles than a mole of bromine atoms is false because, irrespective of difference in their masses, a mole of all substances contain the same number of particles- 6.02 * 10²³.
7. The balanced chemical equation of the reaction of sulfuric acid with hematite (Fe2O3) to produce iron (III) sulfate and water is as follows: .
Fe₂O₃ + 3H₂SO₄ -----> Fe₂(SO₄)₃ + 3H2O
The correct option is (A) 3H₂SO₄
8. Mass of natural gas = 1.14 * 10¹¹ Kg.
Mass of natural gas in grams = 1.14 * 10¹¹ * 10³ g = 1.14 * 10¹⁴ g
Molar mass of CH₄ = 16g/mol
Number of moles of CH₄ in 1.14 * 10¹⁴ g = 1.14 * 10¹⁴ g/ 16 gmol⁻¹ = 7.125 * 10¹² moles
1 mole of CH₄ contains 4 moles of hydrogen atoms.
7.125 * 10¹² moles of CH₄ will contain 4 * 7.125 * 10¹² moles of hydrogen atoms = 2.85 * 10¹³ moles of hydrogen atoms
1 mole of hydrogen atoms contain 6.02 * 10²³ atoms of hydrogen
2.85 * 10¹³ moles of hydrogen atoms will contain 2.85 * 10¹³ * 6.02 * 10²³ atoms of hydrogen = 1.71 * 10³⁷ atoms of hydrogen
Explain why the following chemical equation represents a Lewis acid-base reaction. H+ + NH3 —>NH4+
Answer:
Due to an electron-pair acceptor and donor.
Explanations:
Lewis acid can be defined as an electron-pair acceptor. An example is Hydrogen ion(H+). This is because it is a proton and it distributes positive charge which means that it accepts electrons(negative charge).
Lewis base can be defined as an electron-pair donor. This is because it donates electrons to be accepted by the proton. An example is ammonia(NH3).
How fast are the atoms moving if the temperature of a gas is cold?
A. very, very slowly
B. they are stagnant
C. very, very quickly
Answer:
i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.
Answer:
A. very, very slowly
Explanation:
A is the answer because atoms will move faster in hot gas than in cold gas.
What is the value of ΔG at 25°C when the initial concentrations of A, B, and C are 1 M, 1 mM, and 1 μM, respectively?
Answer:
Explanation:
0,44
The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O 4 (s)+4H 2 (g) 3Fe(s)+4H 2 O(g) When 64.6 grams of Fe 3 O 4 (s) react with excess hydrogenſe) how much heat is absorbed?
Answer:
41.3kJ of heat is absorbed
Explanation:
Based in the reaction:
Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ
1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.
63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:
63.4g Fe₃O₄ × (1mol / 231.533g) = 0.274moles of Fe₃O₄
These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:
0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =
41.3kJ of heat is absorbed
The reaction rate is k[Ce4+][Mn2+] for the following reaction: 2Ce4+(aq) + Tl+(aq) + Mn2+(aq) → 2Ce3+(aq) + Tl3+(aq) + Mn2+(aq What is the catalyst?
Answer:
Manganese (II) ion, Mn²⁺
Explanation:
Hello,
In this case, given the overall reaction:
[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]
Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺
Best regards.
The rate constant for this first‑order reaction is 0.550 s−10.550 s−1 at 400 ∘C.400 ∘C. A⟶products A⟶products How long, in seconds, would it take for the concentration of AA to decrease from 0.690 M0.690 M to 0.220 M?
Answer:
[tex]t=2.08s[/tex]
Explanation:
Hello,
In this case, for first order reactions, we can use the following integrated rate law:
[tex]ln(\frac{[A]}{[A]_0} )=kt[/tex]
Thus, we compute the time as shown below:
[tex]t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s[/tex]
Best regards.
