Answer:
The correct answer is 0.047 mol/L
Explanation:
The atmospheric air is a mixture of gases. We can assume an ideal behavior of the gas and use the ideal gas equation:
PV= nRT
where P is the pressure, V is the volume, n is the number of moles, R is a constant (0.082 L.atm/K.mol) and T is the temperature in K.
We have to first convert the pressure from Torr to atm:
760 Torr= 1 atm
⇒ 718 Torr x 1 atm/760 Torr = 0.945 atm
Then, we convert the temperature from ºC to K:
0ºC = 273 K
⇒ -29ºC+273= 244 K
Finally, we introduce the data in the equation and calculate de densitiy, which is the moles per liters of gas (n/V):
PV = nRT
n/V= P/RT
n/V = (0.945 atm)/(0.082 L.atm/K.mol x 244 K) = 0.047 mol/L
If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be
Answer:
Ca₃(AsO₃)₂
Explanation:
Sodium arsenite, with the chemical formula Na₃AsO₃, is formed by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.
Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.
4. Which of the following statements explains the cause of lanthanide contraction?
A. All lanthanides and actinides are radioactive
B. Protons exhibit a stronger pull on outer f orbitals
C. The d orbitals in lanthanides have unpair electrons
D. The d orbitals in actinides have paired electrons
Answer:
B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS
Explanation:
Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.
Answer:
B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.
Explanation:
What is a double bond?
Cathodic protection of iron involves using another more reactivemetal as a sacrificial anode. Classify each of thefollowing metals by whether they would or would not act as asacrificial anode to iron.
a. Sn
b. Cu
c. Zn
d. Au
e. Pb
f. Ag
g. Mg
An old iron beam was coated with an unknown metal. There is a crackon the coating and it is observed that the iron is rusting at thefracture. The beam is in a structure that experiences high stress,resulting in frequent fractures to the coating.
What was the old metal coating likely made of and what metal youwould use to repair the fractures to avoid further corrosion?
Choices: tin, aluminum, gold
1. The old coating was made of __________________.
2. __________________would be a good choice for repairing thefracture.
Answer:
1.) zinc and aluminum
2.)
a.) The old coating was made of tin.
b.) Aluminum would be a good choice for repairing the fracture.
After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ________ moles of carbon dioxide and ________ mole(s) of ATP (or GTP).A) 2; 2B) 2; 1C) 3; 2D) 3; 1
Answer:
C
Explanation:
Decide which element probably forms a compound with chlorine that has a chemical formula most and least similar to the chemical formula of the compound formed by chlorine and strontium.
Answer:
a. Beryllium,
b. Nitrogen
Explanation:
Most similar to strontium is beryllium, they both are in the 2d group.
Least similar to strontium is nitrogen. Strontium is a metal in the 2d group, Nitrogen is non-metal in the 15th group
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
1) 17.5ml
2) 35ml
3) 52.5ml
4) 3.0ml
Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:
[tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O[/tex]
In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:
[tex]2*n_{acid}=n_{base}[/tex]
And in terms of concentrations and volumes:
[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, we solve for the volume of acid:
[tex]V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL[/tex]
Best regards.
After running a TLC you visualized the TLC plate and observed the distance travelled by the organic compound is 10.0cm and calculated the retention factor (Rf) of that organic compound is 0.83. What is the distance travelled by the eluent?
a. 8.3 cm
b. 9.1cm
c. 10 cm
d. 120 cm
e. None of the above
Gold(ill) hydroxide is used in medicine, porcelain making, and gold plating. It is quite insoluble in aqueous solution. Which of the following substances, if any, can increase the solubility of this compound in water?a) Ca(OH)2(aq) b) NH3(aq) c) NaCl(aq) d) HBr(aq) e) Gold(III) hydroxide is an ionic solid compound; its solubility is not affected by any of the above reagents.
Answer:
NH3(aq)
Explanation:
Gold III hydroxide is an inorganic compound also known as auric acid. It can be dehydrated at about 140°C to yield gold III oxide. Gold III hydroxide is found to form precipitates in alkaline solutions hence it is not soluble in calcium hydroxide.
