Answer:
a) 52.8
Explanation:
M1V1 = M2V2
(7.91 M)(x ml) = (2.13 M) (196.1 ml)
(7.91M) (xml) = 417.693 M.ml
x ml = 417.693/ 7.91
x = 52.8
Answer:
A. 52.8
Explanation:
We make use of the dilution formula to solve this
[tex]M_1\times V_1 =M_2\times V_2[/tex]
Where ([tex]M_1[/tex] and [tex]M_2[/tex] are the initial molarity and final molarity and [tex]V_1[/tex] and [tex]V_2[/tex] are the initial volume and final volume)
Plugging into the values in the formula
[tex]M_1\times V_1 =M_2\times V_2[/tex]
[tex]7.91M \times V_1=2.13M \times 196.1mL[/tex]
[tex]V_1=\frac {(2.13M \times 196.1mL)}{7.91M}[/tex]
= 52.8 mL is the Answer
Which statement about a methyl functional group is correct? 1) a methyl group consists of a carbon bonded of three hydrogen atoms 2) a methyl group is polar 3) a methyl group may be negatively charged
Answer:
Explanation:
A methyl group consists of a carbon bonded to three hydrogen atoms.
A methyl functional group consists of a carbon atom bonded to three hydrogen atoms. Therefore, the correct option is option 1.
The methyl functional group is a basic building block in organic chemistry. It is made up of a carbon atom that is linked to three hydrogen atoms ([tex]CH_3[/tex]). The methyl group is frequently abbreviated as "Me."
Because carbon-hydrogen (C-H) bonds have equal electronegativities, methyl groups are nonpolar in nature. The methyl group's nonpolarity makes it comparatively unreactive in many chemical reactions, especially when contrasted to more polar functional groups.
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Enter your answer in the provided box. Consider the reaction H2(g) + Cl2(g) → 2HCl(g)ΔH = −184.6 kJ / mol If 2.00 moles of H2 react with 2.00 moles of Cl2 to form HCl, what is ΔU (in kJ) for this reaction at 1.0 atm and 25°C? Assume the reaction goes to completion.
Answer : The value of [tex]\Delta E[/tex] of the reaction is, -369.2 KJ
Explanation :
Formula used :
[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]
where,
[tex]\Delta E[/tex] = internal energy of the reaction = ?
[tex]\Delta H[/tex] = enthalpy of the reaction = -184.6 KJ/mole = -184600 J/mole
The balanced chemical reaction is,
[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]
when the moles of [tex]H_2\text{ and }Cl_2[/tex] are 2 moles then the reaction will be,
[tex]2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)[/tex]
From the given balanced chemical reaction we conclude that,
[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 4 - 4 = 0 mole
R = gas constant = 8.314 J/mole.K
T = temperature = [tex]25^oC=273+25=298K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta E=(-184600J/mole\times 2mole)-(0mole\times 8.314J/mole.K\times 298K)[/tex]
[tex]\Delta E=-369200J[/tex]
[tex]\Delta E=-369.2KJ[/tex]
Therefore, the value of [tex]\Delta E[/tex] of the reaction is, -369.2 KJ
Increases in the amount of cytoplasmic calcium required to initiate a muscle contraction are mediated by the coupling between a ________ on the T tubule and a ________ on the membrane of the sarcoplasmic reticulum.
Answer:
1) Dihydropyridine receptor
2) Rynodine receptor
Explanation:
Rynodine receptor: It is a category of interacellular channels of calcium with different forms like neurons and muscles found in animal tissues.
Dihydropyridine receptor: they are present in muscle tissues and are able to sense voltage in skeleton muscles thus can increase or control the release of calcium.
Final answer:
The T-tubule on the T-tubule and the calcium channel on the membrane of the sarcoplasmic reticulum mediate the increases in cytoplasmic calcium required for muscle contraction.
Explanation:
The coupling between a T-tubule on the T-tubule and a calcium channel on the membrane of the sarcoplasmic reticulum mediates the increases in the amount of cytoplasmic calcium required to initiate a muscle contraction. When the action potential reaches the T-tubules, it triggers the opening of calcium channels in the adjacent sarcoplasmic reticulum, causing calcium ions to diffuse out and into the sarcoplasm. This calcium release from the sarcoplasmic reticulum is essential for the contraction of muscle fibers.
Label A-F based on the table using C for concentrated and D for dilute.
A_
B_
C_
D_
E_
F_
Molarity is defined as the number of moles form a certain compound in one liter of solution.
So the higher the molarity the higher the number of moles in one liter of solution, and we say that the concentration is increased. The lower the molarity the lower the number of moles in one liter of solution, and we say that the concentration is decreased.
In a nutshell:
High molarity = concentrated solution
Low molarity = diluted solution
(A) concentrated
(B) dilute
(C) dilute
(D) concentrated
(E) dilute
(F) concentrated
Answer:
A) C
B) D
C) D
D) C
E) D
F) C
hope it helped
ExpLanation:
How many moles of sodium chloride are present in 0.50 kg of a 1.0 m solution of sodium chloride in water? mol sodium chloride
Answer:
=0.5 moles
Explanation:
Let us assume that the sodium chloride solution has a density of 1g/cm³.
