Answer:
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Explanation:
Given that:
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.
If the radiation wavelength is 12.5 cm,
density of water = 1g/cm³
volume of the container = 0.250 L = 250 cm³
density = mass/volume
mass of the water = density × volume
mass of the water = 1g/cm³ × 250 cm³
mass of the water = 250 g
specific heat capacity of water = 4.182 J/g°C
The change in temperature was from 20.0° C to 99° C
ΔT =( 99 -20.0)° C
ΔT = 79.0° C
The heat absorbed in the process is calculated by using the formula,
q = mcΔT
q = 250 g × 4.182 J/g°C × 79.0° C
q = 82594.5 Joules
Recall that the radiation wavelength λ = 12.5 cm = 0.125 m
The amount of energy of one photon of the radiation wavelength is determined by using the formula:
E = hv
since v = c/λ
E = hc/λ
where;
h = Planck's constant = 6.626 × 10⁻³⁴ J.s
c = velocity of light = 3.0 × 10⁸ m/s
∴
E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m
E = 1.59024⁻²⁴ Joules
The total photons required for this radiation = total heat energy/energy of radiation
The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules
The total photons required for this radiation = 5.1938 × 10²⁸ photons
For the following reaction, 3.76 grams of iron are mixed with excess oxygen gas . The reaction yields 4.29 grams of iron(II) oxide . iron ( s ) oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide
Answer:
4.84g of FeO is the theoretical yield
Explanation:
The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:
2Fe(s) + O₂(g) → 2FeO
Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.
To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:
Moles Fe (Molar mass: 55.845g/mol)
Using the molar mass of the compound we can convert grams to moles, thus:
3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe
Moles and mass of FeO
As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.
Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:
0.0673 moles FeO × (71.844g / mol) =
4.84g of FeO is the theoretical yieldWhat is the primary source of energy in most living communities?
Answer:
The sun
Explanation:
The sun is the primary source of energy in most living communities. The producers or the green plants that prepare their own food by the use of sunlight and other natural resources. Carbon dioxide, water, and other minerals are used by the plants to make their food in the presence of chlorophyll. Plants are then consumed by the consumers. This chain helps in forming the food chain and the food web.
Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.
Answer:
Electrical current drives nonspontaneous reactions in electrolytic cells.
Explanation:
Electrochemical cells are cells that produce electrical energy from chemical energy.
There are two types of electrochemical cells; voltaic cells and electrolytic cells.
A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.
Answer:
electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.
Explanation:
Answer via Educere/ Founder's Education
If the theoretical yield of a reaction is 332.5 g and the percent yield for the reaction is 38 percent, what's the actual yield of product in grams? \
A. 8.74 g
B. 12616 g
C. 116.3 g
D. 126.4 g
Answer: D - 126.4g
Explanation:
% Yield = Actual Yield/Theoretical Yield
38% = Actual Yield/332.5
38/100 = Actual Yield/332.5
(.38)(332.5) = 126.35 g = 126.4 g Actual Yield
Answer:
is D. the correct answer
Explanation:
I'm not sure if it is. Please let me know if I'm mistaking.
Draw the major organic product that is expected when cyclopentanecarboxylic acid is treated with each of the following reagents:
a. NaOH
b. [H+]
Answer:
a. Sodium cyclopentanecarboxylate
b. No reaction
Explanation:
In this case, in the cyclopentanecarboxylic acid we have a carboxylic acid functional group. Therefore we have an "acid". The acids by definition have the ability to produce hydronium ions ([tex]H^+[/tex]).
With this in mind, for molecule a. we will have an acid-base reaction, because NaOH is a base. When we put together an acid and a base we will have as products a salt and water. In this case, the products are Sodium cyclopentanecarboxylate (the salt) and water.
For the second molecule, we have the hydronium ion ([tex]H^+[/tex]). This ion can not react with an acid. Because, the acid will produce the hydronium ion also, so a reaction between these compounds is not possible.
See figure 1
I hope it helps!
The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years
Answer:
68%
Explanation:
Since we need a percentage we can use any number we want for our initial value.
5(1/2)^900/1620 = 3.40
(3.40 / 5)*100 = 68%
To make sure lets use a different initial amount
1(1/2)^900/1620 = 0.68
(0.68/1) * 100 = 68%
To solve this question, we'll assume the initial amount of radium-226 to be 1.
Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:
Step 1Determination of the number of half-lives that has elapsed.
Half-life (t½) = 1620 years
Time (t) = 900 years
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]
Step 2:Determination of the amount remaining
Initial amount (N₀) = 1
Number of half-lives (n) = 5/9
Amount remaining (N) =?[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]
N = 0.68Step 3Determination of the percentage remaining.
Initial amount (N₀) = 1
Amount remaining (N) = 0.68
Percentage remaining =?Percentage remaining = N/N₀ × 100
Percentage remaining = 0.68/1 × 100
Percentage remaining = 68%Therefore, the percentage amount of radium-226 that remains after 900 years is 68%
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The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbonding electron pair each nitrogen has one nonbonding electron pair each nitrogen has two nonbonding electron pairs a nitrogen-nitrogen triple bond
Answer:
one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.
Explanation:
H-N=N-H
Find the pH of these buffer solutions using the information provided: 1L solution containing 80g of lactic acid (MW
Answer:
pH of the solution is 2.0
Explanation:
The lactic acid is a weak acid that is in equilibrium with water as follows:
Lactic acid + H2O ⇄ Lactate + H₃O⁺
And Ka for lactic acid: 1.38x10⁻⁴
Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]
Initial concentration of lactic acid is (MW: 112.06g/mol):
80g * (1mol / 112.06g) / 1L = 0.714M
The equilibrium concentration of the species in the equilibrium are:
[Lactate] = X
[H₃O⁺] = X
[Lactic acid] = 0.714-X
Replacing in Ka expression:
1.38x10⁻⁴ = [X] [X] / [0.714-X]
9.8532x10⁻⁵ - 1.38x10⁻⁴X = X²
9.8532x10⁻⁵ - 1.38x10⁻⁴X - X² = 0
Solving for X:
X = -1.0x10⁻². False solution, there is no negative concentrations
X = 9.86x10⁻³M. Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 9.86x10⁻³M
and pH = -log [H₃O⁺] = -log 9.86x10⁻³M
pH = 2.0
pH of the solution is 2.0Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.
A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4
Answer:
B. NH3 < PH3< CH4
Explanation:
Hello,
In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:
B. NH3 < PH3< CH4
Best regards.
Susan was investigating which glue would make the strongest craft stick tower. She
tested rubber cement, Elmer's glue-all, and Super Glue. She will test the strength of the
towers with books as the weight.
What is the independent variable?
Answer:
glue that would make the strongest craft stick tower.
Explanation:
Independent variable: In statistics and research methods, the term "independent variable" is determined as a variable that is being changed, controlled, or altered in an experiment or research by the researcher or the experimenter to see its effect on DV or dependent variable. However, it is said that independent variable directly effect the dependent variable.
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answerβ-1,4- and α-1,6-glycosidicβ-1,4-glycosidicgalactosean unbranchedglucosea branchedfructoseα-1,6-glycosidicAmylose is ......... polymer of ....... units joined by ........ bonds. Amylopectin is ....... polymer of .......units joined by ........ bonds.
The words given are not clear, so the clear question is as follows:
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:
A. β-1,4- and α-1,6-glycosidic
B. α-1,4-glycosidic
C. α-1,4-galactose
D. an unbranched glucose
E. a branched fructose
F. α-1,6-glycosidic
Amylose is ......... polymer of ....... units joined by ........ bonds.
Amylopectin is ....... polymer of .......units joined by ........ bonds.
Answer:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
Explanation:
Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.
Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.
Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.
Hence, the correct answers in the sequential order are:
Amylose:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
Amylopectin:
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
Methyl iodide reacts irreversibly with azide ion with rate = k[CH3I][N3–]. CH3I(aq) + N3–(aq) → CH3N3(aq) + I–(aq) The reaction is carried out with an initial concentration of CH3I of 0.01 M. Which statement about the reaction is correct?
Answer:
(D) The reaction cannot take place in a single elementary step
Explanation:
Statements are:
(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I].
B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M.
(C) The reaction rate is significantly smaller if excess I- is added to the solution.
(D) The reaction cannot take place in a single elementary step.
The rate of the reaction is:
rate = k[CH3I][N3–].
