Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
Which of these species would you expect to have the lowest standard entropy (S°)?
a. CH4(g)
b. H2O(g)
c. NH3(g)
d. HF(g)
Answer:
d. HF(g)
Explanation:
Hello,
In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.
Best regards.
a sample of gas occupies a volume of 2.62 liters at 25 C and 1.00 atm. what will be the volume at 50 C and 2 atm
Answer:2.62 L
Explanation:
A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
What is ideal gas law ?The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.
An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.
By using ideal gas equation,
P₁ V ₁ ÷ T = P₂V₂ ÷ T
1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50
V₂ = 1 × 2.62 × 50 ÷ 25 × 2
V₂ = 2.62 liters.
Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
To learn more about ideal gas law follow the link below;
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The standard entropy of a substance refers to its entropy at:__________.
a. absolute zero and 1 bar
b. 0°C and 1 bar
c. 25 °C and 1 bar
d. 25 °C and 0 bar
Answer:
b. 0°C and 1 bar
Explanation:
Hello,
In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.
Best regards.
A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.
Answer:
Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.
Explanation:
Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.
Small amounts of calcium hydroxide salt, [tex]Ca(OH)_{2}_(s)[/tex] is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.
I hope this explanation is helpful.
A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms
Answer:12.9e-12g or in short 12.9pg
Explanation:as p=1e-12
A sample of radioactive silver contains two isotopes, 108Ag (denoted A) and 110Ag (denoted B). The second of these (B) has a half life of 24 seconds, whereas the first (A) has a half life of 2.3 minutes. If a sample contains equal numbers of each of these isotopes at the beginning of an experiment that runs for an hour, which of the following statements is correct?
A. At the end of the hour, isotope B has a greater decay constant λ than isotope A
B. At the end of the hour, isotope A has the same decay constant λ as isotope B
C. At the end of the hour, isotope A has a greater decay constant λ than isotope B
Answer:
A : At the end of the hour, isotope B has a greater decay constant λ than isotope A
Explanation:
Firstly, we need to understand that radioactive decay follows a first order rate law.
What this means is that we can calculate the radioactive decay constant using the following formula from the half-life
Mathematically;
[tex]t_{1/2}[/tex] = 0.693/λ
where λ represents the radioactive decay constant.
Rearranging the equation, we can have
λ = 0.693/[tex]t_{1/2}[/tex]
Now, to have a fair level playing ground, it is best that the half-life of both isotopes are in the same unit of time(seconds)
For A, the half-life = 2.3 minutes which is same as 2.3 × 60 = 138 seconds
For B, the half-life is 24 seconds
Thus, at the end of the hour, the decay constant for isotope A will be;
λ = 0.693/138 = 0.0050 [tex]s^{-1}[/tex]
For isotope B, the decay constant will be;
λ = 0.693/24 = 0.028875 [tex]s^{-1}[/tex]
We can see that the decay constant of isotope B is higher than that of A at the end of the experiment
You wish to construct a galvanic cell with the anode consisting of a Ni electrode in a 1.0 M Ni(NO3)2 solution. What would be the highest standard cell potential if used as the cathode in this galvanic cell?
Answer:
Au^3+(aq) +3e ------> Au(s). 1.50 V
Explanation:
When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.
E°cell= E°cathode - E°anode
If E°cathode= 1.50 V
E°anode= -0.25 V
E°cell= 1.50 -(-0.25)
E°cell= 1.75 V
Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential
Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the data and mechanism. a. the mechanism for this reaction is E2 b. an increase in 1-butene was observed when t-butoxide was used c. an increase in 1-butene was observed when methoxide was used d. the mechanism for this reaction is E1 e. no significant difference was observed
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
What happens if we put raw eggs in a pot full of hot oil?
Q 13.3: Which of the following is the least stable radical choice? Tertiary radical. B : Allyl radical. C : Secondary radical. D : Methyl radical. E : Primary radical.
Answer:
Methyl radical
Explanation:
A radical is any specie that contains an odd number of electrons. We must note that the greater the number of alkyl groups which are attached to a carbon atom that bears the odd electrons, the more the degree of delocalization of the odd electrons and consequently the more stable we expect the free radical to be.
Hence the order of free radical stability is; Methyl < Primary < Secondary < Tertiary. Hence, we can easily see that the methyl radical is the least stable free radical.
Answer: Methyl radical
Explanation:
If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
Given the following reaction and data, A + B → Products
Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640
Required:
a. What is the rate law of the reaction?
b. What is the rate constant?
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under these conditions is 4.18·J·g−1K−1 . Round your answer to 3 significant digits.
Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb2O E) NaF and SrO
Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
In which pair do both compounds exhibit predominantly ionic bonding?
