Answer:
Lies predominantly to the left.
Explanation:
In the reaction:
NH4⁺(aq) + Br-(aq) ⇄ NH3(aq) + HBr(aq)
Conjugate acid + Ion ⇄ weak base + strong acid
HBr is a strong acid whereas NH3/NH4⁺ are the weak base and its conjugate base. A strong acid as HBr dissociates completely in solution as H⁺ and Br⁻. That means in solution you will never have HBr without dissociation doing the reaction:
lies predominantly to the left.Assume you dissolve 0.235 g of the weak benzoic acid, C6H5CO2H in enough water to make 100.0 mL of the solution and then titrate the solution with 0.108 M NaOH. Benzoic acid is a monoprotic acid.
1. What is the pH of the original benzoic acid solution before the titration is started?
2. What is the pH when 7.00 mL of the base is added? (Hint: This is in the buffer region.)
3. What is the pH at the equivalence point?
Answer:
1. pH = 2.98
2. pH = 4.02
3. pH = 8.12
Explanation:
1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:
0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M
The equilibrium of benzoic acid with water is:
C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)
And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:
Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]
The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:
[C6H5CO2H] = 0.01924 - X
[C6H5O⁻] = X
[H3O⁺] = X
Replacing in Ka:
6.14x10⁻⁵ = [X] [X] / [0.01924 - X]
1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²
1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0
Solving for X:
X = -0.0010→ False solution. There is no negative concentrations
X = 0.0010567M → Right solution.
pH = - log [H3O⁺] and as [H3O⁺] = X:
pH = - log [0.0010567M]
pH = 2.982.
pH of a buffer is determined using H-H equation (For benzoic acid:
pH = pka + log [C6H5O⁻] / [C6H5OH]
pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical
The benzoic acid reacts with NaOH as follows:
C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O
That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)
7.00mL of NaOH 0.108M are:
7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻
And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH
Replacing in H-H equation:
pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]
pH = 4.023. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles
Volume of NaOH to reach equivalence point:
1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:
1.164x10⁻³ moles / 0.111L = 0.01049M
Equilibrium of C₆H₅O⁻ with water is:
C₆H₅O⁻(aq) + H₂O(l) ⇄ C₆H₅OH(aq) + OH⁻(aq)
Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]
Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰
Equilibrium concentrations of the species are:
C₆H₅O⁻ = 0.01049M - X
C₆H₅OH = X
OH⁻ = X
Replacing in Kb expression:
1.63x10⁻¹⁰ = X² / 0.01049- X
1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0
Solving for X:
X = -1.3x10⁻⁶ → False solution
X = 1.3076x10⁻⁶ → Right solution
[OH⁻] = 1.3076x10⁻⁶
as pOH = -log [OH⁻]
pOH = 5.88
And pH = 14 - pOH
pH = 8.12Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
The standard entropy of a substance refers to its entropy at:__________.
a. absolute zero and 1 bar
b. 0°C and 1 bar
c. 25 °C and 1 bar
d. 25 °C and 0 bar
Answer:
b. 0°C and 1 bar
Explanation:
Hello,
In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.
Best regards.
At 2000°C the equilibrium constant for the reaction 9_1.gif is 9_2.gif If the initial concentration of 9_3.gif is 0.200 M, what are the equilibrium concentrations of 9_4.gif and 9_5.gif?
Answer:
[tex][N_2]_{eq}=[H_2]_{eq}=0.09899M[/tex]
[tex][NO]_{eq}=0.00202M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction:
[tex]2NO\rightleftharpoons N_2+O_2[/tex]
We know the equilibrium constant and equilibrium expression:
[tex]Kc=2.4x10^3=\frac{[N_2][O_2]}{[NO]^2}[/tex]
That in terms of the reaction extent [tex]x[/tex] (ICE procedure) we can write:
[tex]2.4x10^3=\frac{x*x}{(0.2M-2*x)^2}[/tex]
In such a way, solving for [tex]x[/tex] by using a quadratic equation or solver, we obtain:
[tex]x_1=0.09899M\\x_2=0.1010M[/tex]
Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:
[tex][N_2]_{eq}=[H_2]_{eq}=x=0.09899M[/tex]
[tex][NO]_{eq}=0.2M-2*0.09899M=0.00202M[/tex]
Regards.
Given the following reaction and data, A + B → Products
Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640
Required:
a. What is the rate law of the reaction?
b. What is the rate constant?
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹Most naturally occurring oxygen (O) isA. oxygen-16 B. oxygen-18 C. oxygen-15 D. oxygen-17
Option A is correct. Oxygen-16 occurs most naturally among these isotopes.