What is the maximum amount of silver (in grams) that can be plated out of 4.7 L of an AgNO3 solution containing 6.3 % Ag by mass
Answer:
296.1g of Ag is the maximum amount of silver
Explanation:
A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:
Mass of solution:
Assuming a density of 1g/mL:
[tex]4.7L \frac{1000mL}{1L} \frac{1g}{mL} = 4700g[/tex]
If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:
4700L × (6.3g Ag / 100g) =
296.1g of Ag is the maximum amount of silverA chemistry graduate student is given of a pyridine solution. Pyridine is a weak base with . What mass of should the student dissolve in the solution
Answer:
34.1g of C₅H₅NHCl the student need to dissolve to the solution.
Explanation:
Full question is:
A chemistry graduate student is given 500.mL of a 0.20M pyridine C5H5N solution. Pyridine is a weak base with =Kb×1.7x10−9 . What mass of C5H5NHCl should the student dissolve in the C5H5N solution to turn it into a buffer with pH =4.76 ?
Using H-H expression for weak bases, it is possible to find pH of a buffer thus:
pOH = pKb + log [BH⁺] / [B]
Where pKb is -log Kb = 8.77, [BH⁺] concentration of C₅H₅NHCl and [B] concentration of C₅H₅N (It is possible to take the moles of both compounds and not its concentration.
As pH the student wants is 4.76, pOH is:
pOH = 14 - pH = 14 - 4.76 = 9.24
Replacing:
9.24 = 8.77 + log [C₅H₅NHCl] / [C₅H₅N]
Moles of C₅H₅N are:
0.500L × (0.20mol / L) = 0.10mol C₅H₅N
Replacing again:
9.24 = 8.77 + log [C₅H₅NHCl] / [0.10mol]
2.9512 = [C₅H₅NHCl] / [0.10mol]
0.29512 moles = [C₅H₅NHCl].
As molar mass of C₅H₅NHCl is 115.56g/mol, mass of 0.29512 moles are:
0.29512 moles C₅H₅NHCl × (115.56g / mol) =
34.1g of C₅H₅NHCl the student need to dissolve to the solution.The substance nitrogen has the following properties: normal melting point: 63.2 K normal boiling point: 77.4 K triple point: 0.127 atm, 63.1 K critical point: 33.5 atm, 126.0 K At temperatures above 126 K and pressures above 33.5 atm, N2 is a supercritical fluid . N2 does not exist as a liquid at pressures below atm. N2 is a _________ at 16.7 atm and 56.5 K. N2 is a _________ at 1.00 atm and 73.9 K. N2 is a _________ at 0.127 atm and 84.0 K.
Answer:
- N2 does not exist as a liquid at pressures below 0.127 atm.
- N2 is a solid at 16.7 atm and 56.5 K.
- N2 is a liquid at 1.00 atm and 73.9 K
- N2 is a gas at 0.127 atm and 84.0 K.
Explanation:
Hello,
At first, we organize the information:
- Normal melting point: 63.2 K.
- Normal boiling point: 77.4 K.
- Triple point: 0.127 atm and 63.1 K.
- Critical point: 33.5 atm and 126.0 K.
In such a way:
- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).
- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.
- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.
- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.
Best regards.
A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.
(1) Calculate the mass of commercial bleaching solution titrated. 1.1 450=559
(2) Determine the percent NaOCl in the commercial bleaching solution.
(3) Calculate the mass of one gallon of the commercial bleaching solution.
llllllllalalalalallalalalallalallalalalalalallalalalallala pls I am really sorry for this I can't take it down
an ideal gas is at a pressure 1.00 x 10^5 N/m^2 and occupies a volume 11.00 m^3. If the gass is compressed to a volume of 1.00 m^3 while the temperature remains constant, what will be the new pressure in the gas.
Answer:
[tex]P_2=1.1x10^6Pa[/tex]
Explanation:
Hello.
In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:
[tex]P_2=\frac{P_1V_1}{V_2}[/tex]
Consider that the given initial pressure is also equal to Pa:
[tex]P_2=\frac{1.00x10^5Pa*11.00m^3}{1.00m^3}\\ \\P_2=1.1x10^6Pa[/tex]
Which stands for a pressure increase when volume decreases.
Regards.
Nitrosyl bromide, NOBr, is formed from NO and Br2.