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indium and cadmium. In(s)|
Answer:
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
Explanation:
Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.
We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.
Oxidation half equation;
3Cd(s) -------> 3Cd^2+(aq) + 6e
Reduction half equation;
2In^3+(aq) + 6e -----> 2In(s)
What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH
Answer:
0.0630
Explanation:
The molar mass of urea = 60 g/mol
we all know that:
[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
Then; the number of moles of urea
= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]
= 0.0667 mol
Similarly; the number of moles of methanol
= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]
= 0.9988 mol
The total number of moles = (0.0667 + 0.9988) mol
= 1.0655 mol
Finally,the mole fraction of urea [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]
[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]
= 0.0630
Given the data: Ag2O(s), = ‑31.1 kJ mol-1, S° = +121.3 J mol-1 K-1 Ag(s), = 0.00 kJ mol-1, S° = +42.55 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = +205.0 J mol-1 K-1 Calculate the temperature at which = 0 for the reaction, Ag2O(s) → 2 Ag(s) + ½ O2(g). Assume that, since the physical states do not change, and are independent of temperature between ‑50.0 °C and 950.0 °C.
Answer:
[tex]T=469.1K\\\\T=195.9\°C[/tex]
Explanation:
Hello,
In this case, for the given decomposition reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):
[tex]\Delta _rH=2\Delta _fH_{Ag}+\frac{1}{2} \Delta _fH_{O_2}-\Delta _fH_{Ag_2O}\\\\\Delta _rH=2*0.00+\frac{1}{2} *0.00-(-31.1)=31.1kJ/mol[/tex]
Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):
[tex]\Delta _rS=2S_{Ag}+\frac{1}{2} S_{O_2}-S_{Ag_2O}\\\\\Delta _rS=2(42.55)+\frac{1}{2} (205.0)-(121.3)=66.3J/(mol*K)[/tex]
Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the unknown temperature, for such Gibbs energy equaling 0, the temperature (in K and °C) turns out:
[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\0=31.1kJ/mol-T(66.3\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\T=\frac{31.1kJ/mol}{0.0663kJ/(mol*K)} =469.1K\\\\T=195.9\°C[/tex]
Which is within the given rank.
Best regards.
Which gas will have the most collisions between its particles?
Answer:
The gas is Methane at 340K
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
A student is using a coffee-cup calorimeter to determine the enthalpy change of the endothermic reaction of two aqueous solutions. After both solutions are added to the cup, the student neglects to put the lid on the cup. This would cause the magnitude of the calculated ΔH° value to be: the answer is: too small, since the solution will absorb heat from the room. But why? Wouldn't depend on if the reaction releases or absorbs heat. Wouldn't it be too large because heat escapes the cup? I'm so confused
Answer:
Explanation:
In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .
The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .
Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.
An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N2(g) + 3 H2(g)<=> 2NH3(g) is 0.29.
What is the concentration of NH3?
Answer:
5.94×10¯³
Explanation:
The following data were obtained from the question:
Concentration of N2, [N2] = 0.036 M
Concentration of H2, [H2] = 0.15 M
Equilibrium constant (Kc) = 0.29 M
Concentration of NH3, [NH3] =.…?
The equation for the reaction is given below:
N2(g) + 3H2(g) <=> 2NH3(g)
Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.
This is illustrated below:
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant for the above equation is given below:
Kc = [NH3]² / [N2] [H2]³
Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:
Kc = [NH3]² / [N2] [H2]³
0.29 = [NH3]²/ 0.036 × 0.15³
Cross multiply
[NH3]² = 0.29 × 0.036 × 0.15³
[NH3]² = 3.5235×10¯⁵
Take the square root of both side
[NH3] = √(3.5235×10¯⁵)
[NH3] = 5.94×10¯³
Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.
For the following reaction, 6.99 grams of oxygen gas are mixed with excess nitrogen gas . The reaction yields 10.5 grams of nitrogen monoxide . nitrogen ( g ) oxygen ( g ) nitrogen monoxide ( g ) What is the theoretical yield of nitrogen monoxide
Answer:
13.11 g.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below :
N2 + O2 —> 2NO
Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32 g.
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO from the balanced equation = 2 x 30 = 60 g.
From the balanced equation above,
32 g of O2 reacted to produce 60 g of NO.
Finally, we shall determine the theoretical yield of NO as follow:
From the balanced equation above,
32 g of O2 reacted to produce 60 g of NO.
Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.
Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.
A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was ________ mm Hg.
Answer:
96 mmHg
[tex]h=96mmHg[/tex]
Explanation:
From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .
The difference of height of the mercury level is given as,
h=9.60cm
h=9.60(10mm/1cm)
[tex]h=96mm[/tex]
But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height
P(gas)=h
P(gas)=96mmHg
This pressure is as a result of the presence of gas.