The three stable isotopes that make up natural oxygen are O-16, O-17, and 0-18, with O-16 having the highest natural abundance (99.762%). The standard atomic weight ranges between [15.99903, 15.99977], depending on the terrestrial source .
Oxygen-16 (16O) is a stable oxygen isotope with 8 neutrons and 8 protons in its nucleus. Its mass is 15.99491461956 u. Oxygen-16 is the most abundant isotope of oxygen, accounting for 99.762% of natural oxygen abundance. 16O has a high relative and absolute abundance because it is a primary product of stellar evolution and a primordial isotope, which means it can be produced by stars that were originally made entirely of hydrogen. The majority of 16O is produced in stars at the end of the helium fusion process; the triple-alpha process produces 12C, which catches an additional 4He to produce 16O. The process of burning neon generates an additional 16O. Oxygen-16 is twice as powerful.
Learn more about oxygen;
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Plssssssssss answer this question
Answer:
Table salt: answer salt
Tea: answer acidic
Carbonated drinks: answer acidic
Baking powder: answer acid and base
Detergent: answer acid and basic
Alum: answer acidic
Explanation:
I hope this helps. Enjoy your day!
You find a clean 100-ml beaker, label it "#1", and place it on a tared electronic balance. You add small amount of unknown solid and place the
beaker with its contents on the balance. The recorded data is:
mass of the empty, clean beaker #1: 74.605 g
mass of the beaker #1 with the white solid: 74.896 g
Using the Law of Conservation of Mass, what is the mass of the unknown solid you placed in beaker #1?
Answer:
the mas is .291 g
Explanation:
the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g
A compound is found to contain 30.45 % nitrogen and 69.55 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 46.01 g/mol. What is the molecular formula for this compound?
Answer:
Empirical formulae is NO2
Molecular Formulae is NO2
What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)
Answer: d=2000 g/L
Explanation:
Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.
[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]
Now that we have grams, we can divide to get density.
[tex]d=\frac{10000g}{5 L}[/tex]
d=2000g/L
Spinocerebellar Ataxia
9. Light is used in photosystem II and photosystem I to
O A. power the Calvin cycle.
B. fix carbon dioxide.
C. make carbon dioxide.
D. excite an electron.
Answer:
A. power the Calvin cycle.
Explanation:
because it helps to run theblife of plants with easily
13. Arrange each group of units from smallest to
largest
a)km, mm, cm, m
b) mg, kg, g
C) L, mL
d) s, ms, min, h
Answer:
A. mm,cm,m,Km
B. mg, g, Kg
C. mL,L
D. ms, s, min, h
Explanation:
Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light? Sort them accordingly. ltems (6 items) (Drag and drop into the appropriate area below)
a. Gamma rays
b. Infrared radiation
c. Ultraviolet liht
d. X-rays
e. Microwaves
f. Radio waves
Answer:
Higher frequency than visible light - Ultraviolet light, X-rays, and Gamma rays
Lower frequency than visible light - Infrared radiation, microwaves, and Radio waves
Explanation:
The frequencies of electromagnetic radiations vary according to their wavelengths. The relationship between the frequency and wavelength of the waves is expressed such that:
λ = c/f, where λ = wavelength, c = speed of light, and f = frequency.
Thus, there is an inverse relationship between the wavelength and the frequency of electromagnetic waves.
The order of the electromagnetic waves based on their frequency from the lowest to the highest is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma-rays
Hence, electromagnetic waves with higher frequencies than visible light include ultraviolet light, X-rays, and Gamma rays while those with lower frequencies include Infrared radiation, microwaves, and Radio waves.
Answer:
need points
Explanation:
(9443+45−9.9) (9443+45−9.9) ×8.4× 10 6
Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O
Answer:
Explanation:
Number of moles of propane:
=Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
= 44 * 6.8181
= 299.9964 grams
A number of moles of propane:
Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
=44 * 6.8181
= 299.9964 grams
What is carbon dioxide useful for?Carbon dioxide is used as a refrigerant, in fireplace extinguishers, for inflating lifestyles rafts and life jackets, blasting coal, foaming rubber and plastics, selling the increased vegetation in greenhouses, and immobilizing animals earlier than slaughter, and in carbonated liquids.
