Answer:
Short wavelength UV radiation ----Oxygen (O2 ), nitrogen (N2) , methane (CH4) , water (H2O) and carbondioxide (CO2).
longer wavelength radiation---- methane (CH4) , water (H2O) and carbondioxide (CO2)
Explanation:
Oxygen (O2 ), nitrogen (N2) , methane (CH4) , water (H2O) and carbondioxide (CO2) are the gases which are transparent for incoming short-wavelength UV radiation while methane (CH4) , water (H2O) and carbondioxide (CO2) are the greenhouse gases which absorb longer wavelength. When the solar radiation comes to the earth, it has short wavelength but after reflection it converts into longer wavelength which can be prevent by greenhouse gases and increases the temperature of atmosphere.
CH4, H2O and CO2 gases absorb longer wavelength terrestrial re-radiation.
UV radiation are short wavelength radiation that reach the earth from outer space. However, certain gases that compose the atmosphere are able to absorb longer wavelength terrestrial re-radiation.
The following gases absorb longer wavelength terrestrial re-radiation;
CH4H2OCO2The following gases are transparent to incoming short-wavelength UV radiation;
O2N2ArLearn more: https://brainly.com/question/6284546
2 points
How many significant figures are there in the measurement 2.30 x 10^5
mm ? *
2
3
4.
5
Answer:
3.
Explanation:
Hello,
In this case, considering the given number:
[tex]2.30x10^5[/tex]
We can see that it does not have zeros to left of the first nonzero digit which is 2, therefore, we just have to consider the 2, 3 and 0 as significant figures because the right-handed zeros are actually considered. Hence, such measurement has 3 significant figures.
At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. What is the composition in mole fractions of a solution that has a vapor pressure of 40. torr at 20°C?
Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]
where, x = mole fraction in solution
[tex]p^0[/tex] = pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2\\[/tex][tex]p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0[/tex]
[tex]x_{benzene}=x[/tex],
[tex]x_{toluene}=1-x_{benzene}=1-x[/tex],
[tex]p_{benzene}^0=75torr[/tex]
[tex]p_{toluene}^0=22torr[/tex]
[tex]p_{total}=40torr[/tex]
[tex]40=x\times 75+(1-x)\times 22[/tex]
[tex]x=0.34[/tex]
Thus (1-x0 = (1-0.34)=0.66
Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66
Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the C-side of L-Arg and L-Lys, but not on the C-side of D-Arg and D-Lys. Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the -side of L- and L-, but not on the -side of D- and D-. All the naturally occurring amino acids are D-amino acids, while active centres of serine proteases are nearly all L. All the naturally occurring amino acids are L-amino acids, while active centres of serine proteases are nearly all D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site. None of the above.
Answer:
D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site
Explanation:
Stereospecificity is the ability to distinguish between stereoisomers of of a particular compound. L- and D- structures of compounds in living organisms are usually present in only one form due to stereospecificity. For example, naturally occuring amino acids in proteins are usually present as L-isomers.
Since enzyme are proteins, their active sites are composed of L-amino acid and they show stereospecificity in the reactions they catalyze. In their binding sites, only substrates complementary in structure can bind in order for catalysis to proceed. Therefore, only amino acids in the L- configuration are complementary to the active site of enzymes.
In the case of serine proteases, The side chains of -Arg and D-Lys will not be positioned properly for binding at the binding site of serine proteases, therefore, no catalysis will occur. On the other hand, L-Arg and L-Lys can bind to the catalytic site of serine proteases since they are complementary fits to the active site of the enzymes.
Magnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.
Magnesium ions produce no emission of absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame
Answer:
Traces of sodium impurity
Explanation:
Metal ions are identified by a characteristic colour imparted to flame by the metal ion solution. Various metals have various colours which they impart to a flame.
The energy of a flame is not sufficient for the excitation of electrons of Mg to a higher energy level. As a result of this, Mg do not give any color in Bunsen flame.
However, a few tinges of yellow-orange colour which is characteristic of sodium metal do appear when magnesium ions are exposed to a flame as a consequence of traces of sodium impurity in the magnesium ion solution.