2 NO(g) + Br2(g) 2 NOBr(g)
Experiments show the reaction is second order in NO and first order in Br2.
(a) What is the rate equation for the reaction?
rate = k[NO]2[Br2]
rate = k[NO][Br2]
rate = k[NO]2[Br2]2
rate = k[NO]2[Br]2
(b) How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
the rate will double
the rate will remain unchanged
the rate will triple
(c) What is the change in the reaction rate if the concentration of NO is changed from 0.0024 mol/L to 0.0048 mol/L?
the rate will increase by a factor of 4
the rate will remain unchanged
the rate will also decrease by a factor of 1/2
Answer:
(a) What is the rate equation for the reaction?
rate = k[NO]2[Br2]
Explanation:
(b) How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
the rate will doubleNitrosyl bromide, NOBr, is formed from NO and Br2.
(a) Rate = [tex]k[NO]2[Br2][/tex]
(b) The rate will triple Nitrosyl bromide, NOBr, is formed from NO and Br2.
(c) The rate will triple increase by a factor of 4
Calculation of Nitrosyl bromide
(a) The reaction is second order in NO ---> [tex]rate , r = k[NO]^2[/tex]
the reaction is first order in Br2 --> rate ,[tex]r = k[Br2]^1[/tex]
Therefore, from the above two total rate equation is , rate , [tex]r = k [NO]^2[/tex][tex][Br2]^1[/tex]
--> [tex]r = k[NO]^2 [Br2][/tex]
(b)The rate law is rate , [tex]r = k[NO]^2 [Br2][/tex]
------> rate , r is directly proportional to [Br2]
----->[tex]r / r’ = [Br2] / [Br2]’[/tex]
r = initial rate
r’ = final rate
[Br2] = initial conc = 0.0022 mol/L
[Br2]’ = final conc = 0.0066 mol/L
Plug the values we get[tex]r/r’ = 0.0022 / 0.0066[/tex]
r / r’ = 0.33
----> r’ = r / 0.33 = 3r
Therefore, the rate will triple
-------------------------------------------------------------------------------
(c) The rate law is rate ,[tex]r = k[NO]^2 [Br2][/tex]
------> rate , r is directly proportional to [tex][NO]^2[/tex]
-----> r / r’ =[tex][NO]^2 / [NO]’^2[/tex]
r = initial rate[tex]r / r’ = [NO]^2 / [NO]’^2[/tex]
r’ = final rate
[NO] = initial conc = 0.0024 mol/L
[NO]’ = final conc = 0.0048 mol/L
Plug the values we get [tex]r/r’ = 0.0024 / 0.0048[/tex]
[tex]r / r’ = (0.0024^2) / (0.0048)^2[/tex]
= [tex](0.5)^2[/tex]
=[tex]0.25[/tex]
----> r’ = [tex]r / 0.25 = 4r[/tex]
Therefore, the rate will triple increase by a factor of 4
Find more information about Nitrosyl bromide here:
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Calculate the pressure exerted by 0.5600 mole of in a 1.1000-L container at 25.0°C. (The gas constant is 0.08206 L·atm/mol·K. Take absolute zero to be –273.2°C.)
Use the ideal gas law.
(Enter your answer to four significant figures.)
Pressure =
atm
Use the van der Waals equation.
(For : a = 1.39 atm L2/mol2, and b = 0.0391 L/mol. Enter your answer to four significant figures.)
Pressure =
atm
Compare the results.
(Enter your answer to two significant figures.)
The
_________
is higher by
atm, or
%.
Answer:
using ideal gas equation =12.4576atm to 4.significant figure
using vander Waals equation = 12.3504
The differences is 0.10atm
At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.22. What is the Ksp of the salt at 22 °C?
