Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V[tex]_1[/tex] ) = 1.50 L,
Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,
Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).
P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],
V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.
Janet observes that bubbles rise inside water when water is heated. Which of the following best names and explains the change that causes bubbles to rise?
Answer:
Boiling
Explanation:
When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.
When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
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
For each of the processes, determine whether the entropy of the system is increasing or decreasing.
a. A snowman melts on a spring day.
b. A document goes through a paper shredder.
c. A water bottle cools down in a refrigerator.
d. Silver tarnishes
e. Dissolved sugar precipitates out of water to form rock candy.
Explanation:
Entropy refers to the degree of disorderliness of a system.
a. A snowman melts on a spring day.
Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.
b. A document goes through a paper shredder.
Entropy increases because random, disorganized bits of paper are left.
c. A water bottle cools down in a refrigerator.
Entropy decreases because temperature is directly proportional to entropy.
d. Silver tarnishes
Entropy increases because random bits of the sliver particles are formed.
e. Dissolved sugar precipitates out of water to form rock candy.
Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.
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
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
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!
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 2.0Enter the complementary strand of this DNA strand. Give your answer as a string.
Answer:
atagcca
Explanation:
t goes with a, and c goes with g
atagcca matches with
ta tcggt
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.
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 yieldMethyl 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.
What 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
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.
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.
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.
Learn more about hydrolysis here:
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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.
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
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.
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.
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
When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
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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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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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
g Solution of barium hydroxide reacts with phosphoric acid to produce barium phosphate precipitate and water. How many mL of 6.50 M calcium hydroxide solution are required to react with a phosphoric acid solution of 45.00 mL that has a concentration of 8.70 M protons (hydrogen ions)
Answer:
30.12 mL.
Explanation:
We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:
Phosphoric acid H3PO4 will dissociate in water as follow:
H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)
From the balanced equation above,
1 mole of H3PO4 produces 3 moles of H+.
Therefore, XM H3PO4 will produce 8.70 M H+ i.e
XM H3PO4 = 8.70/3
XM H3PO4 = 2.9 M.
Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.
Next, we shall write the balanced equation for the reaction. This is illustrated below:
2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O
From the balanced equation above, we obtained the following:
Mole ratio of the acid, H3PO4 (nA) = 2
Mole ratio of the base, Ba(OH)2 (nB) = 3
Data obtained from the question include the following:
Molarity of base, Ba(OH)2 (Mb) = 6.50 M
Volume of base, Ba(OH)2 (Vb) =.?
Molarity of acid, H3PO4 (Ma) = 2.9 M
Volume of acid, H3PO4 (Va) = 45 mL
The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:
MaVa /MbVb = nA/nB
2.9 x 45 / 6.5 x Vb = 2/3
Cross multiply
2 x 6.5 x Vb = 2.9 x 45 x 3
Divide both side by 2 x 6.5
Vb = (2.9 x 45 x 3) /(2 x 6.5)
Vb = 30.12 mL
Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL
Find the standard enthalpy of formation of iodine atoms. (Round your answer to one decimal place.) Standard enthalpy of formation
Answer:
Enthalpy of formation is the energy change when one mole of a substance is formed from its constituent atoms under standard conditions
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.
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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