Answer:
2.81mL
Explanation:
Based on the reaction:
C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl
Benzoyl chloride + ammonia → Benzamide
1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.
Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.
As molar mass of benzoyl chloride is 141g/mol, mg you require are:
mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.
to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:
3.3981g × (1mL / 1.21g) =
2.81mLThe equilibrium between carbon dioxide gas and carbonic acid is very important in biology and environmental science. CO2 ( aq) + H2O ( l) H2CO3 ( aq) Which one of the following is the correct equilibrium constant expression (K c) for this reaction?
a) K =[H2CO3]/ [CO2]
b) K=[CO2]/ [H2CO3]
c) K=[H2CO3]/ [CO2][H2O]
d) K=[CO2][H2O]/ [H2CO3]
e) K=1/[H2CO3]
Answer:
Kc = [H₂CO₃] / [CO₂]
Explanation:
Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.
Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.
Also, each specie must be powered to its reactant coefficient.
For example, for the reaction:
aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)
The equilibrium constant, kc is:
Kc = [D]ⁿ / [B]ᵇ[E]ˣ
You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants
Thus, for the reaction:
CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)
The Kc is:
Kc = [H₂CO₃] / [CO₂]
The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine
Answer:
An alkyl halide can undergo SN2 reaction with an amine
Explanation:
The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.
The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).
This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.
The detailed mechanism of this reaction has been attached to this answer.
Draw the structure of 1,4-hexanediamine.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.
View Available Hint(s)
Answer:
1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.
Explanation:
1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.
The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.
Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].
1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.
The structure of 1,4-hexanediamine is shown below.
what is chemical equation of Braium chloride?
Answer:
BaCl2
Explanation:
Barium = Ba
Chloride => Cl-
Chemical Equation:
Ba + Cl => BaCl2
Note:
The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).
Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.
Answer:
2NaOH(s) → Na₂O(s) + H₂O(g)
Hope that helps.
The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate boiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC
Answer:
Approximately 100 °C.
Explanation:
Hello,
In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:
[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]
We can solve for the temperature as follows:
[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]
Thus, with the proper units, we obtain:
[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]
Hence, answer is approximately 100 °C.
Best regards.
1. In this experiment, the procedure instructs you to dissolve solid potassium hydrogen tartrate (KHT) in two different solvents. What are these two solvents? (2 pts)
Answer:
Water
Explanation:
Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.
The solvent for KHT is water.
suppose you are titrating vinegar, which is an acetic acid solution
Answer:
0.373 M
Explanation:
The balanced equation for the reaction is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, the following were obtained:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Next, we shall write out the data obtained from the question. This include:
Volume of base, NaOH (Vb) = 32.17 mL
Molarity of base, NaOH (Mb) = 0.116 M
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
The molarity of the acid solution can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.116 x 32.17 = 1
Cross multiply
Ma x 10 = 0.116 x 32.17
Divide both side by 10
Ma = (0.116 x 32.17) /10
Ma = 0.373 M
Therefore, the concentration of the acetic acid is 0.373 M.
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride
A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation:
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. Fe3+ (aq)+6CN (aq) Fe(CN) (aq)______is the Lewis acid and_____is the Lewis base. is the Lewis
2. CI- (aq) + AlCl3 (aq) AlCl4-____is the Lewis acid and______is the Lewis base.
3. AlBr3 + NH3 H3NAlBr3______is the Lewis acid and______is the Lewis base.
A. AlCl3
B. CN-
C. AlBr3
D. Cl-
E. NH3
F. Fe3+
Answer:
1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻
2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻
3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃
Hope this helps.
The Lewis acid is chemical substance which possesses an empty orbital and accepts an electron pair from a Lewis base ( donor ), in order to create a Lewis adduct ( molecule created from the bonding of Lewis base and acid ).
The Lewis acid from reaction 1 is Fe₃⁺ while the Lewis base is CN⁻ also the Lewis acid from reaction 2 is AICI₃ while the Lewis base is CI⁻
Hence we can conclude that the Lewis acids and Lewis bases of the reactions in the question are as listed above.
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Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent
Answer:
Lewis acid
Explanation:
In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.
If we look at the formation of PF6^-, the process is as follows;
PF5 + F^- -----> PF6^-
We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.
Hence in the formation of PF6^-, PF5 acts a Lewis acid.
When alkanes react with chlorine in the presence of ultraviolet light, chlorine atoms substitute for one or more alkane hydrogen atoms. What is the number of different chloroalkane compounds that can be formed by the reaction of C2H6 with chlorine?
Answer:
6
Explanation:
Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6
Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For example, the electron configuration of LiLi would be entered in complete form as 1s^22s^1 or in condensed form as [He]2s^1.
Answer:
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Explanation:
Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),
Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of 48 electrons surrounding its nucleus.
The electronic configuration of Cadmium is;
Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
The shorthand notation is given as;
Cd: [Kr] 4d¹⁰5s²
Cd2+ means that it has two less electrons, hence it's electron configuration is given as;
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
The amount of space an object takes up is called _____. gravity weight mass volume
What is titration? Question 1 options: The process of quickly adding one solution to another until a solid is formed. The process of slowly adding one solution to another until the reaction between the two is complete. The process of mixing equal volumes of two solutions to observe the reaction between the two. The process of combining two solids until the reaction between the two is complete.