However, gold III hydroxide forms an inorganic complex with ammonia which makes the insoluble gold III hydroxide to dissolve in ammonia solution. The equation of this complex formation is shown below;
Au(OH)3(s) + 4 NH3(aq) -------> [Au(NH3)4]^3+(aq) + 3OH^-(aq)
Hence the formation of a tetra amine complex of gold III will lead to the dissolution of gold III hydroxide solid in aqueous ammonia.
Arrange these species into isoelectronic groups. It does not matter which group goes in which box, so long as the correct species are grouped. Isoelectronic group A Isoelectronic group B Isoelectronic group C Answer Bank
The species to be arranged are;
Sr2+, N3-, Li+, Ne, Br-, B3+, Al3+, He, Y3+
Answer:
Group A (10 electron species)
Al^3+, Ne, N^3-
Group B (36 electron species)
Sr^2+, Y^3+, Br^-
Group C ( 2 electron species)
He, Li^+, B^3+
Explanation:
Atoms and ions that have the same electron configuration are said to be isoelectronic. The species may not belong to the same group in the periodic table but are connected by the fact that they all have the same number of electrons. Cations and anions may belong to the same group of isoelectronic species provided that they all have the same number of electrons and the same electronic configuration.
Hence, in each group of isoelectronic species, one electronic configuration can be written for all the species and it will accurately represent the number of electrons for all species in the group since they have the same number of electrons.
For instance, all group C members have the electronic configuration, 1s2. This means that they all possess only two electrons.
How unwanted radiation in AES, AFS, AES rejected
Answer:
Reduce the intensity of radiations
Explanation:
Atomic Emission Spectroscopy is the study of electromagnetic radiations that are emitted by matter when it is excited. It is used to analyze intensity of light emitted. To end the radiations of AFS and AES their intensity is reduced and they are gradually rejected. It is not possible to completely eliminate the radiations immediately.
Consider the following reaction where Kp = 2.01 at 500 K: PCl3(g) + Cl2(g) PCl5(g) If the three gases are mixed in a rigid container at 500 K so that the partial pressure of each gas is initially one atm, what will happen? Indicate True (T) or False (F) for each of the following: T 1. A reaction will occur in which PCl5(g) is consumed. T 2. Kp will decrease. F 3. A reaction will occur in which PCl3 is produced. F 4. Q is greater than K. F 5. The reaction is at equilibrium. No further reaction will occur.
Explanation:
Kp remains constant (if T=const.).
If Q<Kp, more reactants are consumed (the direct reaction is in progress). If Q>Kp the reverse reaction is in progress (the products are consumed).
1. A reaction will occur in which PCl5 (g) is consumed
A. If Kp > Qp then 1,2,3 and 5 are F. 4 is T
B. Kc > Qc then 1,3 are T. 2,4,5 are F.
C. Qp > Kp then 1.3.4 are T. 2,5 are F.
Calculate the number of grams in sodium in 8.4g of Na2C6H6O7 (sodium hydrogen citrate) express your answer using two significant figures
please help!
Answer:
2.0 g Na
Explanation:
Stoichiometry.
8.4g sodium hydrogen citrate x (1 mol sodium hydrogen citrate / 192 g sodium hydrogen citrate) x (2 mol Na/1 mol sodium hydrogen citrate) x (23g Na/1 mol Na)
^write it out it makes more sense that way
True/False: ________ To study the effect of sunlight on different plants, I expose the plants to the same amount of sunlight. The independent variable is the sunlight.
Answer: True
Explanation:
The independent variable is the one which can be changed or manipulated in an experiment. The independent variable exerts its influence on the dependent variable. The dependent variable is the result of the experiment.
The amount of sunlight, can be regulated or changed in an experiment, thus it is an independent variable. The effect of sunlight on different plants is the dependent variable.
Analyze: The metallic character of an element is determined by how readily it loses electrons. Elements that lose electrons most easily have the greatest metallic character
A. Which group has the greatest metallic character?
B. Which group has the lowest metallic character?
C. What is the relationship between metallic character and ionization energy?
Answer:
Group 1 or akali metals have the greatest metallic property.
Group 17 has the lowest metallic character.
C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.
Explanation:
Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.
Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.
What does a complete ionic equation look like?