Identify if the following are chemical or physical properties: "CP" for
chemical property and "PP" for physical property
1. Oxygen is odorless and colorless
2. Copper turns green when exposed to the environment
3. The piece of metal is magnetic
4. The density of water is 1.0 g/cm3
5. Diamonds are a very hard substance
6. The tree is 8 meters high
Answer:
1.pp
2.cp
3.pp
4.pp
5.pp
6.pp
Which lists metric units, in order, from smallest to largest?
A) milligram, centigram, gram
B) kilogram, gram, centigram
C) kilogram, hectogram, decagram
D) decagram, hectogram, milligram
Answer:
milligram, centigram, decigram
Hope this answer correct :)
Answer:
C
Explanation:
hope it helps.
Draw the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.Draw the starting amine. Be sure to draw nonbonding electron pairs.
Draw the starting carboxylic acid. Be sure to draw nonbonding electron pairs.
Answer:
Explanation:
Amidation is a reaction with the formation of an amide. It involves a process where series of reaction between amine and carboxylic acid takes place.
The main objective of this question is to draw with the aid of a diagram.
the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.
But before that,
we are to draw the starting amine and to be sure to include the nonbonding electron pairs.
Also, to draw starting carboxylic acid and to be sure to include the nonbonding electron pairs.
The whole series of the diagrammatic expression for the amine formation can be found in the diagram attached below.
Which of the following can't undergo aldol condensation.
(CH3)3CCHO
CH3CH2CHO
PhCOCH3
CH3COCH3
Answer:
(CH3)3CCHO
Explanation:
Carbonyl compounds that contain alpha hydrogen atoms are the only ones that can undergo aldol condensation reaction.
Aldol condensation is one of the condensation reactions in organic chemistry that occurs when an enol or an enolate ion reacts with another carbonyl compound to yield a β-hydroxyaldehyde or β-hydroxyketone, this is immediately followed by the dehydration of the product to yield a conjugated enone.
(CH3)3CCHO lacks alpha hydrogen atoms hence it does not undergo the aldol condensation.
A coefficient is a constant by which a variable is multiplied. In chemistry, a coefficient is the number in front of a formula. (The 2 in 2CH4 is the coefficient.) The coefficients in a chemical equation represent the proportion of A) products to reactants. B) the amount of atoms in a solid phase to the amount in a liquid phase. C) mass to volume. D) molecules or atoms relative to other molecules or atoms.
Answer:
Below
Explanation:
The coefficients in a chemical reaction are called stochiometric coefficients
they represent the number of atoms or molecules they are multiplied to.
A chemical reaction has generally this form:
Reactants ⇒ productsAnswer:
Products to reactants
Which of the following conversions involves an oxidation of the underlined element?
A. H3PO3 → H2P03
B. 03 → 02
C. S03 → S042-
D. BrO2 → BrO-
E. CIO3 → ClO4
Of the following solutions, which has the greatest buffering capacity?
a. 0.365M HC2H3O2 and 0.497 M NaC2H3O2
b. 0.521 M HC2H3O2 and 0.217 M NaC2H3O2
c. 0.821 M HC2H3O2 and 0.713 M NaC2H3O2
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )[/tex]
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a. [tex]\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36[/tex]
b. [tex]\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417[/tex]
c. [tex]\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868[/tex]
d. [tex]\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959[/tex]
Therefore, the d. solution has the best buffering capacity.
Regards.
What volume of a 0.124 M KOH solution neutralizes 23.4 mL of 0.206 M HCl solution?
A) 15.9 mL
B) 38.9 mL
C) 31.8 mL
D) 1.00 × 104 mL
E) 5.00 × 102 mL
The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.
Neutralization reaction has resulted in the formation of salt and water with the reaction of acid and base.
In the neutralization reaction, the strength of the acid and base can be given by:
Molarity of acid [tex]\times[/tex] Volume of acid = Molarity of base [tex]\times[/tex] Volume of base
Given, the molarity of KOH base = 0.124 M
The volume of acid (HCl) = 23.4 ml
Molarity of acid (HCl) = 0.206 M.