Answer:
The process of slowly adding one solution to another until the reaction between the two is complete.
Explanation:
When you perform a titration, you are slowly adding one solution of a known concentration called a titrant to a known volume of another solution of an unknown concentration until the reaction reaches neutralization, in which the reaction is no longer taking place. This is often indicated by a color change.
Hope that helps.
A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?
Answer:
Molarity of the sulfuric acid solution is 4.61M
Explanation:
The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:
2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻
That means, 2 moles of base react with 1 mole of sulfuric acid.
If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:
0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄
As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:
0.200 moles H₂SO₄ / 0.0434L = 4.61M
Molarity of the sulfuric acid solution is 4.61MWhat states can electrons exist in? A. Electron clouds or energy levels B. Positive and negative C. Up and down spin D. In phase and out of phase
Answer:
A. Electron clouds or energy levels
Explanation:
Electrons can exist in two states:
Stablized in electronic orbitalsFreely movingElectrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.
Hence, the correct answer is A.
Answer:
Up and DOWN spin
Explanation:
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Answer:
1.3 mL
Explanation:
First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:
1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.
It should have 2 significant figures, because 1.2kg has 2 and we are dividing.
The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.
What is volume?Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.
Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.
This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.
Here, it is given that mass of olive oil is 1.2kg.
We know that,
Density of olive oil = 0.917kg/l.
Volume = mass/density
Volume = 1.2/0.917.
Volume = 1.30 lit.
Volume = 1300mL.
Thus, the volume of olive oil will be 1300 mL.
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A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
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A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?
Answer:
El volumen del gas era 12.95 L
Explanation:
Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:
“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”
La ley de Boyle se expresa matemáticamente como: P*V=k
Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:
[tex]\frac{V}{T}=k[/tex]
Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:
[tex]\frac{P}{T}=k[/tex]
Combinado las mencionadas tres leyes se obtiene:
[tex]\frac{P*V}{T} =k[/tex]
Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:
P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °KReemplazando:
[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]
Resolviendo:
[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]
V2= 12.95 L
El volumen del gas era 12.95 L
Interpret the following equation for a chemical reaction using the coefficients given:
Cl2(g) + F2(g) 2ClF(g)
On the particulate level:
________ of Cl2(g) reacts with ______ of F2(g) to form______ of ClF(g).
On the molar level:
______ of Cl2(g) reacts with______ of F2(g) to form______ of ClF(g).
Answer and Explanation:
Given the following chemical equation:
Cl₂(g) + F₂(g) ⇒ 2ClF(g)
The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:
On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).
On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).
Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm
please.
Answer:
3.00 L
Explanation:
Convert the pressure to Pascals.
P = 82 atm × (101325 Pa/atm)
P = 8,308,650 Pa
Convert temperature to Kelvins.
T = 27°C + 273
T = 300 K
Use ideal gas law:
PV = nRT
(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)
V = 0.00300 m³
If desired, convert to liters.
V = (0.00300 m³) (1000 L/m³)
V = 3.00 L
Answer:
[tex]\large \boxed{\text{3.0 L}}[/tex]
Explanation:
[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]
Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50 A. Q2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO3 solution, what was the current used in amperes
Answer:
[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]
Explanation:
Q1. Mass of Cu
(a) Write the equation for the half-reaction.
Cu²⁺ + 2e⁻ ⟶ Cu
The number of electrons transferred (z) is 2 mol per mole of Cu.
(b) Calculate the number of coulombs
q = It
[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]
(c) Mass of Cu
We can summarize Faraday's laws of electrolysis as
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]
Note: The answer can have only two significant figures because that is all you gave for the time.
Q2. Current used
(a) Write the equation for the half-reaction.
Ag⁺ + e⁻ ⟶ Ag
The number of electrons transferred (z) is 1 mol per mole of Ag.
(a) Calculate q
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]
(b) Calculate the current
t = 3 h = 3 × 3600 s = 10 800 s
[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]
Note: The answer can have only one significant figure because that is all you gave for the time.
The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?
Answer:
T2 = 500K
Explanation:
Given data:
P1 = 1atm
V1 = 300ml
T1= 27 + 273 = 300K
T2 = ?
V2 = 1.00ml
P2 = 500atm
Apply combined law:
P1xV1//T1 = P2xV2/T2 ...eq1
Substituting values into eq1:
1 x 300/300 = 500 x 1/T2
Solve for T2:
300T2 = 500 x 300
300T2 = 150000
Divide both sides by the coefficient of T2:
300T2/300 = 150000/300
T2 = 500K
How long should you hold the iron on the hair to heat the strand and set the base ?
A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute
Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene
Answer:
Explanation:
a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .
C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O
O is provided by KMnO₄
b ) In this reaction isophthalic acid is formed .
C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂
c)
4-Propyl-3-t-butyltoluene
In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .
The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )
Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)
Answer:
[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].
One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.
Explanation:
Formula for each of the speciesStart by finding the formula for each of the compound.
Both chlorine [tex]\rm Cl[/tex] and bromine [tex]\rm Br[/tex] are group 17 elements (halogens.) Each On the other hand, potassium [tex]\rm K[/tex] is a group 1 element (alkaline metal.) EachTherefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between
The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].
Balanced equation for the reactionWrite down the equation using these chemical formulas.
[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.
Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:
[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.
There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.
Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.
In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.
One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].
Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.
One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].
Hence:
[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:
[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:
[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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