A. All substances are written as ionic compounds bonded together.
B. All substances are labeled with the oxidation states of the atoms.
C. All ionic substances are written as separate ions in solution.
D. All ionic substances are written with the state symbol (eo) after it.
Answer:
All ionic substances are written as separate ions in solution
All ionic substances are written as separate ions in solution in a complete ionic equation. Therefore, option (C) is correct.
What is the ionic equation?A complete ionic equation can be described as a particular chemical equation where charged atoms such as ions are expressed in a given solution. The complete ionic equations always contain all ions that are formed or act during a particular chemical reaction.
The net ionic equation can be described as an equation that provides information about ions that exists in an aqueous medium. Salts get dissolved in polar solvents such as water which are present as cations and anions in their dissolved state.
The ionic equation shows the chemical species that undergo a chemical change. The ions which are present on both sides of the equation are considered to be spectator ions. Therefore, in order to obtain the net ionic equation we can eliminate them.
Learn more about complete ionic equations, here:
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glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate ATP+glucose⟶ADP+glucose 6‑phosphate K′eq2=890 K′eq2=890 Using this information for equilibrium constants determined at 25∘C,25∘C, calculate the standard free energy of hydrolysis of ATP. standard free energy:
Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
6. If concentrations measured in moles per liter, and time in seconds, what are the units of the rate constant for (a) a first order reaction, (b) a second-order reaction, and (c) a third-order reaction
Answer:
(a) a first order reaction = s^-1.
(b) a second-order reaction = L(mol·s)
(c) a third-order reaction = s⁻¹M⁻²
Explanation:
Okay, this question is a question that has to do with kinetics that is how reaction occurs.
So, the RATE LAW is one of the concept that is being used in determining conditions of chemical reactions. The relationship between the the reactants molarity and the reaction rate is known as rate law. That is to say;
Rate law = K × Molarity.
So, In this question we are given that the concentrations measured in moles per liter, and time in seconds.
The FIRST ORDER REACTION has its unit for rate constant as per seconds(s^-1).
That is from; Rate law = k
Molarity/s = k × Molarity.
k = s^-1
(b) a second-order reaction;
Rate law= k × (molarity)^2.
Molarity/s = k × (Molarity)^2.
k = L(mol·s)
(c) A third-order reaction = s⁻¹M⁻²
Same thing applies.
Answer:
The rate constant has units of
a) s⁻¹ or /s for a first order reaction.
b) /sM or (s⁻¹M⁻¹) or (L/s.mol) or (L.s⁻¹mol⁻¹) for a second order reaction.
c) /sM² or s⁻¹M⁻² or (L²/s.mol²) or (L².s⁻¹mol⁻²) for a third order reaction.
Explanation:
The rate of a chemical reaction is defined as the amount (in concentration terms) of reactant used up or products formed per unit time. It's units generally is given in the units of concentration per unit time.
Rate = M/s or written extensively as mol/L/s = mol/L.s
And for all types of orders of reactions, the rate of reaction is given as
Rate = kCⁿ
k = Reaction rate constant
C = concentration of specie involved in the reaction
where n is the order of reaction.
Note that the units of the rate of the reaction still has to be mol/L.s or M/s regardless of its order.
M = mol/L
a) For a first order reaction
Rate = kC
M/s = k × M
K = (1/s)
Hence, the rate constant has a unit of /s or s⁻¹
b) Second order reaction
Rate = kC²
M/s = k × M²
k = (1/Ms)
Units of /sM or s⁻¹M⁻¹ or (L/s.mol)
c) Third order reaction
Rate = kC³
M/s = k × M³
k = (1/sM²)
Units of /sM² or s⁻¹M⁻² or (L²/s.mol²)
Hope this Helps!!!
Using appropriate chemical equation distinguish between cation and anion hydrolysis
Answer:
HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)
Explanation:
The equation to distinguish between cation and anion hydrolysis is given below :
HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)
The important thing to remember is their origin. The anions can react with water and can produce hydroxide ions while hydroxide ions make a solution basic.
AgNO3 is added to a solution containing Cl- and CrO42- in order to separate the ions. If the Cl- and CrO42- concentrations are 0.020 and 0.010 M, respectively, what are the minimum Ag+ concentrations required to precipitate out the anions?
Answer: The minimum [tex][Ag^{+}][/tex] concentrations required to precipitate out the anions is [tex]9 \times 10^{-9}[/tex] M.