That means rate depends of concentration of CH₃I as much as N₃⁻ concentration
(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. FALSE. The reaction rate depends of N₃⁻ as much as CH₃I
B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. FALSE. Reaction is second-order. Half-life is 1/K[A]₀. If initial concentration is 0.1M, to a concentration of 0.005M it takes:
1/K*0.1. If initial concentration is 0.005M it takes 1/K*0.005. That means it takes half to decrease from 0.005M to 0.0025 as it does for it to decrease from 0.01M to 0.005M.
(C) The reaction rate is significantly smaller if excess I- is added to the solution. FALSE. Reaction rate is independent of I⁻
(D) The reaction cannot take place in a single elementary step. TRUE. As this reaction is a single-replacement reaction implies the formation of 1 C-N bond. But also the rupture of the C-I bond is impossible to explain this kind of reaction in a single elementary step.
A student completed the experiment but found that the total amount of material recovered weighed more than the original sample. What is the most likely source of error and how may it be corrected?
Answer:
This is due to the water moisture present in the recovered sample.
Explanation:
The total amount of material recovered isn’t meant to weigh more than the original sample. However when this happens then it means there is the presence of water moisture in the recovered sample.
The recovered samples however needs to be heated to make it dry and eliminate the water moisture through evaporation.
Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)
Explanation:
C5H10O2 + NaOH = C2H5COONa + C2H5OH
your result are : sodium propanoate and ethanol
A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.
What is hydrolysis?Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.
The base hydrolysis of methyl butanoate is shown as,
C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH
Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).
Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.
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The total kinetic energy of a body is known as:
A. Thermal energy
B. Convection
C. Potential energy
D. Temperature
The total kinetic energy of a body is known as Thermal energy. Option A
What is thermal energy?Thermal energy is the direct sum of all the available random kinetic energies of molecules.
Also note that thermal energy is directly proportional to temperature in Kelvin.
Thus, the total kinetic energy of a body is known as Thermal energy. Option A
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Answer:
A.) Thermal energy
Explanation:
I got it correct on founders edtell
An enzyme is discovered that catalyzes the chemical reaction:SAD -------->HAPPY
<-------A team of motivated researchers sets out to study the enzyme which they call Happyase. They find that the Kcat for happyase is 600s-1. They carry out several experiments. When [Et]=20 nM and [SAD]=40 \mu M, the reaction velocity, Vo is 9.6 \mu Ms^{-1} . Calculate the Km for the substrate SAD.
Answer: Km = 10μM
Explanation: Michaelis-Menten constant (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: Turnover number (Kcat), which is the number of time an enzyme site converts substrate into product per unit time.
Enzyme veolcity is calculated as:
[tex]V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}[/tex]
where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;
To calculate Km:
[tex]V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate][/tex]
[tex]K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}[/tex]
[tex]K_{m} = \frac{0.02*600*40-9.6*40}{9.6}[/tex]
Km = 10μM
The Michaelis-Menten for the substrate SAD is 10μM.
In the reaction of Zn(s) + 2HCl (aq) Imported Asset ZnCl2 (aq) + H2 (g), if [HCl] increases from 2.6 M to 8.2 M:
The rate at which Zn disappears decreases.
The rate at which H2 appears decreases.
The rate at which ZnCl2 appears increases.
The concentration of Zn (s) also increases.
Answer:
The rate at which ZnCl2 appears increases.
Explanation:
Hello,
In this case, the reaction is:
[tex]Zn(s) + 2HCl (aq) \rightarrow ZnCl_2 (aq) + H_2 (g)[/tex]
Therefore, the law of rate proportions is:
[tex]\frac{1}{-1}r_{Zn}= \frac{1}{-2}r_{HCl}= \frac{1}{1}r_{ZnCl_2}= \frac{1}{1}r_{H_2}}[/tex]
In such a way, since the concentration of hydrochloric acid is increasing The rate at which ZnCl2 appears increases, because the addition of a reactant is directly related with the products formation due to the fact that more reactant will yield more product.
Best regards.
16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
acid. A 30.0 mL sample of KOH is neutralized by 42.7 mL of 0.498 M HNO3. What is the
concentration of the potassium hydroxide solution?
Answer:
[tex]M_{base}=0.709M[/tex]
Explanation:
Hello,
In this case, since the reaction between potassium hydroxide and nitric acid is:
[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]
We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:
[tex]n_{acid}=n_{base}[/tex]
That in terms of molarities and volumes is:
[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, solving the molarity of the base (KOH), we obtain:
[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]
Regards.
f a substance has a half-life of 8.10 hr, how many hours will it take for 75.0 g of the substance to be depleted to 3.90 g?