Therefore the volume of the 0.5 kg of solution will be calculated as follows.
0.5kg into grams=0.5 kg×1000g/kg
=500g
volume= mass/density
=500g/1g/cm³
=500cm³
The solution is 1.0 M which means that 1.0 moles are in 1000 cm³
500cm³ will have:
(500 cm³×1.0 moles)/1000 cm³
=0.5 moles
Answer: 0.5
Explanation: edge 2021
The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq) (thiosulfate ion). The determination is based on the net ionic equation 2S2O2−3(aq)+I3(aq)⟶S4O2−6(aq)+3I−(aq) Given that it requires 29.6 mL of 0.260 M Na2S2O3(aq) to titrate a 30.0 mL sample of I−3(aq), calculate the molarity of I−3(aq) in the solution.
0.128 M of iodide solution was reacted with thiosulphate.
The equation of the reaction is;
[tex]2S2O2^-3(aq) + I3^-(aq)------->S4O2^-6(aq) + 3I^- (aq)[/tex]
Number of moles of S2O2^-3- = 29.6/1000 × 0.260 M
= 0.0077 moles
Since 2 moles of thiosulphate reacts with 1 mole of iodide
0.0077 moles of thiosulphate reacts with 0.0077 moles × 1 mole/ 2 moles
= 0.00385 moles of iodide.
Since;
Number of moles = concentration × volume
concentration of iodide = Number of moles/volume
Volume of iodide = 30/1000 = 0.03 L
Concentration of iodide = 0.00385 moles/0.03 L
= 0.128 M
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The density of pure copper is 8.96 at 20°C. If 5.00 g of pure copper pellets is added to a graduated cylinder containing 14.6 mL of water, to what volume level will the water in the cylinder rise?
Final answer:
The volume of water in the cylinder will rise to approximately 0.558 mL.
Explanation:
To find the volume to which the water in the cylinder will rise, we need to calculate the volume of the copper pellets. The density of pure copper is given as 8.96 g/cm³. From the given mass of the copper pellets (5.00 g), we can calculate the volume using the formula:
Volume = Mass / Density
Substituting the values, we get:
Volume = 5.00 g / 8.96 g/cm³ = 0.558 g/cm³
Therefore, the water in the cylinder will rise to a volume of 0.558 mL.
The solubility of Cd(OH)2 can be increased through formation of the complex ion CdBr2−4 (Kf=5×103). If solid Cd(OH)2 is added to a NaBr solution, what would the initial concentration of NaBr need to be in order to increase the molar solubility of Cd(OH)2 to 1.0×10−3 moles per liter?
Answer:
Concentration of sodium bromide required = 2.38 M (around 2.4 M)
Explanation:
The equilibrium representing the complex ion formation is:
[tex]Cd^{2+} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-} .....Kf =5*10^{3}[/tex]-----(1)
where K(f) = formation equilibrium
The equilibrium representing the dissolution of Ca(OH)2 is:
[tex]Cd(OH)_{2}\rightleftharpoons Cd^{2+}+2OH^{-}.....Ksp = 2.5*10^{-14}[/tex]---(2)
where Ksp = solubility product
adding Equation (1) and equation(2) gives the net reaction:
[tex]Cd(OH)_{2} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-}+2OH^{-}[/tex]
[tex]K = K_{f}*K_{sp} = 5*10^{3}*2.5*10^{-14}=\frac{[CdBr_{4}^{2-}][OH^{-}]^{2}}{[Br{-}]^{4}}[/tex]
[tex]12.5*10^{-11} =\frac{1*10^{-3} *[2*10^{-3}]^{2}}{[Br-]^{4} }\\[/tex]
[tex][Br-] = 2.38 M[/tex]
The study of chemicals is called chemistry. when the amount of reactant or product gets equal is said to be an equilibrium state.
The correct answer is 2.38M.
What is equilibrium constant?The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.The data is given in the question is as follows:-
[tex]Kf =5*10^3\\Ksp =2.5*10^{-14}[/tex]
The reaction in the given question is as follows:-
[tex]Cd(OH)_2 +4Br^- +[CdbBr_4]^{2-} +2OH^-[/tex]
The formula we used to solve the question is as follows:-
[tex]K =\frac{{[cdbr_4^2-]}[OH]^2}{{Br^-}^2}[/tex]After placing the value, the correct answer for the bromine is 2.38M.
Hence, the correct answer is 2.38M
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The structural level of a protein least affected by a disruption in hydrogen bonding is the
Answer:
The structural level of a protein least affected by a disruption in hydrogen bonding is the primary level.
Which statement about van der Waals forces is true?
a)When the forces are weaker, a substance will have higher volatility.
b)When the forces are stronger, a substance will have lower viscosity.
c)When the forces are weaker, the boiling point of a substance will be higher.
d)When the forces are stronger, the melting point of a substance will be lower.