Explanation:
We know that,
[tex]K_{sp}[/tex] for AgCl is [tex]1.8 \times 10^{-10}[/tex]
and, [tex]K_{sp}[/tex] for [tex]Ag_{2}CrO_{4}[/tex] is [tex]9 \times 10^{-12}[/tex]
Now, we will calculate the concentration of at which these ions precipitate out are as follows.
For AgCl :
[tex][Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]}[/tex]
= [tex]\frac{1.8 \times 10^{-10}}{0.02}[/tex]
= [tex]9 \times 10^{-9}[/tex] M
For [tex]Ag_{2}CrO_{4}[/tex] :
[tex][Ag^{+}]^{2} = \frac{K_{sp}}{CrO^{2-}_{4}}[/tex]
= [tex]\frac{9 \times 10^{-12}}{0.01}[/tex]
= [tex]9 \times 10^{-10}[/tex]
[tex][Ag^{+}] = \sqrt{(9 \times 10^{-9})}[/tex]
= [tex]3 \times 10^{-5}[/tex] M
This shows that concentration of ions in AgCl is less than the concentration of AgCl will precipitate first.
The heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its temperature. This is one of the reasons why desert region, although very hot during the day, are bitterly cold at night. The heat capacity of air at room temperature and pressure is appoximately 21 J/K*mol. How much energy is required to raise the temperature of a room of dimensions 5.5m x 6.5m x 3.0m by 10 degrees Celsius? If losses are neglected, how long will it take a heater rated at 1.5 kW to achieve that increase given that 1 W = 1 J/s?
Answer:
[tex]Q=9.2x10^5J[/tex]
[tex]t=614s=10.2min[/tex]
Explanation:
Hello,
In this case, we can compute the energy by using the following formula for air:
[tex]Q=nCp\Delta T[/tex]
Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:
[tex]n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol[/tex]
Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:
[tex]Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J[/tex]
Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:
[tex]t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min[/tex]
Regards.
Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net ionic equation to show why the solubility of AgBr (s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Ag(NH3)2+ , Kf = 1.6×107 . Use the pull-down boxes to specify states such as (aq) or (s).
Answer:
- [tex]AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]
- [tex]K=1.2x10^{-5}[/tex]
Explanation:
Hello,
In this case, by considering the dissolution of silver bromide:
[tex]AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}[/tex]
And the formation of the complex:
[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7[/tex]
We obtain the balanced net ionic equation by adding the aforementioned equations:
[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]
Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:
[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}[/tex]
So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:
[tex]K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}[/tex]
Best regards.
help asappppppppp please
Answer:
c
Explanation:
it goes in lowest energy orbital
Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl2(g)+3F2(g)→2ClF3(g) A 2.05 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 730 mmHg .
Answer:
2.4 grams of ClF3
Explanation:
First let us determine the moles of Cl2 and F2,
Cl2 = ( ( 337 )( 2.05 L ) / ( 0.082 )( 298 K ) ) * ( 1 atm / 780 ),
Cl2 = ( 690 / 24.436 ) * ( 1 / 780 ),
Cl2 = ( About ) 0.036 moles of Cl2
_________________________________________________
F2 = ( ( 729 )( 2 L ) / ( 0.082 )( 298 K ) ) * ( 1 atm / 780 ),
F2 = ( 1458 / 24.436 ) * ( 1 / 780 )
F2 = ( About ) 0.078 moles of F2
Now let us identify the limiting reactant, considering the ratio between ClF3 and Cl2 / F2. In this case F2 is the limiting reactant, as it forms a smaller molar ratio;
The theoretic yield is thus performed with the limiting reactant F2,
0.078 * ( 2 / 3 ) * ( 92.45 / 2 ) = ( About ) 2.4 grams of ClF3
An 8.5 mL sample of gasoline has a mass of .75 g. What is the density of the gasoline?
Answer:
density = 8.824g/mL
Explanation:
given
mass = 75g
volume = 8.5mL
density = mass/volume
density = 75g/8.5mL
density = 8.824g/mL
Answer:0.088g/ml
Explanation:
Density=mass/volume
d=0.75g/8.5ml
d=0.088g/ml
When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 1.76 g copper(I) sulfide. What is the percent yield?