Answer:
2.29x10⁻¹² is Ksp of the salt
Explanation:
The Ksp of the metal hydroxide is:
M(OH)₂(s) ⇄ M²⁺ + 2OH⁻
Ksp = [M²⁺] [OH⁻]²
As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:
pOH = 14- pH
pOH = 14 - 10.22
pOH = 3.78
pOH = -log[OH⁻]
1.66x10⁻⁴ = [OH⁻]
And [M²⁺] is the half of [OH⁻], [M²⁺] = 8.30x10⁻⁵
Replacing in Ksp formula:
Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²
Ksp = 2.29x10⁻¹² is Ksp of the salt
What is the mass percent of vitamin C in a solution made by dissolving 5.20 g of vitamin C, C6H8O6, in 55.0 g of water
Answer:
The correct answer is 8.4 %
Explanation:
The mass percent of a compound in a solution is calculated as follows:
mass percent = mass of solute/mass of solution x 100
The solute is vitamin C, so its mass is:
mass of solute = 5.20 g
The solvent is water, and its mass is 55.0 g. The mass of the solution is given by the sum of solute + solvent:
mass of solution= 5.20 g + 55.0 g = 60.2 g
Finally we calculate the mass percent:
mass percent = 5.20 g/60.2 g x 100 = 8.64%
If 40.0 g of molten iron(II) oxide reacts with 10.0 g of mag-nesium, what is the mass of iron produced
Answer:
[tex]m_{Fe}=23.0gFe[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]FeO+Mg\rightarrow Fe+MgO[/tex]
Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:
[tex]n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO[/tex]
Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:
[tex]n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO[/tex]
Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:
[tex]m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\ \\m_{Fe}=23.0gFe[/tex]
Regards.
1. How many hydrogen atoms are in an alkane that has 7 carbon atoms?
2. An alkane contains 18 H atoms. Name the compound.
Answer:
1- there are 16 hydrogen atoms. ( heptane C7H16)
2- The compound is octane (C8H18)
explanation
to determine any atom in an alkane, this formula is used Cn H 2n +2, where n is the number of carbon atoms in the molecule.
to find the compound in Q2 we'll apply the numbers:
C nH₂n+2
Calculate the pH of a solution that has [H3O+]=4.3 x 10^-5 M
Answer:
pH = 4.4
Explanation:
pH = -log[H₃O⁺]
= -log(4.3 × 10⁻⁵)
pH = 4.37 = 4.4
I'm actually learning this in my chemistry class right now lol. Hope this helps though. :)
Select the correct answer
In an experiment, chemists heated a solid, red substance in a test tube. Vapors from the heated red substance condensed as a metallic liquid on
the side of the tube. The red substance eventually disappeared, leaving only the metallic liquid and producing a gas that escaped from the tube.
Which statement best describes what happened to the red substance in this experiment?
OA The experiment demonstrates the formation of plasma from the red substance.
B. The experiment demonstrates a physical property of the red substance.
OC The experiment shows that the red substance experienced a chemical change.
OD. The experiment shows that the red substance can be a solid, liquid, or gas.
CE. The experiment shows that the red substance is an element.
Answer:
C The experiment shows that the red substance experienced a chemical change.
Explanation:
Apparently, adding heat caused the red substance to decompose into a gas and a metallic liquid. If it were simply a phase change, the original red substance could be expected to return when the temperature cooled. Because the substance apparently decomposed, it is clearly not an element. At no point in the experiment is there any evidence of a plasma being formed.
The observed decomposition is a chemical change.
which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene
The question is incomplete as the options are missing, however, the correct complete question is attached.
Answer:
The correct answer is option A. ( check image)
Explanation:
The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.
Thus, the correct answer is option A. ( check image)
How many moles of RNA are found in 250mL of a 0.0125 M solution? Group of answer choices 3.1 moles 0.031 moles 0.0031 moles 1.0 moles
Answer:
0.0031 moles
Explanation:[tex]Molarity=\frac{molSolute}{LitreSolution}\\ 0.0125M=\frac{molRNA}{0.25L} \\molRNA=0.0125*0.25=0.0031 mol[/tex]
One brand of laundry bleach is an aqueous solution containing 4.00% sodium hypochlorite (NaOCl) by mass. You may want to reference (Pages 552 - 557) Section 13.5 while completing this problem. Part A What is the molarity of this solution
Answer:
molarity of the solution = 0.548 mol/L
Note: Additional information about the question is given as follows;
One brand of laundry bleach is an aqueous solution containing 4.00% sodium hypochlorite (NaOCl) by mass
What is the molarity of this solution? (Assume a density of 1.02 g/mL .)