Answer:
A
Explanation:
Van der Waals forces are the weak electric forces of attraction between molecules and their strength is dependent on the distance between the molecules. The longer the distance between the molecules the weaker the forces. Weaker Van der Waals forces mean that molecules can easily escape from the liquid - hence meaning higher volatility.
Answer:
A!!!
Explanation:
Which question can be answered using the scientific process?APEX
A.) Should people be made to reuse all bags?
B.) What is the effect of plastic bags on birds?
C.) Is the government doing enough to fight pollution ?
D.) Is it right to make people stop using plastic bags?
Answer: B.) What is the effect of plastic bags on birds?
Explanation:
A scientific process is a detailed sequential process in which answer of the scientific question can be derived on the basis of the implementation of the scientific methodology. The scientific methodology exhibit the direct observation and experimentation process.
B is the correct option this is because of the fact that this can be answered by direct observation and experimental trails which are the parts of scientific process.
How many moles of aluminum will be required to produce 0.45 moles of copper metal?
Answer:
0.3 moles of aluminum
Explanation:
The reaction of Aluminium with copper sulfate reacts to give aluminium sulfate and copper metal ,
The balanced chemical reaction is as follows -
2Al + 2 CuSO4 ----> Al2(SO4)3 + 3CuFrom the above balanced equation ,considering the stoichiometry , 3 mole of Copper is produced , by using 2 moles of Aluminium,using unitary method ,hence, 1 mole of copper is produced , by using 2/3 moles of Aluminum ,(from the question , 0.45 moles of copper )Therefore , 0.45 mole of copper is produced , by using 2/3 * 0.45 mole Aluminium,Solving, 2/3 * 0.45 mole = 0.3 molHence, 0.3 moles of aluminum will be required to produce, 0.45 moles of copper metal.To produce 0.45 moles of copper metal, 0.30 moles of aluminum are required.
To determine how many moles of aluminum (Al) are needed to produce 0.45 moles of copper (Cu) metal, we need to refer to the balanced chemical reaction between aluminum and copper (II) chloride (CuCl₂).
The balanced equation is:
2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu
This means that 2 moles of aluminum produce 3 moles of copper. We can set up a ratio to find out how many moles of aluminum are needed to produce 0.45 moles of copper:
(2 moles Al / 3 moles Cu) = (x moles Al / 0.45 moles Cu)
Solving for x,
x = (2/3) x 0.45 = 0.30 moles of Al
Therefore, 0.30 moles of aluminum will be required to produce 0.45 moles of copper metal.
An NaOH solution contains 3.6 mol of NaOH, and its concentration is 0.804 M. What is its volume?
Select one:
a. 4.48
b. 2.89
c. 179
d. 116.01
Answer:
A. 4.48
Explanation:
3.6/0.804 = 4.48
Answer:
a. 4.48 L is the Answer
Explanation:
Molarity (M), Molality (m), Normality (N), Mass %, Parts per million(ppm), billion(ppb), thousands(ppt) are some of the terms we use to represent the concentration of the solution that is to represent the amount of solute present in a solvent.
Molarity is moles of solute present in 1L of the solution. The formula to find Molarity is
[tex]Molarity = \frac {(moles solute)}{(volume of solution in L)}[/tex] and its unit is mol/L
Rearranging the formula
We get
Moles = Molarity × Volume
or
[tex]volume= \frac {moles}{Molarity}[/tex]
Plugging in the values
[tex]volume=\frac {3.6mol}{0.804M}[/tex]
[tex]=\frac {3.6mol}{(0.804 mol/L)}=4.48 L[/tex]
(Answer)
Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 mol dm-3 NaOH. The temperature rose from 298 K to 317.4 K. The specific heat capacity is the same as water, 4.18 J/K g.
A. -101.37 kJ
B. -7.05 kJ
C. 7055 kJ
D. 10,1365 kJ
Answer : The correct option is, (A) -101.37 KJ
Explanation :
First we have to calculate the moles of HCl and NaOH.
[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole[/tex]
The balanced chemical reaction will be,
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH
Thus, the number of neutralized moles = 0.1392 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water = [tex]87ml+87ml=174ml[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat absorbed = ?
[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
m = mass of water = 174 g
[tex]T_{final}[/tex] = final temperature of water = 317.4 K
[tex]T_{initial}[/tex] = initial temperature of metal = 298 K
Now put all the given values in the above formula, we get:
[tex]q=174g\times 4.18J/g^oC\times (317.4-298)K[/tex]
[tex]q=14110.008J=14.11KJ[/tex]
Thus, the heat released during the neutralization = -14.11 KJ
Now we have to calculate the enthalpy of neutralization.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy of neutralization = ?
q = heat released = -14.11 KJ
n = number of moles used in neutralization = 0.1392 mole
[tex]\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole[/tex]
Therefore, the enthalpy of neutralization is, -101.37 KJ
When generating equal amounts of energy, which of the following is true? (A) It is unknown how much carbon dioxide would be produced if burning coal or natural gas. (B) Burning coal produces more carbon dioxide than burning natural gas. (C) Burning natural gas produces more carbon dioxide than burning coal. (D) Burning natural gas produces the same amount of carbon dioxide as burning coal
Answer: Option (B) is the correct answer.