Substituting the values:
0.206 [tex]\times[/tex] 23.4 = 0.124 × Volume of base (KOH)
Volume of base (KOH) = 38.874 ml.
Volume of KOH = 38.9 ml
The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.
For more information about neutralization reaction, refer to the link:
https://brainly.com/question/9691296
what mass of TiCl4 must react with an excess of water to produce 50.0g of TiO2 if the reaction has a 78.9% yield
Answer:
\large \boxed{\text{150 g TiCl}_{4}}
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 189.68 79.87
TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl
m/g: 50.0
To solve this stoichiometry problem, you must
Convert the actual yield to the theoretical yield Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂ Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄ Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄1. Theoretical yield of TiO₂
[tex]\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}[/tex]
2. Moles of TiO₂
[tex]\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}[/tex]
3, Moles of TiCl₄
The molar ratio is 1 mol TiO₂:1 mol TiCl₄.
[tex]\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}[/tex]
4. Mass of TiCl₄
[tex]\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}[/tex]
The mixing of which pair of reactants will result in a precipitation reaction?
A. AsI(aq) + NaOH(aq)
B. NaNO3(aq) + NH4Cl(aq)
C. K2SO4(aq) + Ba (NO3)2(aq)
D. HCl(aq) + Ca(OH)2(aq)
E. None of the above.
Answer:
B
Explanation:
2. Which of these is an extensive property?
a. Density
b. Melting point
Temperature
d. Volume
Volume ..........is the answer
9.57: How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4(s) at standard state conditions?
Answer:
459.6kj
Explanation:
the chemical equation =
3mn(s) + 2CO2(g) --- Mn3O4(s)
we have change in H to be -1378.83
we are required to find the amount of heat that is liberated for a mole of manganese.
3 moles of manganese = -1378.83
1 mole of manganese = ?
when we cross multiply, we will have a ratio that we will use to find this amount
1378.83/3 = 459.6kj
A student weighed 0.550 g of lithium chloride, LiCl, to use in a reaction. How many moles is this? a. 5.11 b. 42.39 c. 23.31 d. 77.08 e. 0.0130
Answer:
The answer is option eExplanation:
To find the number of moles we must first find the molar mass of LiCL
From the question
mass = 0.550g
M(Li) = 6 Cl = 35.5
Molar mass of LiCL = 6 + 35.5 = 41.5 g/mol
Then we use the formula
[tex]n = \frac{m}{M} [/tex]
where
n is the number of miles
m is the mass
M is the molar mass
So we have
[tex]n = \frac{0.550}{41.5} [/tex]
We have the final answer as
0.0130 molHope this helps you
If the toxic quantity is 1.5 gg of ethylene glycol per 1000 gg of body mass, what percentage of ethylene glycol is fatal
Answer:
The percentage of ethylene glycol that is fatal is 0.15 %
Explanation:
If the toxic quantity of ethylene glycol in a 1kg or 1000 g body weight is 1.5 g then the percentage of ethylene glycol that is fatal is
[tex]\frac{1.5}{1000}[/tex] ˣ 100 = 0.15%
Hence, the percentage of ethylene glycol that is toxic for any body weight is 0.15%. This percentage is very important in various aspects of science including drug discovery and food production/processing
The complete combustion of copper(I) sulphide is according to the following equation:
2Cu2S(s) + 302(g) →→2Cu2O(s) + 2SO2(g)
If the mass of Cu2S in the mixture is 14.0 g, calculate
(a) the number of molecules of oxygen gas reacted.
(b) the mass of SO2 gas produced.