Answer:
Percent yield = 22.8 %
Explanation:
Step 1: Data given
Numbers of moles copper = 0.0970 moles
Mass of copper(I) sulfide = 1.76 grams
Step 2: The balanced equation
2Cu + S ⇒ Cu2S
Step 3: Calculate moles of Cu2S
For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S
For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles
Step 4: Calculate mass of Cu2S
Mass Cu2s = moles Cu2S * molar mass Cu2S
Mass Cu2S = 0.0485 moles * 159.16 g/mol
Mass Cu2S = 7.72 grams
Step 5: Calculate percent yield
Percent yield = (actual yield/ theoretical mass) * 100%
Percent yield = (1.76 grams / 7.72 grams)*100%
Percent yield = 22.8 %
The percentage yield of the experiment obtained by the reaction of 0.0970 mole of copper with excess sulfurs is 22.8%
We'll begin by calculating the number of mole of Cu₂S produced from the reaction. This can be obtained as follow:
2Cu + S —> Cu₂S
From the balanced equation above,
2 moles of Cu reacted to produce 1 mole of Cu₂S.
Therefore,
0.0970 mole of Cu will react to produce = [tex]\frac{0.0970}{2}[/tex] = 0.0485 mole of Cu₂S.
Next, we shall determine the theoretical yield by calculating the mass of 0.0485 mole of Cu₂S.
Molar mass of Cu₂S = (63.5×2) + 32 = 159 g/mol
Mole of Cu₂S = 0.0485 mole
Mass of Cu₂S =?Mass = mole × molar mass
Mass of Cu₂S = 0.0485 × 159
Mass of Cu₂S = 7.7115 gThus, the theoretical yield of Cu₂S is 7.7115 g
Finally, we shall determine the percentage yield of Cu₂S.
Actual yield = 1.76 g
Theoretical yield = 7.7115 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{1.76}{7.7115} * 100\\\\[/tex]
Percentage yield = 22.8%Therefore, the percentage yield of the experiment is 22.8%
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under the same conditions carbon (iv) oxide,propane and nitrogen (i) oxide diffuse at the same rate.Explain
Answer:
Rate of diffusion is same .
Explanation:
As we know that Rate of the diffusion is directly proportional to the [tex]\frac{1}{\sqrt{M} }[/tex] .They have same mass if there is same rate and similar condition therefore the mass of carbon (iv) oxide,propane and nitrogen (i) oxide will be similar.
The mass is directly proportional to the Rate of the diffusion.Therefore the rate of diffusion is similar in all carbon (iv) oxide,propane and nitrogen (i) oxide .Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of micrograms. How many milligrams are in a dose labeled 125 µg? View Available Hint(s) Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of micrograms. How many milligrams are in a dose labeled 125 µg? 1.25 x 105 mg 0.125 mg 1.25 x 10?4 mg 1.25 x 102 mg
Answer:
0.125 mg
Explanation:
The correct answer would be 0.125 mg
According to the conversion factor, one milligram of a sample is equivalent to one thousand micrograms of the same sample.
milligram = [tex]10^{-3}[/tex]
microgram = [tex]10^{-6}[/tex]
Hence,
1 milligram = 1000 micrograms or 1 microgram = [tex]10^{-3}[/tex] milligram
Therefore, 125 micrograms will be:
125/1000 = 0.125 milligram
Activity: A Ferris wheel with a diameter of 60.0 m is moving at a speed of 2.09 m/s. What is the centripetal
acceleration?
Answer:
Centripetal acceleration of the wheel is [tex]0.145\ m/s^2[/tex].
Explanation:
We have,
Diameter of a Ferris wheel is 60 m
Radius of the wheel is 30 m
Speed of the wheel is 2.09 m/s
It is required to find the centripetal acceleration of the wheel. The formula of centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2.09)^2}{30}\\\\a=0.145\ m/s^2[/tex]
So, the centripetal acceleration of the wheel is [tex]0.145\ m/s^2[/tex].
In the process of making soap, I poured some of the cooked mixture through some muslin fabric, in order to separate the solid particles from liquid. What am I doing to this mixture?
A) Serrating it
B) Decanting it
C) Mixing it
D) Filtering it
Answer:
filtering
Explanation:
you're pouring the mixture through muslin cloth to keep the particles and bigger peaces out of the soap.