Explanation:
Natural gas is defined as the gas which is formed naturally beneath the surface of Earth that mostly contains methane and small amounts of ethane, propane etc.
Since, natural gas upon burning produces water and carbon dioxide resulting in the release of a clean gas as compared to other fuels. Carbon dioxide produced upon burning of natural gas is 50-60% lesser in amount as compared to release of carbon dioxide upon burning of coal.
Therefore, we can conclude that burning coal produces more carbon dioxide than burning natural gas.
Answer:
B is your answer
Explanation:
i took a test got it wrong and thats how i found out so your welcome
What is a photon?
A.) Part of a chloroplast
B.) A carbon dioxide molecule
C.) A Light particle
D.) Part of a ribosome
the correct (answer) is (c.) a light particle
Vitamin K is involved in normal blood clotting. When 0.802 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point of the solution is lowered by 2.69 °C. The freezing point and Kf constant for camphor can be found here. Calculate the molar mass of vitamin K.
Answer:
Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]
Explanation:
The freezing point of camphor = 178.4 ⁰C
the Kf of camphor = 37.7°C/m
where : m = molality
the relation between freezing point depression and molality is
Depression in freezing point = Kf X molality
Where
Kf = cryoscopic constant of camphor
molality = moles of solute dissolved per kg of solvent.
putting values
2.69°C = 37.7°C/m X molality
molality = 0.0714 mol /kg
[tex]molality=\frac{molesofvitaminK}{massofcamphor(kg)}=\frac{moles}{0.025}[/tex]
moles of vitamin K = 0.0714X0.025 = 0.00178 mol
we know that moles are related to mass and molar mass of a substance as:
[tex]moles=\frac{mass}{molarmass}[/tex]
For vitamin K the mass is given = 0.802 grams
therefore molar mass = [tex]\frac{mass}{moles}=\frac{0.802}{0.00178}=450.56\frac{g}{mol}[/tex]
The molar mass of vitamin K, involved in normal blood clotting, is calculated to be 450 g/mol using the freezing point depression method with camphor as the solvent.
Vitamin K is involved in normal blood clotting. When 0.802 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point of the solution is lowered by 2.69 °C. To calculate the molar mass of vitamin K, we will use the formula for freezing point depression:
ΔTf = Kf * m, where:
ΔTf is the freezing point depression (2.69 °C),Kf is the cryoscopic constant for camphor (37.7 °C/m),m is the molality of the solution.First, we rearrange the formula to solve for molality:
m = ΔTf / Kf = 2.69 °C / 37.7 °C/m = 0.0713 mol/kg
Next, calculate the moles of solute (vitamin K) using the mass of the solvent (camphor):
Moles of solute = m * mass of solvent (kg) = 0.0713 mol/kg * 0.0250 kg = 0.001783 mol
Finally, we find the molar mass (M) of vitamin K by dividing the mass of the solute by the moles:
M = mass of solute / moles = 0.802 g / 0.001783 mol = 450 g/mol
Thus, the molar mass of vitamin K is 450 g/mol.
Which is an example of a chemical change?
water boiling
a cake baking
sugar dissolving
butter melting
Answer:
A Cake Baking
Explanation:
All of the other options are physical changes because you can always bring it bake to it's original state unlike a cake baking you can't separate the flour, sugar, and eggs as it was before.
Answer: a cake baking
Explanation:
Chemical change can be define as a change in which the substance combines with the another substances so as to form a new substance, this is called as the chemical synthesis. The chemical decomposition can be define as the break down of one substance into different substances. The chemical change brings the change in the chemical composition of substances.
A cake baking is the example of the chemical change which occurs due to the combination various substances like flour, sugar, butter and others so as to form the cake on baking.
Bromine is one of only two elements that is a liquid at room temperature. Bromine has a heat of vaporization of 30.91 kJ/mol and its boiling point is 59 °C. What is the entropy of vaporization for bromine?
A. -301 J/(mol∙K)
B. -93.1 J/(mol∙K)
C. 10.7 J/(mol∙K)
D. 93.1 J/(mol∙K)
The element of bromine has entropy of vaporization as -93.1 J/(mol∙K) which is calculated as S=-ΔH/T.
What is an element?It is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.
Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.
The entropy of vaporization for bromine is calculated by the formula S=-ΔH/T,substitution in given formula gives,
S=-30.91/332=-93.1 J/(mol∙K)
Thus, the element of bromine has entropy of vaporization as -93.1 J/(mol∙K) .
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Final answer:
The entropy of vaporization for bromine is 93.1 J/(mol·K), which means the correct answer is D. 93.1 J/(mol·K).