Answer:
35 hrs
Explanation:
half life of the substance [tex]t_{1/2 }[/tex] = 8.1 hr
initial amount [tex]N_{0}[/tex] = 75 g
The final amount [tex]N[/tex] = 3.9 g
The time elapsed [tex]t[/tex] = ?
we use the relationship
[tex]N[/tex] = [tex]N_{0}[/tex] [tex](\frac{1}{2} )^{\frac{t}{t_{1/2} } }[/tex]
substituting values, we have
3.9 = 75 x [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]
0.052 = [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]
take the log of both side
log 0.052 = log [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]
log 0.052 = [tex]\frac{t}{8.1}[/tex] log 1/2
-1.284 = [tex]\frac{t}{8.1}[/tex] x -0.301
1.284 = 0.301t/8.1 =
1.284 = 0.0372t
t = 1.284/0.037 = 34.5 ≅ 35 hrs
If a salt is formed by combining NH3 (Kb=1.8×10−5) and CH3COOH (Ka=1.8×10−5), an aqueous solution of this salt would be:
Answer:
Neutral
Explanation:
pKa of acid = -log Ka
= -log (1.8 x 10^-5)
= 4.74
pKb of base = -log Kb
= 4.74
pKa of acid = pKb of base
salt pH formula : pH = 7 + 1/2 [pKa -pKb ]
here pKa = pKb
so pH = 7
the salt it is CH3COONH4 exactly neutral solution .
If a salt is formed by combining NH₃ (Kb=1.8×10⁻⁵) and CH₃COOH (Ka=1.8×10⁻⁵), an aqueous solution of this salt would be neutral.
What information does pH convey?pH of any solution tells about the acidity or basicity or neutral nature of the solution.
pH of any solution is directly proportional to the acid dissociation constant value (Ka) and base dissociation constant (Kb). In the question it is given that,
Value of Kb for NH₃ = 1.8×10⁻⁵
Value of Ka for CH₃COOH = 1.8×10⁻⁵
Ka & Kb values for the base and acid is same means it dissociates with same extent. So the aqueous solution of this acid and base is a neutral in nature as they have same number of acid and base ions in it.
Hence resultant solution will be a neutral solution .
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The ΔHvap of nitrous oxide is 16.53 kJ · mol−1 and its ΔSvap is 89.51 J · mol−1 · K−1. What it the boiling point of nitrous oxide?
Answer:
[tex]T_b=-88.48\°C[/tex]
Explanation:
Hello,
In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:
[tex]\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}[/tex]
Solving for the boiling point of nitrous oxide, we obtain:
[tex]T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K[/tex]
Which in degree Celsius is also:
[tex]Tb=184.67-273.15\\\\T_b=-88.48\°C[/tex]
Best regards.
what is non metal?
help meh
The element which can not loose electron easily and having electronagtive character is called non-metal it has following property-1. it can not conduct heat and electricity1. it is netiher ductile not malleable3. it is not lsuturous and also not sonorous
Answer:
Non-metals are the elements which form negative ions by accepting or gaining electrons. Non-metals usually have 4, 5, 6 or 7 electrons in their outermost shell. Non-metals are those which lack all the metallic attributes. They are good insulators of heat and electricity. They are mostly gases and sometimes liquid.
When the nuclide carbon-14 undergoes beta decay: The name of the product nuclide is . The symbol for the product nuclide is
Answer:
Explanation:
The equation for the decay is given as;
¹⁴₆C --> X + ⁰₋₁e
For conservation of matter, the mass number and atomic number has to be the same in both the reactant and product side f he equation;
Mass number;
14 = x + 0
x = 14
Atomic Number;
6 = x + (-1)
x = 6 + 1 =7
¹⁴₆C --> ¹⁴₇N + ⁰₋₁e
The name of the product nuclide with atomic number of 7 is Nitrogen. The symbol is; ¹⁴₇N
Gaseous indium dihydride is formed from the elements at elevated temperature:
In(g)+H2(g)⇌InH2(g),Kp=1.48 at 973 K
The partial pressures measured in a reaction vessel are
PIn =0.0540atm
PH2= 0.0250atm
PInH2 =0.0780atm
Calculate Qp and give equal partial pressure for In, H2, and InH2.
