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 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 silveran 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.
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]
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.
Phosphoric acid is a polyprotic acid, with p K values of 2.14, 6.86, and 12.38. Which ionic form predominates at pH 9.3
Answer:
HPO₄⁻² predominates at pH 9.3
Explanation:
These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ pKa 2.14
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺ pKa 6.86
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺ pKa 12.38
The H₂PO₄⁻ works as amphoterous, it can be a base and acid, according to these equilibriums.
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺
H₂PO₄⁻ + H₂O ⇄ H₃PO₄ + OH⁻
pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:
HPO₄⁻ + H₂O ⇄ H₂PO₄⁻ + OH⁻
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺
The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:
(6.86 + 12.38) /2 = 9.62
Above this pH, [PO₄⁻³] > [HPO₄⁻²].
In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]
Of the following, only ________ has sp2 hybridization of the central atom. Of the following, only ________ has sp2 hybridization of the central atom. ICl3 PBr3 SiH2Br2 HCN BF3
Answer:
BF3
Explanation:
Hybridization can be defined as the mixing of two or more atomic pure orbitals. ( s, p , and d) to produce two or more hybrid atomic orbitals that are similar and identical in shape and energy e.g sp,sp²,sp³ ,sp³d, sp³d². Usually , the central atom of a covalent molecules or ion undergoes hybridization.
in BF3; Boron is the central atom. Here, A 2s electron is excited from the ground state of boron ( 1s²2s²2p¹) to one empty orbitals of 2p.
The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals tat are trigonally arranged in the plane in order to minimize repulsion . Each of the three hybrid orbitals overlaps with p-orbital of fluorine atom to form three bonds of equal strength and with bond angles of 120⁰.
Energy
B ⇵ ║ ⇅ ║ ↑ -----------> *B ⇵ ║ ↑ ║ ↑ ║ ↑
1s 2s 2p 1s 2s 2p
Ground state Excited State
The above shows an illustrative example of how electrons move from the ground state to the excited state.
Name the compound Ga S3
Answer:
Gallium(III) sulfideCalculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation of the solute. Express your answer in atmospheres to three significant figures. Pvap = atm
The question is incomplete, the solute was not given.
Let the solute be K₂CrO₄ and the solvent be water
Complete Question should be like this:
The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.
Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.
Pvap = ________atm
Answer:
Pvap (of water above the solution) = 0.0306 atm
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
Explanation:
Given
volume of solution = 1 Litre = 1000 mL of the solution
density of the solution = 1.063 g/mL
concentration of the solution= 0.438M
temperature of the solution= 298 K
vapour pressure of pure water = 0.0313atm
Recall: density = mass/volume
∴mass of solution = volume x density
m = 1000 x 1.063 = 1063 g
To calculate the moles of K₂CrO₄ = volume x concentration
= 1 x 0.438 = 0.438 mol
Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g
Mass of water = mass of solution - mass of K₂CrO₄
= 1063 - 85.055 = 977.945 g
moles of water = mass/molar mass
∴ moles of water = 977.945/18.02 = 54.27 mol
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution
moles of ions = 3 x moles of K₂CrO₄
= 3 x 0.438 = 1.314 mol
Vapor pressure of solution = mole fraction of water x vapor pressure of water
= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm
which element will have higher electronegativity
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
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.
If the particles of matter that make up a substance are relatively far apart and can move freely, the substance is in what state?
gaseous
liquid
solid
Answer:
Gaseous
Explanation:
Gasses can move freely and do not form the shape of their containers
Liquids are more free than solids, but they conform to the shape of their container
Solids are not free
Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺; E°´ = -0.320 V
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; E°´ = -0.219 V
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
A 0.1-L unbuffered solution needs the pH adjusted from 3.5 to 1. How many microliters of a 6 molar HCl solution need to be added to adjust the pH
Answer:
1661μL of a 6M HCl you need to add
Explanation:
pH is defined as -log[H⁺] ([H⁺] =10^{-pH}), the initial and final concentrations of [H⁺] you need are:
Initial [H⁺] = 10^{-3.5} = 3.16x10⁻⁴M H⁺
Final [H⁺] = 10^{-1} = 0.1M H⁺
In moles, knowing volume of the solution is 0.1L:
Initial [H⁺] = 0.1L ₓ (3.16x10⁻⁴mol H⁺ / L) = 3.16x10⁻⁵moles H⁺
Final [H⁺] = 0.1L ₓ (0.1mol H⁺ / L) = 0.01 moles H⁺.