(c)the volume of SO2 gas at STP
Answer:
a) 7.94 x 10²³ molecules, b) 5.62 g SO2, c) 1.97 L
Explanation:
a) We need to first convert Cu2S to moles. Since molar mass of Cu2S is 159.14 g/mol, 14.0 g = 0.0880 mol. Using molar ratios (3 mol O2/2 mol Cu2S, 0.0880 mol of Cu2S = 0.132 mol O2. Since 1 mol contains 6.02 x 10²³ molecules, 0.132 mol O2 = 7.94 x 10²² molecules O2.
b) Since the molar ratios of Cu2S to SO2 is 1:1, 0.0880 mol of Cu2S produces 0.0880 mol SO2. To convert mol to grams, we use the molar mass of SO2 (64.06 g/mol) to figure out that 0.0880 mol SO2 = 5.63 g SO2.
c) At STP, 1 mol occupies 22.4 liters. 0.0880 mol SO2 x 22.4 L/1 mol = 1.97 L
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Calculate the molality of a 15.0% by mass solution of MgCl 2 in H 2O. The density of this solution is 1.127 g/mL. Group of answer choices
Answer:
1.86 m
Explanation:
Step 1: Calculate the mass of solute and solvent in 100 grams of solution
We have a 15.0% by mass solution of MgCl₂ (solute) in H₂O (water). In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.
Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂
The molar mass of MgCl₂ is 95.21 g/mol.
[tex]15.0g \times \frac{1mol}{95.21 g} = 0.158mol[/tex]
Step 3: Convert the mass of water to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]85.0g \times \frac{1kg}{1,000g} = 0.0850kg[/tex]
Step 4: Calculate the molality of the solution
[tex]m = \frac{0.158mol}{0.0850kg} = 1.86m[/tex]
The molality is 1.86 m.
Given:
Mass solution of MgCl₂ in H₂O= 15%
Density= 1.127g/mL
To find:
Molality=?
Let's solve this question step by step:
Step 1: Calculate the mass of solute and solvent in 100 grams of solution
As it is given that 15.0% by mass solution of MgCl₂ (solute) in H₂O (water).
In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.
Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂
The molar mass of MgCl₂ is 95.21 g/mol.
Number of moles is given mass divided by the molar mass.
[tex]\text{Number of moles}= \frac{15g}{95.21g/mol} =0.158mol[/tex]
Step 3: Convert the mass of water to kilograms.
As we know, 1 kg = 1,000 g.
[tex]=\frac{85}{1000} =0.085kg[/tex]
Step 4: Calculate the molality of the solution.
Molality is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.
[tex]\text{Molality}=\frac{0.158\text{mol}}{0.085\tetx{kg}} =1.86 m[/tex]
Learn more:
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What is the speed 0.36
km/h in cm/s
Answer:
It's 10 cenimeters per second
Which of the following has the longest half-life?
a. 0-19
b. H-3
c. Zn-71
d. Rb-85
Answer:
b. H-3
Explanation:
The half life of the following isotopes is given as;
a. 0-19
26.88 seconds
b. H-3
12 years
c. Zn-71
2.4 minutes
d. Rb-85
This is a stable isotope. It does not decomposes.
Comparing the half lives, the isotope with the longest half life is Hydrogen - 3. With half life of 12 years.
In an electrolytic cell, the magnitude of the standard cell potential is the:______
a. minimum voltage that must be supplied for a redox reaction to occur
b. maximum voltage that willallow the redox reaction to occur
c. always equal to Eanode - Ecathode
d. none of the above
Answer:
a. minimum voltage that must be supplied for a redox reaction to occur
c. always equal to Eanode - Ecathode
Explanation:
In an electrolytic cell; The electromotive force(the maximum standard potential difference) of the cell formed by the system is defined as the standard electrode potential of the right handed electrode minus the standard electrode potential of the left hand electrode. (i.e [tex]\mathbf{E^{\theta}_{cell}=E_{anode} - E_{cathode}}[/tex] )
As we all known that the process by which chemical energy is being converted to electrical energy is called the Electrochemical cell. It consists of two half cells , an oxidation half cell reaction and a reduction half-cell reaction.The overall redox reaction results in a flow of electrons in an electric current which is produced by a minimum voltage.