Answer:
The reaction given is:
In (g) + H₂ (g) ⇔ InH₂ (g), the Kp is 1.48 at 973 K.
The partial pressures measured in the reaction vessel is Partial pressure of In is 0.0540 atm, partial pressure of H₂ is 0.0250 atm, and the partial pressure of InH₂ is 0.0780 atm. By using the table given in the attachment below, the value of PInH₂ is (0.078-x), PIn is (0.054 + x), and the value of PH2 is (0.025 + x).
Kp = PInH₂/PIn × PH₂ = (0.078 - x) / (0.054 +x) (0.025 + x)
1.48 = (0.078 - x) / (0.054 +x) (0.025 + x)
x = 0.06689
Now the partial pressures of In, H₂ and InH₂ will be,
PH₂ = 0.025 + x = 0.025 + 0.0668 = 0.0918 atm
PIn = 0.054 + 0.0668 = 0.1208 atm
PInH₂ = 0.078 - 0.0668 = 0.0112 atm
Now the Qp or the reaction quotient will be,
Qp = (0.078) / (0.054) (0.025) = 57.78.
Calculate ΔS∘rxn for the balanced chemical equation 2H2S(g)+3O2(g)→2H2O(g)+2SO2(g) Express the entropy change to four significant figures and include the appropriate units.
Answer:
-170.65
188.8+ 256.8-205.8-(2x205.2)
-170.65 is the entropy change.
What is Entropy Change?Entropy trade is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic gadget. It is associated with the conversion of heat or enthalpy completed in work. A thermodynamic device that has extra randomness means it has high entropy.
Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each extended by using their suitable stoichiometric coefficients, to reap ΔS° for the reaction.
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You have a saturated solution of BaSO4, a slightly soluble ionic compound. What happens if you add Ba(OH)2, NaNO3, and CuSO4 to this solution
Answer:
- Addition of Ba(OH)2: favors the formation of a precipitate.
- Undergo a chemical reaction forming soluble species.
- Addition of CuSO4 : favors the formation of a precipitate.
Explanation:
Hello,
In this case, since the dissociation reaction of barium sulfate is:
[tex]BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)[/tex]
We must analyze the effect of the common ion:
- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
- By adding sodium nitrate, the following reaction will undergo:
[tex]BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)[/tex]
So the precipitate will turn into other soluble species.
- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
All of this is supported by the Le Chatelier's principle.
Best regards.
When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration
Answer:
B) hydroxide concentration
Explanation:
Hello,
In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.
After that, due to the fact that the pH is related with the pOH as shown below:
pH=14-pOH
We can directly compute the pH.
Best regards.
Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 428.7 pm, what is the density of CsBr in g/cm3.
Answer:
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
Explanation:
Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.
The density d of CsBr in g/cm³ can be calculated by using the formula:
[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]
where;
z = 1 mole of CsBr
edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm
molar mass of CsBr = 212.81 g/mol
avogadro's number = 6.023 × 10²³
[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]
[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. [H3O+] = 2.6 x 10-8 M
Answer:
To calculate the [OH-] in the solution we must first find the pOH
That's
pH + pOH = 14
pOH = 14 - pH
First to find the pH we use the formula
pH = - log [H3O+]From the question
[H3O+]= 2.6 × 10^-8 M
pH = - log 2.6 × 10^-8
pH = 7.6
pH = 8
So we pOH is
pOH = 14 - 8 = 6
To find the [OH-] we use the formula
pOH = - log [OH-]6 = - log [OH-]
Find antilog of both sides
[OH-] = 1.0 × 10^-6 MThe solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7
Hope this helps you
After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which side of the reaction do they appear?
MnO41- (aq) + Cl1- (aq) → Mn2+ (aq) + Cl2 (g)
a. 2 moles of H2O on the reactant side
b. 2 moles of H2O on the product side
c. 4 moles of H2O on the product side
d. 8 moles of H2O on the product side
e. 10 moles of H2O on the reactant side
Answer:
d. 8 moles of H2O on the product side
Explanation:
Hello,
In this case, we need to balance the given redox reaction in acidic media as shown below:
[tex]MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\[/tex]
Then, we add the half reactions:
[tex]2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0[/tex]
Thereby, we can see d. 8 moles of H2O on the product side.
Best regards.