That means, moles of H⁺ you need to add to the solution is:
0.01mol - 3.16x10⁻⁵moles = 9.9684x10⁻³ moles of H⁺.
A solution of HCl dissociates in H⁺ and Cl⁻ ions, that means moles of HCl added are equal to moles of H⁺. As you need to add 9.9684x10⁻³ moles of H⁺ = 9.9684x10⁻³ moles of HCl:
9.9684x10⁻³ moles of HCl ₓ (1L / 6mol) = 1.6614x10⁻³L
In μL:
1.661x10⁻³L × (1x10⁶μL / 1L) =
1661μL of a 6M HCl you need to addOne 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
Calculate Ecell at 80 ºC for a voltaic cell based on the following redox reaction: H2(g, 1.25 atm) + 2AgCl(s) → 2Ag(s) + 2H+(aq, 0.10 M) + 2Cl–(aq, 1.5 M) The standard cell potential Eºcell = +0.18 V at this temperature.
Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - [tex]\frac{0.059}{n}[/tex] logQ
= 0.18 - [tex]\frac{0.056}{1}[/tex] log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
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
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 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
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.
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
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)
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 sample of a compound is made up of 57.53 g C, 3.45 g H, and 39.01 g F. Determine the empirical formula of this compound.
Answer:
C7H5F3
Explanation:
The following data were obtained from the question:
Mass of Carbon (C) = 57.53g
Mass of Hydrogen (H) = 3.45g
Mass of Fluorine (F) = 39.01g
The empirical formula of the compound can be obtained as follow:
C = 57.53g
H = 3.45g
F= 39.01g
Divide each by their molar mass
C = 57.53/12 = 4.79
H = 3.45/1 = 3.45
F = 39.01/19 = 2.05
Divide each by the smallest
C = 4.79/2.05 = 2.3
H = 3.45/2.05 = 1.7
F = 2.05/2.05 = 1
Multiply through by 3 to express in whole number
C = 2.3 x 3 = 7
H = 1.7 x 3 = 5
F = 1 x 3 = 3
Therefore, the empirical formula for the compound is C7H5F3
What is the density of a gas with a molar mass of 28.26g/mol at 3.510atm and 237k?
Answer:
density = 5.1g/m³
Explanation:
use ideal gas equation
Pv=nRT
28.26 x 3.51 = m/v x 0.08206 x 237
m/v = (28.26 x3.51)/(237 x 0.08206) = 5.1g/m³
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.
How many moles of hydrogen gas will be produced when 12 g of Mg will react completely with excess of an acid according to the following reaction? 2 Mg + 2 HCI — 2 MgCl2 + H,
Answer:
0.49 mol
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCI ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 12 g of Mg
The molar mass of Mg is 24.31 g/mol.
[tex]12g \times \frac{1mol}{24.31g} = 0.49mol[/tex]
Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.
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
The gas evolved in the metal carbonate reaction with acid turns limewater milky The milky substance formed is
Answer:
The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of formation of white precipitate of calcium carbonate.
Hope it helps you!
The milky substance formed is CO₂ gas.
What leads to the formation of white precipitate of calcium carbonate ?The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas.
This happens because of formation of white precipitate of calcium carbonate.
Lime water turns milky when the gas liberated from an acidified carbonate solution is passed into it.
The liberated CO₂ reacts with lime water to give calcium bicarbonate as the precipitate.
Hence, the milky substance formed is CO₂ gas.
Learn more about white precipitate, here:
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The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2 atm. What is the temperature of the air in the tire? a. 272.72 K b. 330 K c. 0.014 K d. 175 K
Hey there!
For this we can use the combined gas law:
[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]
We are only working with pressure and temperature so we can remove volume.
[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]
P₁ = 2 atm
T₁ = 27 C
P₂ = 2.2 atm
Plug these values in:
[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]
Solve for T₂.
[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]
[tex]2atm * T_{2}={2.2atm}*27C[/tex]
[tex]T_{2}={2.2atm}\div2atm*27C[/tex]
[tex]T_{2}=1.1*27C[/tex]
[tex]T_{2}=29.7C[/tex]
Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.
Hope this helps!