Explanation:
To find the entropy of vaporization (ΔSvap) for bromine, we can use the formula ΔSvap = ΔHvap / Tb, where ΔHvap is the heat of vaporization and Tb is the boiling point in Kelvin. Given that the heat of vaporization (ΔHvap) for bromine is 30.91 kJ/mol and its boiling point is 59 °C, we first need to convert these units appropriately. The boiling point in Kelvin is 59 °C + 273.15 = 332.15 K.
Converting the heat of vaporization to J/mol (since 1 kJ = 1000 J), we have 30.91 kJ/mol = 30910 J/mol.
Now, we can calculate the entropy of vaporization as follows: ΔSvap = 30910 J/mol / 332.15 K = 93.1 J/(mol·K).
A compound weighing 0.458 g is dissolved in 30.0 g of acetic acid. The freezing point of the solution is found to be 1.50 K below that of the pure solvent. Calculate the molar mass of the compound.
Answer:
Molar mass of compound = 38.17 g/mol
Explanation:
The mass of compound dissolved = 0.458 g
The mass of acetic acid taken = 30.0g = 0.03 kg
the depression in freezing point =1.50 K or 1.50 ⁰C
the relation between depression in freezing point and molality is:
Depression in freezing point = Kf X molality
Where Kf= cryoscopic constant = 3.90 ⁰C Kg/mol
Putting values
1.50 = 3.90 X molality
[tex]molality=\frac{1.50}{3.90}=0.385[/tex]
molality is moles of solute per Kg of solvent
[tex]molality=\frac{moles}{massofsolvent}=\frac{moles}{0.03}=0.385[/tex]
moles = 0.385 X 0.03 = 0.012
[tex]moles=\frac{mass}{molarmass}[/tex]
[tex]molarmass=\frac{mass}{moles}=\frac{0.458}{0.012}= 38.17g/mol[/tex]
Answer: The molar mass of the compound is 39.69 g/mol
Explanation:
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
To calculate the depression in freezing point, we use the equation:
[tex]\Delta T_f=iK_fm[/tex]
Or,
[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
[tex]\Delta T_f[/tex] = Depression in freezing point = 1.50 K = 1.50°C (Change remains constant)
i = Vant hoff factor = 1 (For non-electrolytes)
[tex]K_f[/tex] = molal freezing point elevation constant = 3.90°C/m
[tex]m_{solute}[/tex] = Given mass of solute = 0.458 g
[tex]M_{solute}[/tex] = Molar mass of solute (glucose) = ? g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (acetic acid) = 30.0 g
Putting values in above equation, we get:
[tex]1.50^oC=1\times 3.90^oC/m\times \frac{0.458\times 1000}{\text{Molar mass of solute}\times 30.0}\\\\\text{Molar mass of solute}=\frac{1\times 3.90\times 0.458\times 1000}{1.50\times 30.0}=39.69g/mol[/tex]
Hence, the molar mass of the compound is 39.69 g/mol
What does stressing an equillibrium system mean? How is stress applied?
Answer:
Stressing an equilibrium system means altering physical conditions to favor either side of the reaction in progress.
Explanation:
The stress of equilibrium is achieved either by increasing physical properties or decreasing them eg. properties like pressure,volume or temperature of a substance.
These properties affect the direction of a reaction for a system in equilibrium.
Final answer:
Stressing an equilibrium system means causing a disruption in its balance, which can be done by changing reactant/product concentrations, pressure, or temperature. According to Le Châtelier's principle, the system will react by shifting in a direction to minimize the effects of the stress and re-establish equilibrium.
Explanation:
Le Châtelier's Principle
Stressing an equilibrium system means causing a change in conditions that disturbs the dynamic balance of a system where forward and reverse reactions occur at equal rates. Stress can be applied in several ways, including changes in the concentration of reactants or products, changes in pressure or volume (for gaseous reactions), or changes in temperature. These actions can push the system out of equilibrium, and as summarized by Le Châtelier's principle, when such stress is applied, the equilibrium will shift in a direction that helps to counteract or minimize the stress.
For example, if additional reactant is added to a system at equilibrium, the equilibrium will shift toward forming more products, decreasing the concentration of the added reactants. Conversely, if additional product is added, the equilibrium will tend to shift toward forming more reactants, to reduce the concentration of the added product. This is how Le Châtelier's principle allows us to predict the response of a stressed equilibrium.
A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of the compound were dissolved in 100mL of water, what weight of compound could be extracted by THREE sequential 10-mL portions of benzene?
Answer:
It could be extracted 0.512 g of solute
Explanation:
The equation that relates the [tex]K_{D}[/tex] and the volumes of organic and aqueous phases is:
[tex]q_{solute-aq} =\frac{V_{aq} }{K_{D}xV_{org} + V_{aq} }[/tex]
Where q_{solute-aq} refers to the fraction of solute remaining in the aqueous phase, V_{aq} is the aqueos phase volume, V_{org} is the organic phase volume and K_{D} is the partition coefficient of the solute in the solvents.