Therefore, the pressure of the argon gas in the container is 96mmHg.
The pressure of the argon in the container was 96mmHg.
We were told that the manometer has closed ends which means that the
height will not be affected by atmospheric pressure.
The height which is the difference in mercury level is
h=9.60cm
We can convert it to millimeter by multiplying it by 10
h=9.60 × 10 = 96mm
The pressure of the closed end manometer will be equal to the height
P(gas)=h
P(gas)=96mmHg
The pressure of the argon gas in the container is 96mmHg.
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How many moles of NaOH is needed to neutralize 45.0 ml of 0.30M H2SeO4? Question 2 options: A) 0.00675 B) 27.0 C) 0.027 D) 0.0135
Answer:
C) 0.027
Explanation:
In this case we can start with the reaction between [tex]NaOH[/tex] and [tex]H_2SeO_4[/tex], so:
[tex]H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O[/tex]
We have an acid ([tex]H_2SeO_4[/tex]) and a base ([tex]NaOH[/tex]), therefore we will have an acid-base reaction in which a salt is produced ([tex]Na_2SeO_4[/tex]) and water ([tex]H_2O[/tex]).
Now we can balance the reaction:
[tex]H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O[/tex]
If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.3~M~=~\frac{mol}{0.045~L}[/tex]
[tex]mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4[/tex]
In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of [tex]H_2SeO_4[/tex] 2 moles of [tex]NaOH[/tex] are consumed), with this in mind we can calculate the moles of NaOH:
[tex]0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH[/tex]
I hope it helps!
The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M
Answer:
[tex][NH_2NO_2]=0.0868M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction, the first-order rate law is:
[tex]r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2][/tex]
Which integrated is:
[tex][NH_2NO_2]=[NH_2NO_2]_0exp(-kt)[/tex]
Thus, the concentration after 31642.0 s for a 0.384-M solution is:
[tex][NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)\\[/tex]
[tex][NH_2NO_2]=0.0868M[/tex]
Best regards.
Answer:
[A] = 0.0868 M
Explanation:
Rate constant = 4.70×10-5 s-1
First order reaction
Initial concentration, [A]o = 0.384 M
Final concentration, [A] = ?
Time, t = 31642.0 s
All these variables are related by the following equation;
[A] = [A]o e^(-kt)
[A] = 0.384 e^(-4.70×10-5 x 31642.0)
[A] = 0.384 e^(-1.4872)
[A] = 0.384 * 0.2260
[A] = 0.0868 M
A student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) Br2(g) When she introduced 0.143 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.108 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
Answer:
1.84 × 10⁻³
Explanation:
Step 1: Write the balanced equation
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
Step 2: Calculate the initial concentration of NOBr
0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:
M = 0.143 mol / 1.00 L = 0.143 M
Step 3: Make an ICE chart
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
I 0.143 0 0
C -2x +2x +x
E 0.143-2x 2x x
Step 4: Find the value of x
The equilibrium concentration of NOBr(g) was 0.108 M. Then,
0.143-2x = 0.108
x = 0.0175
Step 5: Calculate the concentrations at equilibrium
[NOBr] = 0.108 M
[NO] = 2x = 0.0350 M
[Br₂] = x = 0.0175 M
Step 6: Calculate the equilibrium constant (Kc)
Kc = [0.0350]² × [0.0175] / [0.108]²
Kc = 1.84 × 10⁻³
Sulfur dioxide reacts with oxygen to form sulfur trioxide. What change in hybridization of the sulfur occurs in this reaction ? g
Answer:
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Explanation:
g If the titration of a 10.0-mL sample of sulfuric acid requires 28.15 mL of 0.100 M sodium hydroxide, what is the molarity of the acid
Answer:
[tex]M_{acid}=0.141M[/tex]
Explanation:
Hello,
In this case, the reaction between sulfuric acid and hydroxide is:
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
We can notice a 1:2 molar ratio between the acid and the base respectively, therefore, at the equivalence point we have:
[tex]2*n_{acid}=n_{base}[/tex]
And in terms of volumes and concentrations:
[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]
So we compute the molarity of sulfuric acid as shown below:
[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} =\frac{0.100M*28.15mL}{2*10.0mL}\\ \\M_{acid}=0.141M[/tex]
Best regards.
An object has a mass of 4.9g and a volume of 14.ml.what is the density of the object?
Answer:
0.35 g/mL
Explanation:
Use the formula D = [tex]\frac{m}{v}[/tex], where D is density, m is mass, and v is volume.
D = 4.9/14
D = 0.35
D = 0.35 g/mL