Explanation:
A 4.00 percentage by mass composition of sodium hypochlorite (NaOCl) solution means that 100 g of the solution contains 4.00 g NaOCl.
Thus, a 1000 g of the solution contains 40.0 g NaOCl
Density of solution = 1.02 g/mL
Therefore, the volume occupied by 1000 g solution = mass/density
volume of 1000 g solution = 1000 g/1.02 g/ml = 980.4 mL
Molar mass of NaOCl = 74.5 g/mol
Number of moles = mass/molar mass
Number of moles of NaOCl = 40.0 g/74.5 g/mol
Number of moles of NaOCl = 0.537 mole
Therefore, molarity = number of moles / volume(L)
volume of solution in litres = 980.4/1000 = 0.9804 L
Molarity = 0.537/0.9804 = 0.548 mol/L
Therefore, molarity of the solution = 0.548 mol/L
which conditions make snow
Answer:
It depends on the weather.
Explanation:
Like rain and hail, snow comes from the water vapor that forms the clouds, but obviously its formation is different: snow forms when the temperature is low in the atmosphere. In these conditions the water vapor drops freeze and when they collide form tiny crystals that join together to form snowflakes, which fall to Earth when they are heavy enough.
Rank the following in terms of increasing atomic radius Mg, Sc, Ca
Answer:
Mg, Sc, Ca
Explanation:
To figure out increasing atomic radii, we use Periodic Trends applied to the Elements of the Periodic Table to help us out. We know that the trend for atomic radii is increasing left and down. Since Ca is the furthest down and left of the 3, it has the largest atomic radius. Since Sc is next element to Ca, it would be the 2nd largest atomic radius of the 3. Since Mg is above Ca, it has the smallest atomic radius of the 3.
A gas originally at 27 °C and 1.00 atm pressure in a 3.3 L flask is cooled at constant pressure until the temperature is 11 °C. The new volume of the gas is ________ L.
Answer:
THE NEW VOLUME OF THE GAS IS 3.124 L
Explanation:
Initial Temperature = 27°C = 27 + 273 K = 300 K
Pressure = constant
Initial Volume = 3.3 L
Final Temperature = 11 °C = 11+ 273 K = 284 K
Final volume = unknown
Using Charles' law, at constant pressure, the volume of a given gas is directly proportional to the temperature.
Mathematically,
V1 / T1 = V2 / T2
Re-arranging the formula by making V2 the subject of the formula
V2 = V1 T2 / T1
V2 = 3.3 * 284 / 300
V2 = 3.124 L
The new volume of the gas is therefore 3.124 L at 11°C
The new volume is 3.1 L
The parameters given in the question are
T1= 27°c
convert to kelvin
= 273 + 27
= 300k
T2= 11°c
convert to kelvin
= 273+11
= 284k
Volume 1= 3.3 L
V1/T1= V2/T2
3.3/300= V2/284
Cross multiply
284×3.3= 300×V2
937.2= 300V2
V2= 937.2/300
= 3.1
Hence the new volume is 3.1 L
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The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions
as AgCl and PbCl2. The total mass of the resulting precipitate is 62.37 g. Find the mass of PbCl, and AgCl in the precipitate.
Answer:
The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.
Explanation:
As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,
Moles of KCl = 0.275 L × 1.62 M = 0.445 moles
Now the reaction will be,
Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺
Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.
Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams
Now the mass of PbCl2 formed will be,
y × molecular mass = y × 278.1 grams = 278.1 y grams
Now the total precipitate will be,
62.37 grams = 143.32 x + 278.1 y -----------(i)
Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,
x + 2y = total mass of KCl
x + 2y = 0.445 moles ------ (ii)
On solving equation (i) and (ii) we get,
x as 0.115 and y as 0.165
Now the mass of AgCl will be,
143.32 × 0.115 = 16.48 grams
The mass of PbCl₂ will be,
278.1 × 0.165 = 45.88 grams.