Moreover,for the three consecutive extractions of the same volume of organic phase we can write:
[tex]q_{solute-aq} =(\frac{V_{aq} }{K_{D}xV_{org} + V_{aq} })^{ 3}[/tex]
So, plugging the values given into the equation we get:
[tex]q_{solute-aq} =(\frac{100 mL }{2.7x10 mL + 100 mL })^{ 3}[/tex]
[tex]q_{solute-aq} =0.488[/tex]
The result obtained indicates that a fraction of 0.488 of solute remains in the aqueous phase.
Taking in account that the fraction formula is:
[tex]q_{solute-aq} = \frac{mass- of- solute- aq}{initial-mass- of -solute}[/tex]
[tex]0.488= \frac{mass- of- solute- aq}{1.0 g}\\\\0.488 x 1.0 g= {mass- of- solute- aq}\\0.488 g= {mass- of- solute- aq}\\[/tex]
Finally we substract the solute in the aqueous phase form the initial to get the amount in the organic phase:
[tex]1.0g - 0.488g = 0.512 g[/tex]
Final answer:
The weight of compound that can be extracted by three sequential 10mL portions of benzene can be calculated based on the compound's distribution coefficient.Therefore, a weight of approximately 0.05g of compound can be extracted by three sequential 10mL portions of benzene.
Explanation:
The distribution coefficient (K) represents the ratio of the concentration of a compound in one solvent to the concentration in another solvent. In this case, the compound distributes between benzene and water with a K value of 2.7. If 1.0g of the compound is dissolved in 100mL of water, we can calculate the weight of compound that can be extracted by three sequential 10mL portions of benzene.
Since K = 2.7, the ratio of compound in benzene to water is 2.7:1. This means that for every 2.7 parts of compound in water, there is 1 part in benzene. We can use this ratio to calculate the amount of compound that can be extracted.
Starting with 1.0g of compound in water, the first 10mL portion of benzene can extract (1.0g/2.7) = 0.37g of compound. The second 10mL portion can extract (0.37g/2.7) = 0.14g of compound. Similarly, the third 10mL portion can extract (0.14g/2.7) = 0.05g of compound. Therefore, a weight of approximately 0.05g of compound can be extracted by three sequential 10mL portions of benzene.
What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 23 ∘C , V1= 1.69×103 L ) to a volume of 3.38×103 L ? Assume no change in pressure or the amount of gas in the balloon.
Answer:
319.15^{o}C[/tex]
Explanation:
When all other variables are constant, we are allowed to use the formula
[tex]\frac{T_{2} }{V_{2} } = \frac{T_{1} }{V_{1} } \\
Which can be rewritten as T_{2} = \frac{T_{1} V_{2} }{V_{1} }
if you make T2 the subject of the formula. This formula is true only if temperature is in Kelvin not degrees Celsius so T1 must be converted to Kelvin
Now to calculate T2
[tex]T_{2}= \frac{296.15K*3.38.10^{3}L }{1.69.10^{3}L }= 592.3K[/tex] = [tex]319.15^{o} C[/tex]
A sample of an ideal gas at 1.00 atm1.00 atm and a volume of 1.81 L1.81 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 40.0 atm,40.0 atm, what was the volume of the sample? Assume that the temperature was held constant.
By applying Boyle's Law to the given conditions, the volume of the ideal gas compresses to 0.04525 L when the pressure increases to 40.0 atm, assuming constant temperature.
Explanation:The problem involves calculating the volume of a sample of ideal gas under constant temperature when the pressure changes. This scenario is perfectly described by Boyle's Law, which states that the product of the initial pressure and volume is equal to the product of the final pressure and volume, given by the formula P1V1 = P2V2, where P is pressure and V is volume. In this problem, the initial pressure (P1) is 1.00 atm, the initial volume (V1) is 1.81 L, and the final pressure (P2) is 40.0 atm. Our goal is to find the final volume (V2).
Applying Boyle's Law:
1.00 atm × 1.81 L = 40.0 atm × V2
To find V2, we rearrange the formula to solve for V2:
V2 = (1.00 atm × 1.81 L) / 40.0 atm
This gives us:
V2 = 0.04525 L
Therefore, when the pressure increases to 40.0 atm, the volume of the gas compresses to 0.04525 L.
The diagram below shows DNA molecules. What do the broken lines connected to hydrogen atoms represent in this diagram? the weakest intermolecular forces the weakest intramolecular forces the strongest intermolecular forces the strongest intramolecular forces
Hydrogen bonds are the broken lines between the DNA molecules. These are the strongest intermolecular forces, so C.
Answer:
c
Explanation:
c
Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.
(a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)
(b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)
(c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)
Answer: (a) [tex]\Delta G^0=-432.25kJ[/tex] , (b) [tex]\Delta G^0=55.96kJ[/tex] and (c) [tex]\Delta G^0=-171.74kJ[/tex]
Explanation: (a) Oxidation half reaction for the given equation is:
[tex]Mg(s)\rightarrow Mg^2^+(aq)+2e^-E^0=2.37V[/tex]
The reduction half equation is:
[tex]Pb^2^+(aq)+2e^-\rightarrow Pb(s)E^0=-0.13V[/tex]
[tex]E^0_c_e_l_l=E^0_r_e_d_u_c_t_i_o_n+E^0_o_x_i_d_a_t_i_o_n[/tex]
[tex]E^0_c_e_l_l=-0.13V+2.37V[/tex]
[tex]E^0_c_e_l_l=2.24V[/tex]
[tex]\Delta G^0=-nFE^0_c_e_l_l[/tex]
where n is the number of moles of electrons transferred and F is faraday constant.
2 moles of electrons are transferred in the cell reaction which is also clear from both the half equations.
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*2.44V[/tex]
[tex]\Delta G^0=-432252.8J[/tex]
or [tex]\Delta G^0=-432.25kJ[/tex]
(b) Oxidation half reaction for the given equation is:
[tex]2Cl^-(aq)\rightarrow Cl_2(g)+2e^-E^0=-1.36V[/tex]
Reduction half equation is:
[tex]Br_2(l)+2e^-\rightarrow 2Br^-E^0=1.07V[/tex]
[tex]E^0_c_e_l_l=1.07V+(-1.36V)[/tex]
[tex]E^0_c_e_l_l=1.07V-1.36V[/tex]
[tex]E^0_c_e_l_l=-0.29V[/tex]
Now, we can calculate the [tex]\Delta G^0[/tex] same as we did for part a.
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*(-0.29V)[/tex]
[tex]\Delta G^0=55961.3J[/tex]
or [tex]\Delta G^0=55.96kJ[/tex]
(c) Oxidation half reaction for the given equation is:
[tex]Cu(s)\rightarrow Cu^2^+(aq)+2e^-E^0=-0.34V[/tex]
reduction half equation is:
[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^2^+(aq)+2H_2O(l)E^0=1.23V[/tex]
[tex]E^0_c_e_l_l=1.23V+(-0.34V)[/tex]
[tex]E^0_c_e_l_l=0.89V[/tex]
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*0.89V[/tex]
[tex]\Delta G^0=-171743.3J[/tex]
or [tex]\Delta G^0=-171.74kJ[/tex]
To calculate the standard Gibbs free energy change (∆G°) for each reaction at 25°C in kJ, we can use tabulated electrode potentials. By writing half-cell reactions and summing the electrode potentials, we can determine the overall reaction's standard potential (E°). Then, using the formula ∆G° = -nFE°, we can calculate the standard Gibbs free energy change.
Explanation:Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.
a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)First, we need to write half-cell reactions for the given equation:Pb2+ + 2e- ➝ Pb (E° = -0.126 V)Mg2+ + 2e- ➝ Mg (E° = -2.37 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Pb) - E°(Mg)E° rxn = -0.126 V - (-2.37 V) = 2.244 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(2.244 V)b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)First, we need to write half-cell reactions for the given equation:Br2 + 2e- ➝ 2Br- (E° = 1.07 V)Cl2 + 2e- ➝ 2Cl- (E° = 1.36 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(2Br-) - E°(2Cl-)E° rxn = 2(1.07 V) - 2(1.36 V) = -0.640 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(-0.640 V)c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)First, we need to write half-cell reactions for the given equation:MnO2 + 4H+ + 2e- ➝ Mn2+ + 2H2O (E° = 1.23 V)Cu2+ + 2e- ➝ Cu (E° = 0.34 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Mn2+) - E°(Cu)E° rxn = 1.23 V - 0.34 V = 0.89 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(0.89 V)Learn more about Calculating standard Gibbs free energy change here:https://brainly.com/question/34263086
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A certain drug is made from only two ingredients: compound A and compound B. There are 7 milliliters of compound A used for every 5 milliliters of compound B. If a chemist wants to make 1116 milliliters of the drug, how many milliliters of compound A are needed?
Answer:
First step:
7 ml + 5 ml = 12 ml
Second step:
% of A = 7/12 x 100 = 58.33%
% of B = 5/12 x 100 = 41.67%
Third step:
In 1116 ml
compound A = 1116 x (58.33/100) = 651 ml
compound B = 1116 x (41.67/100) = 465 ml
Explanation:
In the 1st step: with what is given, the total volume is 12 ml
In the 2nd step: Find the percentage of each compound in the drug according to what is given.
In the 3rd step: calculate the volume of each compound separately in the new total volume of 1116 ml using the percentage composition.
volume of compound A will therefore be 651 milliliters
By setting up a proportion based on the ratio of 7 milliliters of A for every 5 milliliters of B, solving the subsequent equations yields that 650 milliliters of compound A are needed to make 1116 milliliters of the drug.