Therefore, option a and c are both correct.
what is 31/50000 If the toxic quantity is 1.5 g of ethylene glycol per 1000 g of body mass, what percentage of ethylene glycol is fatal
Answer:
[tex]\%m=0.15\%[/tex]
Explanation:
Hello,
In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed with the shown below formula:
[tex]\%m=\frac{m_{ethylene \ glycol}}{m_{ethylene \ glycol}+m_{body\ mass}}*100\%[/tex]
We can use the given masses to obtain:
[tex]\%m=\frac{1.5g}{1.5g+1000g}*100\%\\ \\\%m=0.15\%[/tex]
Best regards.
Which type(s) of matter would need to be separated by chemical methods?
Answer:
solids, liquids, gases, or plasma
Explanation:
In chemistry, a chemical substance is a form of matter that has constant chemical composition and characteristic properties. It cannot be separated into components without breaking chemical bonds. Chemical substances can be solids, liquids, gases, or plasma
What mass (in grams) of aspirin (C₉H₈O₄) is produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield from the reaction below? C₇H₆O₃ (s) + C₄H₆O₃ (s) → C₉H₈O₄ (s) + HC₂H₃O₂ (aq).
Answer:
71.3 g
Explanation:
molar mass of C₇H₆O₃ = 138.13 g
molar mass of [tex]C_{9}H_{8}O_{4}[/tex] = 180.17 g
find the moles of reactant: C₇H₆O₃
C₇H₆O₃ that reacted = mass/molar mass = 57.6 g C₇H₆O₃ / 138.13 g C₇H₆O₃
= 0.417 mol [tex]C_{7}H_{6}O_{3}[/tex]
From the reaction equation, 1 mole of C₇H₆O₃ yields one mole of aspirin
find theoretical yield of aspirin:
0.417 mol C7H6O3 x 180.17 g C9H8O4 / 1 mol C9H8O4[tex]\frac{0.417 mol C_{7}H_{6}O_{3}}{} x \frac{180.17 g C_{9}H_{8}O_{4}}{1 mol C_{9}H_{8}O_{4}}[/tex]
= 75.1 g C9H8O4
Actual yield= % yield × theoretical yield/100
Actual yield = 95.0 × 75.1/100
Actual yield = 71.3 g
Taking into account the reaction stoichiometry and actual yield, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.
The balanced reaction is:
22 C₇H₆O₃ (s) + 32 C₄H₆O₃ (s) → 24 C₉H₈O₄ (s) + 33 HC₂H₃O₂ (aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₇H₆O₃: 22 moles C₄H₆O₃: 32 moles C₉H₈O₄: 24 moles HC₂H₃O₂: 33 molesThe mass molar of each compound is:
C₇H₆O₃: 138 g/moleC₄H₆O₃: 102 g/mole C₉H₈O₄: 180 g/moleHC₂H₃O₂: 60 g/moleThen by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:
C₇H₆O₃: 22 moles× 138 g/mole= 3036 gramsC₄H₆O₃: 32 moles× 102 g/mole= 3264 grams C₉H₈O₄: 24 moles× 180 g/mole= 4320 gramsHC₂H₃O₂: 33 moles× 60 g/mole= 1980 gramsThen you can apply the following rule of three: if by stoichiometry 3036 grams of C₇H₆O₃ produce 3264 grams of C₉H₈O₄, 57.6 grams of C₇H₆O₃ will produce how many mass of C₉H₈O₄?
[tex]mass of C_{9} H_{8} O_{4} =\frac{57.6 grams of C_{7} H_{6} O_{3} x3264 gramsof C_{9} H_{8} O_{4} }{3036 grams of C_{7} H_{6} O_{3}}[/tex]
mass of C₉H₈O₄= 60.03 grams
On the other side, actual yield is the amount of product actually obtained from a reaction. Assuming 95.0% yield and considering the previously calculated mass the maximum amount of product that can be produced in the reaction, you can apply the following rule of three: If 100% equals 60.03 grams of the compound produced, 95% equals how much mass of the compound?
