Answer:
0.56 mL
Explanation:
Volume = mass ÷ density
Volume = 1.5 ÷ 2.7 g/mL
Volume = 0.5555555556 = 0.56 mL
The volume of the aluminum block is 0.56 mL.
Hope this helps. :)
The volume of aluminum block is 0.556 mL.
The density of a substance is its mass per unit volume.
It given by formula:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} [/tex]
Given:
Density = 2.7 g/mL
Mass= 1.5 g
To find:
Volume=?
On substituting the values in the above formula:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}} \\\\\text{Volume}=\frac{\text{Mass}}{\text{Density}} \\\\\text{Volume}=\frac{1.5}{2.7} \\\\\text{Volume}=0.556mL[/tex]
Thus, the volume of the aluminum block is 0.556mL.
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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
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]
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)
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
Name the compound Ga S3
Answer:
Gallium(III) sulfideIf a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength, therefore, will be?
Answer:
if a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one-half
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
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
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.
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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 silverAccording to Le Châtelier's principle, how would a change in pressure affect a
gaseous system in equilibrium?
Answer:
Le Châtelier's principle states that when a chemical system at equilibrium is distributed by a change in conditions, the equilibrium position will shift in a direction that tends to counteract the change.
Therefore, when there is a change in pressure, the equilibrium will counteract the change by reducing/increasing the pressure through adjusting the no. of moles of gas.
Note: At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.
For example, when there's an increase in total pressure, the equilibrium position will shift to the side with a smaller no. of moles of gas so as to reduce the pressure.
Answer:
The equilibrium would shift to reduce the pressure change
Explanation:
what is the chemical formula for deionized water?
Answer:
Formula: H2O Formula Weight: 18.02 CAS No.: 7732-18-5 Density: 1.000 g/mL at 3.98 °C(lit.)
Explanation:
What is the name of the compound Br8P4? A. Bromine tetraphosphate B. Octaboron tetrapotassium C. Boron tetraphosphide D. Octabromine tetraphosphide
Answer:
Option D.
Explanation:
This compound has in its formula:
- Eight bromines
- Four phosphorous
8 → octa prefix
4 → tetra prefix
Right answer is Octabromine tetraphosphide
It can not be option A, tetraphosphate is P4O7
It can not be B and C, we do not have B (boron)
Answer:
Octabromine tetraphosphide
Explanation:
Answer via Educere/ Founder's Education
which element will have higher electronegativity
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 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
A 5.00 gram sample of an oxide of lead PbxOy contains 4.33 g of lead. Determine simplest formula for the compund
Answer: The empirical formula is [tex]PbO_2[/tex]
Explanation:
Mass of Pb = 4.33 g
Mass of O = (5.00-4.33) g = 0.67 g
Step 1 : convert given masses into moles
Moles of Pb =[tex]\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Pb = [tex]\frac{0.021}{0.021}=1[/tex]
For O = [tex]\frac{0.042}{0.021}=2[/tex]
The ratio of Pb O= 1: 2
Hence the empirical formula is [tex]PbO_2[/tex]
Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Write a balanced equation depicting the formation of one mole of SO3(g) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
For SO3(g) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
For NaBr(s) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
Write a balanced equation depicting the formation of one mole of Pb(NO3)2(s) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
For Pb(NO3)2(s) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
Answer:
Check Explanation.
Explanation:
Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.
NaBr (s)
The Standard formation reaction is
Na (s) + (1/2)Br₂ (g) → NaBr (s)
Using appendix C, the standard heat of formation of NaBr(s) is
ΔH∘f = -359.8 kJ/mol.
SO₃ (g)
The Standard formation reaction is
S (s) + (3/2) O₂ (g) → SO₃ (g)
Using appendix C, the standard heat of formation of SO₃(g) is
ΔH∘f = -395.2 kJ/mol.
Pb(NO₃)₂ (s)
The Standard formation reaction is
Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)
Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is
ΔH∘f = -451.9 kJ/mol.
Hope this Helps!!!
Answer:
Explanation:
Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.
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%
g If 1.00 mol of ethane gas and 5.00 mol of oxygen gas react, what is the limiting reactant and how many moles of water are produced from the reaction
Answer:
First, write a balanced equation of the reaction.
A hydrocarbon reacting with oxygen usually gives out carbon dioxide and water (combustion)
2C2H6 + 7O2 ---> 4CO2 + 6H2O
From the equation, we can see the mole ratio of ethane : oxygen is 2:7, meaning 2 moles of ethane reacts with 7 moles of oxygen.
If there's 1 mole of ethane gas, we need y moles of oxygen gas for complete reaction.
[tex]\frac{2}{7} =\frac{1}{y}[/tex]
y= 3.5 moles
Since 5 moles of oxygen is more than the required 3.5 moles, we can deduce oxygen gas is in excess, meaning ethane is limiting.
(You can get the same results too if you take y as ethane gas required).
The no. of moles of product produced ALWAYS depend on the no. of moles of the limiting reactant, since they are the ones which reacts completely.
So, from the equation, since the mole ratio of ethane : water = 2:6,
hence, (take the no. of moles of water produced as z),
[tex]\frac{2}{6} =\frac{1}{z} \\z= 3[/tex]
Therefore, 3 moles of water is produced.
Ethane is the limiting reagent in the given reaction. The number of moles of water produced from the reaction is 3 moles.
What is a limiting reagent?A limiting reagent can be defined as that reactant in the chemical reaction which is consumed first among the other reactants during the completion of a chemical reaction.
The limiting reagent is the reactant which decides the yield of the product when the quantity of the reactants is not taken in stoichiometry.
Given, chemical reaction of ethane and oxygen is:
[tex]C_2H_6 + \frac{7}{2} O_2 \longrightarrow 2CO_2 +3H_2O[/tex]
Given, the number of moles of ethane = 1 mol
The number of moles of oxygen = 5 mol
One more of ethane reacts with moles of oxygen = 3.5 mol
Therefore, ethane is the limiting reagent in this reaction as the oxygen is given in excess.
From the balanced equation, one mole of ethane produces moles of water equal to 3 moles.
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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.
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.
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
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 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.
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!
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.
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.
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.
Of the following species, ________ will have bond angles of 120°. A) PH3 B) ClF3 C) NCl3 D) BCl3 E) All of these will have bond angles of 120°.
Answer:
D. BCl₃
Explanation:
BCl₃ molecular geometry is trigonal planar and it has a bond angle of 120°.
Hope that helps.
According to the molecular geometry, BCl₃ has trigonal planar geometry and a bond angle of 120°.
What is molecular geometry?Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.
It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.
They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information about the distance between nuclei and electron density.
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