Answer: The net ionic equation is [tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex](NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)[/tex]
The equation can be written in terms of ions as:
[tex]2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)[/tex]
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The ions which are present on both the sides of the equation are ammonium and chlorate ions and hence are not involved in net ionic equation.
Hence, the net ionic equation is :
[tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]
g Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.
Answer:
1.42 × 10⁻⁹ M
Explanation:
Step 1: Given data
Concentration of Ba²⁺ ([Ba²⁺]): 0.0758 MSolubility product constant of BaSO₄ at 25°C: 1.08 × 10⁻¹⁰Step 2: Write the reaction for the solution of BaSO₄
BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)
Step 3: Calculate the concentration of SO₄²⁻
We will use the following expression.
Ksp = 1.08 × 10⁻¹⁰ = [Ba²⁺] × [SO₄²⁻]
[SO₄²⁻] = 1.08 × 10⁻¹⁰ / [Ba²⁺] = 1.08 × 10⁻¹⁰ / 0.0758 = 1.42 × 10⁻⁹ M
What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.28 M? Express your answer to two significant figures and include the appropriate units.
Answer:
0.56M
Explanation:
Molar concentration is defined as the ratio between moles of solute and volume in liters of solution.
In a 0.28M H₂ there are 0.28moles of H₂ per liter of solution.
Now, in 1 molecule of H₂ there are 2 atoms of H. Following this idea, in 0.28 moles of H₂ there are 0.28*2 = 0.56 moles of H atoms.
Thus, molar concentration of H atoms in a 0.28M H₂ is 0.56M
If the heat of combustion for a specific compound is −1380.0 kJ/mol and its molar mass is 44.53 g/mol, how many grams of this compound must you burn to release 822.00 kJ of heat?mass:g
Answer:
Mass = 26.53 g
Explanation:
Heat of combustion = −1380.0 kJ/mol
This means 1 mol of the compound releases 1380 kJ
Molar mass = 44.53 g/mol
This means 1 mol of the compound has a mass of 44.53 g
How many grams would release 822kJ..?
First, we have to obtain the number of moles
1 mol = 1380
x = 822
x = 0.5957 moles
Moles = Mass / Molar mass
Mass = Molar mass * moles
Mass = 44.53 * 0.5957
Mass = 26.53 g
the equilibrium concentrations were found to be [H2O]=0.250 M , [H2]=0.600 M , and [O2]=0.800 M . What is the equilibrium constant for this reaction?
Answer:
Keq = 0.217
Explanation:
Let's determine the equilibrium reaction.
In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.
2H₂(g) + O₂(g) ⇄ 2H₂O (g) Keq
Let's make the expression for the equilibrium constant
Products / Reactants
We elevate the concentrations, to the stoichiometry coefficients.
Keq = [H₂O]² / [O₂] . [H₂]²
Keq = 0.250² / 0.8 . 0.6² = 0.217
g If attempting to dissolve both silver bromide and silver chloride in aqueous solution through complex ion formation, which data will be the most relevant?
Answer:
Kf
Explanation:
The stability constant Kf of a given complex specie is an equilibrium constant that represents the formation of that particular complex specie in solution. It measures the strength of the interaction between the ligands and metal that form the particular complex specie. The magnitude of Kf shows how easily a complex specie is formed in solution.
Hence if I want to dissolve the bromides or chlorides of silver which are ordinarily insoluble in water by means of complex formation, the magnitude of the stability constant for each particular complex specie is important as it gives information regarding the thermodynamic feasibility of the process.
The reaction, 2 NO(g) + O2(g) → 2 NO2(g), was found to be first order in each of the two reactants and second order overall. The rate law is therefore
Answer:
[tex]r=k[NO][O_2][/tex]
Explanation:
Hello,
In this case, rate laws allows us to compute how fast a chemical reaction is carried out by means of the change in the concentration of the species affecting the rate. In such a way, since the statement says that the reaction was found to be first order to both nitrogen monoxide and oxygen, it means that their concentrations are powered to first power by separated. It also implies that the overall order is second-order since the specific orders are added (powers properties). Therefore, the rate law is:
[tex]r=k[NO][O_2][/tex]
Whereas k is the rate constant and we find the concentration of the reactants to the first power each one.
Best regards.
Which of the following is the correct equation for the reaction below?