[tex]mass of C_{9} H_{8} O_{4} =\frac{95 percentx60.03gramsof C_{9} H_{8} O_{4} }{100 percent}[/tex]
mass of C₉H₈O₄= 57.0285 grams
Finally, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.
Learn more:
brainly.com/question/16487206?referrer=searchResults brainly.com/question/14446695?referrer=searchResults brainly.com/question/11564309?referrer=searchResults brainly.com/question/4025026?referrer=searchResults brainly.com/question/18650135?referrer=searchResultsThere is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+.
A. Write an equation for the half-reaction occurring at each electrode. Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.
B. What minimum voltage is necessary to drive the reaction? Vmin =
Answer:
See explanation
Explanation:
Anode;
Sn(s) ------> Sn^2+(aq) + 2e
Cathode;
Mn^2+(aq) + 2e ------> Mn(s)
The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;
E°cell= E°cathode - E°anode
E°cell= -1.19 - (-0.14)
E°cell= -1.05 V
Use the balanced chemical equation below. How many grams of the product are formed when 2.34 g of sulfur is completely reacted with fluorine? S8 + 16F2(g) → 8SF4
Answer:
7.89 g
Explanation:
Step 1: Write the balanced equation
S₈ + 16 F₂(g) → 8 SF₄
Step 2: Calculate the moles corresponding to 2.34 g of S₈
The molar mass of S₈ is 256.52 g/mol.
[tex]2.34g \times \frac{1mol}{256.52g} = 9.12 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of SF₄ produced from 9.12 × 10⁻³ mol of S₈
The molar ratio of S₈ to SF₄ is 1:8. The moles of SF₄ produced are 8/1 × 9.12 × 10⁻³ mol = 0.0730 mol
Step 4: Calculate the mass corresponding to 0.0730 moles of SF₄
The molar mass of SF₄ is 108.07 g/mol.
[tex]0.0730 mol \times \frac{108.07 g}{mol} = 7.89 g[/tex]
How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?
Answer:
0.088 mole of Al.
Explanation:
First, we shall determine the number of mole in 23.6 g of AlBr₃.
This is illustrated below:
Mass of AlBr₃ = 23.6 g
Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol
Mole of AlBr₃ =.?
Mole = mass/Molar mass
Mole of AlBr₃ = 23.6 / 267
Mole of AlBr₃ = 0.088 mol
Next, we shall writing the balanced equation for the reaction.
This is given below:
2Al(s) + 3Br₂(l) → 2AlBr₃(s)
From the balanced equation above,
2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.
Finally, we shall determine the number of mole of Al needed for the reaction as follow:
From the balanced equation above,
2 moles of Al reacted to 2 moles AlBr₃.
Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.
0.085 moles of Al are required to form 23.6 g of AlBr₃.
Let's consider the following balanced equation for the synthesis reaction of AlBr₃.
2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)
First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).
[tex]23.6 g \times \frac{1mol}{266.69g} = 0.0885 mol[/tex]
The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885 moles of AlBr₃ are:
[tex]0.0885molAlBr_3 \times \frac{2molAl}{2molAlBr_3} = 0.0885molAl[/tex]
0.085 moles of Al are required to form 23.6 g of AlBr₃.
You can learn more about stoichiometry here: https://brainly.com/question/22288091
You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the following statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.
Answer: option (a) is the correct answer
Explanation:
The complete questions says;
You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the following statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.
The Answer:
(a). This reaction will become spontaneous in forward direction at some temperature below 298K
Explanation: First of all, we can acknowledge that the reaction seen here is an exothermic one, i.e energy is released in the process outwardly and as a result temperature is reduced during this process of energy loss.
Having understood that scenario, say we reduce it's temperature by ourself than forward reaction favors and after reaching at particular temperature, therefore we can confirm this to be a spontaneous reaction.
Let us use this to confirm what we have been saying.
Given;
ΔG = ΔH - TΔS
here ΔH is negative
it is non spontaneous, which means ΔG is positive so we continuously decraeses it's temperature than at a particular temperature.
The Entropy change becomes positive and reaction becomes spontaneous and ΔG become negative.
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