To determine how many milliliters of compound A are needed to make 1116 milliliters of the drug, we can set up a proportion based on the ratio given. With 7 milliliters of compound A used for every 5 milliliters of compound B, we get the following equation:
Compound A / Compound B = 7 / 5
Let's let x be the amount of compound A and y be the amount of compound B needed to make 1116 milliliters of the drug, where:
x + y = 1116 mL
Furthermore, we have:
x / y = 7 / 5
From the second equation, we solve for y:
y = 5/7 x
Substituting this into the first equation:
x + 5/7 x = 1116
Multiplying every term by 7 to clear the fraction, we get:
7x + 5x = 1116 × 7
12x = 7804
Dividing both sides by 12:
x = 7804 / 12
x = 650.333...
We round this to the nearest milliliter, since we are dealing with a measurable quantity. Therefore:
x = 650 mL
Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of disappearance of Cl2 is 4.24 × 10–2 M/s, what is the rate of formation of NO? 2 NO(g) + Cl2(g) → 2 NOCl(g)
Answer : The rate of formation of [tex]NOCl[/tex] is, [tex]8.48\times 10^{-2}M/s[/tex]
Explanation : Given,
Rate of disappearance of [tex]Cl_2[/tex] = [tex]4.24\times 10^{-2}M/s[/tex]
The given rate of reaction is,
[tex]2NO(g)+Cl_2(g)\rightarrow 2NOCl[/tex]
The expression for rate of reaction :
[tex]\text{Rate of disappearance}=-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[Cl_2]}{dt}[/tex]
[tex]\text{Rate of formation}=\frac{1}{2}\frac{d[NOCl]}{dt}[/tex]
From this we conclude that,
[tex]\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}[/tex]
[tex]\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}[/tex]
[tex]\frac{d[NOCl]}{dt}=2\times \frac{d[Cl_2]}{dt}[/tex]
Now put the value of rate of disappearance of [tex]Cl_2[/tex], we get:
[tex]\frac{d[NOCl]}{dt}=2\times (4.24\times 10^{-2}M/s)=8.48\times 10^{-2}M/s[/tex]
Therefore, the rate of formation of [tex]NOCl[/tex] is, [tex]8.48\times 10^{-2}M/s[/tex]
The rate of reaction decides the direction in which the reaction goes. It decides the rate of flow of conversion.
The correct rate of the reaction is [tex]8.48*10^{-2[/tex]
The rate of the reaction of a given element is as follows:-
Formation =[tex]-\frac{1}{2}\frac{d[NO]}{dt} =-\frac{1}{2} \frac{dCL_2}{dt}[/tex]Disappearance =[tex]\frac{1}{2}\frac{d[NOCL]}{dt}[/tex]After solving it from the equation,:-
[tex]\frac{d[NOCL]}{dt} = 2*\frac{d[CL_2]}{dt}[/tex]
After solving it, the value we get is
[tex]2 * 4.24*10^{-2}\\=8.48*10^{-2[/tex]
Hence, the correct answer is [tex]8.48*10^{-2[/tex]
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2-butanone is converted into 3-methyl-3-hexanol using a grignard reagent prepared from 1-bromopropane and magnesium metal in thf solution. List the procedural steps required to collect the alcohol product by microdistillation.
Answer:
Here's what I get.
Explanation:
At the end of the reaction you will have a solution of the alcohol in THF.
The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.
Transfer the solution to a conical vial. Add a boiling stone. Attach a Hickman head (shown below) and condenser. Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer. Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly. Use a Pasteur pipet to withdraw the THF as needed. When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).Rank the SN2 reaction rate of the following species, from fastest to slowest.
CH3CH2OH
CH3CH2I
CH3CH2Cl
Answer:
CH₃CH₂I > CH₃CH₂Cl > CH₃CH₂OH
Explanation:
SN₂ reaction -
It is the nucleophilic reaction bimolecular , where the nucleophile replaces the leaving group present in the reaction , this is a one step reaction .
The rate of the reaction depends on -
how good is the leaving group , as a good leaving group will immediately leave and the nucleophile can readily attack to form the product . how strong is the nucleophile , as the stronger nucleophile can efficiently attack replaces the leaving group to form the product .From the question ,
I is the best leaving group , then is Cl and least is OH .
Hence ,
The fastest to slowest rate of reaction is as follows -
CH₃CH₂I > CH₃CH₂Cl > CH₃CH₂OH
The order of reactivity in an SN2 reaction for the given species, from fastest to slowest, is CH3CH2I > CH3CH2Cl > CH3CH2OH. Iodine is a better leaving group than chlorine, and chlorine is better than hydroxyl.
Explanation:In SN2 reactions, the rate of reaction is largely determined by the leaving group. The better the leaving group, the faster the reaction proceeds. In the given species, the order of reactivity in an SN2 reaction, from fastest to slowest, is CH3CH2I > CH3CH2Cl > CH3CH2OH. This is because iodine (I) is a better leaving group than chlorine (Cl), and chlorine is a better leaving group than hydroxyl (OH).
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