A. CH (g) + O2 (g) CO (g) + H2O (g)
B. CH (g) + 2O (g) CO (g) + 2HO (g)
C. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
D. CH4 (g) + 2O2 (g) CO2 (g) + H2O (g)
Answer:
(c) is the correct answer
CH4(g)+2O2(g)⟶CO2(g)+2H2O(l)
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Some metal oxides, such as Sc2O3, do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc2O3 to react when the solution becomes acidic or when it becomes basic?
Write a balanced chemical equation to support your answer.
Answer:
[tex]Sc_2O_3[/tex] reacts with an acidic solution
Explanation:
Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is a compound that contains only two elements, one of which is oxygen .
The objective of this question is to Write a balanced chemical equation to support your answer.
The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:
Assuming the acidic solution to be HCl
[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]
The ionic equation :
[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]
In this experiment, you will analyze your sample by TLC by spotting pure benzophenone and your product and eluting with 5:1 hexanes/ethyl acetate. Choose the statement that BEST describes what you should observe for a successful experiment.
A) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value.
B) The benzophenone in the standard and reaction samples should travel the least and will have the same Rf value. No unreacted benzhydrol should be observed at a higher Rf value.
C) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. Unreacted benzhydrol should be observed at a lower Rf value.
D) The benzophenone in the standard and reaction samples should travel the farthest and will have the different Rf values. No unreacted benzhydrol should be observed at a lower Rf value.
Answer:
The correct answer is statement A.
Explanation:
In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.
For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.
The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.
What is TLC ?In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.
The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.
TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.
Thus, option A is correct.
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Which one of the following reactions must be carried out in an electrolytic cell rather than in an electrochemical cell?
a. Zn2+ + Ca → Zn + Ca2+
b. Al3+ + 3Br– → Al + (3/2)Br2
c. 2Al + 3Fe2+ → 2Al3+ + 3Fe
d. H2 + I2(s) → 2H+ + 2I–
Answer:
b. Al3+ + 3Br– → Al + (3/2)Br2
Explanation:
If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.
However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.
b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]
Electrolytic Cell v/s Electrochemical Cell:
Electrochemical cells convert chemical energy into electrical energy or vice versa while an electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy.Electrochemical cells consist of a cathode (+) and an anode(-) while Electrolytic cells consist of a positively charged anode and a negatively charged cathode.Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex] is carried out in an electrolytic cell rather than in an electrochemical cell.
As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells
one is a reduction half-cell, the other is an oxidation half-cell.In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.
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Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
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 )
interpret the electron configuration
Answer:
Ca for calcium
20 electrons
2-2s electron
A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0630 M in Pb(NO3)2 and 0.0103 M in NaBr. What is the value of Q for the insoluble product? Express the reaction quotient to three significant figures.
Answer:
Q = 6.68x10⁻⁶
Explanation:
The initial solutions are Pb(NO₃)₂ and NaBr. As sodium and nitrates salts are soluble, the insoluble product must be the formed from the other ions, PbBr₂.
The soluble product equilibrium is written as:
PbBr₂(s) ⇄ Pb²⁺(aq) + 2Br⁻(aq)
Where Q is defined as:
Q = [Pb²⁺] [Br⁻]²
As:
[Pb²⁺] = 0.0630M
[Br⁻] = 0.0103M
Q = [Pb²⁺] [Br⁻]²
Q = [0.0630M] [0.0103M]²
Q = 6.68x10⁻⁶Which is the correct way to write 602,200,000,000 ,000,000,000,000 in scientific notation
Answer:
6.022 × 10^23
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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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Calculate the ratio of moles of H2O to moles of anhydrous CuSO4 in CuSO4.
1H2O, CuSO4, 3H2O, CuSO4, 5H2O, CuSO4, 7H2O, and CuSO4, 9H2O.
(Mol wt. of H2O- 18.02g; Mol wt. of CuSO4 - 159.61g).
Answer:
CuSO₄.H₂O = 1
CuSO₄.3H₂O = 3
CuSO₄.5H₂O = 5
CuSO₄7.H₂O = 7
CuSO₄.9H₂O = 9
Explanation:
Some salts as CuSO₄ are presented in the hydratated form to give some stability in their uses.
Ratio of moles represents moles of H₂O / moles of CuSO₄.
In CuSO₄.H₂O you have 1 mole of water per mole of CuSO₄, Ratio is 1/1 = 1.
For CuSO₄.3H₂O are 3 moles of water per mole of CuSO₄. Ratio is 3/1 = 3
For CuSO₄.5H₂O are 5 moles of water per mole of CuSO₄. Ratio is 5/1 = 5
For CuSO₄.7H₂O are 7 moles of water per mole of CuSO₄. Ratio is 7/1 = 7
For CuSO₄.9H₂O are 9 moles of water per mole of CuSO₄. Ratio is 9/1 = 9
At what pressure would 11.1 moles of a gas occupy 44.8 L at 300 K?
Answer:
[tex]P=6.10atm[/tex]
Explanation:
Hello,
In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:
[tex]PV=nRT[/tex]
Thus, since we are asked to compute the pressure y simply solve for it as follows:
[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]
Best regards.
consider the following reaction
N2 + 3H2=2NH3
if 30 dm3 of each reactant react together then the volume of mixture at the end is
a 10 dm3
b 20 dm3
c 30 dm3
d 40 dm3
Answer:
d, 40 dm3.
Explanation:
According to Avogadro's law, the mole ratio of chemicals in a reaction is equal to the ratio of volumes of chemicals reacted (for gas).
From the equation, the mole ratio of N2 : H2 : NH3 = 1 : 3 : 2, meaning 1 mole of N2 reacts completely with 3 moles of H2 to give 2 moles of NH3, the ratio of volume required is also equal to 1 : 3 : 2.
Considering both N2 and H2 have 30dm3 of volume, but 1 mole of N2 reacts completely with 3 moles of H2, so we can see H2 is limiting while N2 is in excess. Using the ratio, we can deduce that 10dm3 equals to 1 in ratio (because 3 moles ratio = 30dm3).
With that being said, all H2 has reacted, meaning there's no volume of H2 left. 2 moles of NH3 is produced, meaning the volume of NH3 produced = 10 x 2 = 20 dm3. (using the ratio again)
1 mole of N2 has reacted, meaning from the 30dm3, only 10 dm3 has reacted. This also indicate that 20 dm3 of N2 has not been reacted.
So at the end, the mixture contains 20dm3 of NH3, and 20 dm3 of unreacted N2. Hence, the answer is d, 40 dm3.
what is radiologist
Radiologists are medical doctors that treat injuries using medical imaging (radiology)
Answer:
a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.
Explanation:
What's the electron configuration of a Ca+2 ion?
A. [Kr]
B. [Ar]
C. [Ne]
D. [He]
Answer:
B
Explanation:
The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.
The element you hit is Argon.
The answer is B
Answer:
B ar
Explanation:
pen foster answer
Given the initial rate data for the reaction being A + B + C --> D determine the rate expression for the reaction and the (k) rate constant. (The units of [A] [B] and [C] are all moles/liter and the units of IRR is moles/liter seconds). If the [A]=[B]=[C]=.30M, what would the IRR be? [A] [B] [C] IRR 0.20 0.10 0.40 .20 0.40 0.20 0.20 1.60 0.20 0.10 0.20 .20 0.20 0.20 0.20 .80
Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
Select the correct answer. A certain reaction has this form: aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the half-life of this reaction? A. 23.33 seconds B. 0.043 minutes C. 0.0043 seconds D. 23.33 minutes E. 1680 minutes
Answer:
[tex]\large \boxed{\text{D. 23.34 min}}[/tex]
Explanation:
1. Find the order of reaction
Use information from the graph to find the order.
If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.
2. Find the half-life
[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]
The half life of the reaction is 23.33 minutes.
We know that for a first order reaction;
ln[A]t = ln[A]o - kt
A plot of ln[A]t against time (t) will yield a straight line graph with a slope of -k.
From the question, the slope is -2.97 x 10-2 min-1.
So, -2.97 x 10-2 min-1 = - k
k = 2.97 x 10-2 min-1
The half life of a first order reaction is obtained from;
t1/2 = 0.693/k
t1/2 = 0.693/2.97 x 10-2 min-1
t1/2 = 23.33 minutes
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It is known that the kinetics of recrystallization for some alloy obey the Avrami equation andthat the value of n in the exponential is 2.5. If, at some temperature, the fraction recrystallized is0.40 after 200 min, determine the rate of recrystallization at this temperature.
Answer:
rate of recrystallization = 4.99 × 10⁻³ min⁻¹
Explanation:
For Avrami equation:
[tex]y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ k = \dfrac{-In(1-y)}{t^n}[/tex]
To calculate the value of k which is a dependent variable for the above equation ; we have:
[tex]k = \dfrac{-In(1-0.40)}{200^{2.5}}[/tex]
[tex]k = 9.030 \times 10 ^{-7}[/tex]
The time needed for 50% transformation can be determined as follows:
[tex]y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ t =[ \dfrac{-In(1-y)}{k}]^{^{1/n}}[/tex]
[tex]t_{0.5} =[ \dfrac{-In(1-0.4)}{9.030 \times 10^{-7}}]^{^{1/2.5}}[/tex]
= 200.00183 min
The rate of reaction for Avrami equation is:
[tex]rate = \dfrac{1}{t_{0.5}}[/tex]
[tex]rate = \dfrac{1}{200.00183}[/tex]
rate = 0.00499 / min
rate of recrystallization = 4.99 × 10⁻³ min⁻¹
Need help finding major products
Answer:
Explanation:
RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻
C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅
C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻
In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺
CH₃CH₂CH₂CH₂CH₂⁺ ⇒ CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )
CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃
Hence option D is correct .
b )
In the second case carbocation produced is
CH₃CH₂CH₂CH⁺CH₃
CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃
The product formed is same as in case of first
Option B is correct
Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm L / mole K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.
Answer:
1. 0.98 atm
Explanation:
The following data were obtained from the question:
Mass of unknown liquid (m) = 0.3175 g
Temperature (T) = 99 °C
Pressure (P) = 748.2 mmHg
Volume (V) = 145.0 mL
Gas constant (R) = 0.08206 atm.L/Kmol
1. Determination of the pressure in atm.
760 mmHg = 1 atm
Therefore,
748.2 mmHg = 748.2/760 = 0.98 atm
Therefore, the pressure in atm is 0.98 atm.
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=3.0×10−8 M?
Answer:
[tex][OH^-]=3.33x10^{-7}M[/tex]
Explanation:
Hello,
In this case, for the given concentration of hydronium, we can compute the pH as shown below:
[tex]pH=-log([H^+])=-log(3.0x10^{-8})=7.52[/tex]
Now, given the relationship between pH and pOH we can compute the pOH which is directly related with the concentration of hydroxyl in the solution:
[tex]pOH=14-pH=14-7.52=6.48[/tex]
Then, the concentration of hydroxyl turns out:
[tex][OH^-]=10^{-pOH}=10^{-6.48}[/tex]
[tex][OH^-]=3.33x10^{-7}M[/tex]
Best regards.
The overall process that uptakes energy-poor molecules (CO2 and H2O) from their reservoirs in nature and converts them into energy-rich molecules is
Answer:
Photosynthesis
Explanation:
Photosynthesis is a process by which photosynthetic organisms use the energy of captured sunlight to convert energy-poor molecules such as carbon (iv), CO₂ and water (H₂O) into energy-rich organic molecules sch as carbohydrates e.g. glucose.
Photosynthesis occurs in a variety of bacteria and algae as well in vascular plants. The overall equation for the reaction of photosynthesis is as follows:
CO₂ + H₂O + light-------> (CH₂O) + O₂
It is a redox reaction in which water donates electrons (as hydrogen) for the reduction of CO₂ to carbohydrate, (CH₂O).
Carbohydrates are energy-rich molecules which serves as energy sources for many living organisms.
All of the following reactions can be described as displacement reactions except:____________.
a.) Zn(s) + FeCl2(aq) → ZnCl2(aq) + Fe(s).
b.) C6H6(l) + Cl2(g) → C6H5Cl(l) + HCl(g).
c.) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g).
d.) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s).
e.) CuSO4(aq) + Fe(s) → Cu(s) + FeSO4(aq).
Answer:
b
Explanation:
The reaction that is not a displacement reaction from all the options is [tex]C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}[/tex]
In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.
Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.
Correct option = b.
The reaction given in Option A is not a displacement reaction. In Displacement reaction functional group of one reactant is replaced by the functional group of the another reactant.
Displacement reaction:
In this reaction functional group of one reactant is replaced by the functional group of the another reactant.
[tex]\bold { Zn(s) + FeCl_2(aq) \rightarrow ZnCl_2(aq) + Fe(s).}[/tex]
In the above reaction Zinc does not any functional group to exchange with iron chloride.
Therefore, the reaction given in Option A is not a displacement reaction.
To know more about Displacement reaction,
https://brainly.com/question/4339209
Why was it important to establish the Clean Air Act?
Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.
Explanation:
Answer:
Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.