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tính chất hóa học của propylen
Answer:
Propilen là một chất khí không màu với mùi giống như dầu mỏ. Propilen nhẹ hơn nước và tan rất ít trong nước 0.61 g/ . Không hòa tan trong các dung môi phân cực như nước, chỉ tan trong dung môi không phân cực hay ít phân cực.. Propilen không có tính dẫn điện.
what is the molality of a solution
Answer: The number of moles of a solute per kilogram of solvent
Explanation:
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K. Assuming the ideal gas model for the gas and negligible kinetic and potential energy effects, evaluate the work, in kJ.
Answer:
W = -120 KJ
Explanation:
Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.
Thus; T1 = T2 = 400K
change in entropy; ΔS = −0.3 kJ/K
Formula for change in entropy is written as;
ΔS = Q/T
Where Q is amount of heat transferred.
Thus;
Q = ΔS × T
Q = -0.3 × 400
Q = -120 KJ
From the first law of thermodynamics, we can find the workdone from;
Q = ΔU + W
Where;
ΔU is Change in the internal energy
W = Work done
Now, since it's an ideal gas model, the change in internal energy is expressed as;
ΔU = m•C_v•ΔT
Where;
m is mass
C_v is heat capacity at constant volume
ΔT is change in temperature
Now, since it's an isothermal process where temperature is constant, then;
ΔT = T2 - T1 = 0
Thus;
ΔU = m•C_v•ΔT = 0
ΔU = 0
From earlier;
Q = ΔU + W
Thus;
-120 = 0+ W
W = -120 KJ
For each of the following reactions calculate the mass (in grams) of both the reactants that are required to form 15.39g of the following products.
a. 2K(s) + Cl2(g) → 2Cl(aq)
b. 4Cr(s) + 302(g) → 2Cr2O3(s)
c. 35r(s) + N2(g) → SraNa(s)
Answer:
a.
[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]
b.
[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]
c.
[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]
Explanation:
Hello,
In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:
a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]
[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]
b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]
[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]
c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]
[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]
Regards.
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. [H3O+] = 2.6 x 10-8 M
Answer:
To calculate the [OH-] in the solution we must first find the pOH
That's
pH + pOH = 14
pOH = 14 - pH
First to find the pH we use the formula
pH = - log [H3O+]From the question
[H3O+]= 2.6 × 10^-8 M
pH = - log 2.6 × 10^-8
pH = 7.6
pH = 8
So we pOH is
pOH = 14 - 8 = 6
To find the [OH-] we use the formula
pOH = - log [OH-]6 = - log [OH-]
Find antilog of both sides
[OH-] = 1.0 × 10^-6 MThe solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7
Hope this helps you
What is the correct IUPAC name for Ir(NO₂)₄
Answer
Iridium(IV)Nitrite
The correct IUPAC name of the Ir(NO₂)₄ compound is Iridium(IV)Nitrite.
What is the IUPAC name?Whether it's in a continuous chain or just a ring, the largest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature.
What is a compound?
A chemical compound would seem to be a substance that contains numerous similar molecules made of atoms from different elements joined by chemical bonds.
The given compound is Ir(NO₂)₄. It can be seen that 4 nitro group is attached with Ir and its coordination number is 4. Hence, the IUPAC name will be Iridium(IV)Nitrite.
To know more about IUPAC nomenclature.
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Assume that 33.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.How many moles of have been added at the equivalence point?n = ? mol
How many grams of Br are in 445 g CaBr2 ?
Explanation:
The mass of bromine is 79.904, and since there are two of them in the Calcium Bromide molecule, we'll multiply it by 2 to get 159.808. Dividing that by the full mass of the molecule then multiply it by 100 will give you the answer.
Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?
Answer:
A precipitate will form, BaCO₃
Explanation:
When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:
Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]
Where the concentrations of the ions are the concentrations in equilibrium
For actual concentrations of a solution, you can define Q, reaction quotient, as:
Q = [Ba²⁺] [CO₃²⁻]
If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.
Actual concentrations of Ba²⁺ and CO₃²⁻ are:
[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M
[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M
100.0mL is the volume of the mixture of the solutions
Replacing in Q expression:
Q = [Ba²⁺] [CO₃²⁻]
Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]
Q = 1.5x10⁻⁷
As Q > Ksp
A precipitate will form, BaCO₃
A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determined by using an indicator. Based on 20.00mL HNO3 being used to reach the endpoint, what was the mass of the Mg * (OH) in the antacid tablet? * 0.0583 g 0.583 5.83 g 58.3 g
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
11. How many oxygen atoms are there in one formula unit of Ca2+?
Answer:
NO3 has 3 atoms of oxygen in it per molecule (indicated by the subscript of 3) and (NO3)2 means you have 2 NO3 molecules, meaning you double the subscript to get 6 oxygen atoms.
Explanation: