To calculate the equilibrium constant (Kc), we use the law of mass action. The equation for the reaction is N2 + 3H2 ⇌ 2NH3. By plugging in the given equilibrium concentrations into the equation and solving, we find that Kc is approximately 0.05927.
Explanation:To calculate the equilibrium constant, Kc, we need to use the law of mass action. The equation for the reaction is:
N2 + 3H2 ⇌ 2NH3
Kc is calculated by taking the concentrations of the products (NH3) raised to their stoichiometric coefficients and dividing by the concentrations of the reactants (N2 and H2) raised to their stoichiometric coefficients.
Kc = ([NH3]²) / ([N2][H2]³)
Plugging in the given equilibrium concentrations ([NH3] = 0.75 M, [N2] = 0.31 M, [H2] = 1.51 M) into the equation, we get:
Kc = (0.75²) / (0.31 × 1.51³)
Kc = 1.125 / (0.468 × 3.448)³
Kc = 1.125 / (0.468 × 40.7855)
Kc = 1.125 / 18.986
Kc ≈ 0.05927
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Equilibrium is established in a reversible reaction when:
a) the [product] = [reactants]
b) rate of reaction of products = rate of reaction of reactants
c) all the reactants dissolve or dissociate
d) when product are no longer produced
Equilibrium in a reversible reaction is established when the rate of the forward reaction (reactants transforming into products) equals the rate of the backward reaction (products reconverting into reactants). This indicates that the amounts of products and reactants are no longer changing over time.
Explanation:In a reversible reaction, equilibrium is established when option (b) is correct: the rate of the forward reaction (reactants transforming into products) becomes equal to the rate of the backward (or reverse) reaction (products converting back into reactants). This does not necessarily mean the concentrations of the products and reactants are equal. Rather, it means the amounts of products and reactants are no longer changing over time, demonstrating a state of dynamic equilibrium. The equilibrium can shift depending on external factors and conditions such as temperature, pressure, or concentration changes, which is described by Le Chatelier's Principle.
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Compared to a solution with a pH value of 7 a solution with a thousand times greater hydronium ion concentration has a pH value of what
The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.
The pH has been defined as the negative log of hydronium ion concentration in the solution. The pH has been expressed as:
[tex]\rm pH=-\;log\;[H_3O^+][/tex]
Computation for the pH of the solutionThe pH of the given solution has been 7. The hydronium ion concentration has been given as:
[tex]\rm pH=\;-log\;[H_3O^+]\\\\ 7=\;-\;log[H_3O^+]\\\\ H_3O^+=10^-^7\;M[/tex]
The concentration of the new solution has been 1000 times greater than the previous solution.
The previous solution has hydronium ion concentration of [tex]\rm 10^-^7\;M[/tex]. The concentration of new solution will be:
[tex]\rm New\;solution=10^-^7\;\times\;1000\;M\\ New\;solution=10^-^4\;M[/tex]
The concentration of the new solution has been [tex]\rm 10^-^4\;M[/tex]. The pH of the solution has been given as:
[tex]\rm pH=-log\;[H_3O^+]\\ pH=-log\;[10^-^4]\\ pH=4[/tex]
The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.
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Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.030 m in naclo at 25 °c?
Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq)
+ H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
c(NaClO) = 0,030 M.
[ClO⁻] =
0,03 M - x.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
[OH⁻]
= [HClO] = x.
2,5·10⁻⁷ = x² / (0,03 M
-x).
Solve quadratic equation: x = [OH⁻] = 0.0000893 M.
pOH = -log(0.0000893 M) = 4.05.
pH = 14 - 4.05.
pH = 9.95.
A compound consisting of atoms of large atomic mass is more likely to require:
Answer:
A) a higher temperature to liquefy
BRAINLIST?
Explanation:
Answer:
The answer is
A: a higher temperature to liquefy
elaborate on the difference in natural occurrences between nuclear fission and nuclear fusion reactions. A) Neither nuclear fission nor nuclear fusion reactions naturally occur on earth or outer space. B) Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars. C) Both nuclear fission and nuclear fusion reactions readily naturally occur in the pressure of the earth's core. D) Nuclear fusion reactions do not naturally occur while nuclear fission reactions occur in the volcanoes.
Option B: Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.
Nuclear fission reaction is defined as formation of two or more atoms by splitting of large atoms. On the other hand, nuclear fusion reaction is formation of large atom from small atoms.
The splitting of large atoms into small atoms (nuclear fission) generally does not occur naturally because it requires high speed neutrons and critical mass of the substance undergoing fission.
Nuclear fusion reaction occurs in stars such as sun because it requires extremely high energy to bring two more proton closer to each other by overcoming electronic repulsion.
Therefore, nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.
What are the boiling point and freezing point of a 3.47 m solution of naphthalene in benzene? (the boiling point and freezing point of benzene are 80.1°c and 5.5°c respectively. the boiling point elevation constant for benzene is 2.53°c/m, and the freezing point depression constant for benzene is 5.12°c/m.)?
The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.
The correct answer is -12.266.
What is the boiling point?The boiling point of a liquid varies according to the applied pressure; the normal boiling point is the temperature at which the vapor pressure is equal to the standard sea-level atmospheric pressure (760 mm [29.92 inches] of mercury). At sea level, water boils at 100° C.Elevation in boiling point is mathematically expressed as
[tex]Tb = Kb * m[/tex]
Where
Kb = boiling point elevation constant = 2.53°c/m (for benzene)m = molality of solution = 3.47 m (given)ΔTb[tex]= 2.53 * 3.47 = 8.779 oC[/tex]
But, the boiling point of benzene = 80.1 oC
Boiling point of solution = 88.879 oC
Now, Depression in freezing point = ΔTf[tex]= Kf * m[/tex]
where,
Kf = freezing point depression constant = 5.12°c/m (for benzene)ΔTf =[tex]5.12 X*3.47 = 17.766 oC[/tex]
But the freezing point of benzene = 5.5 oC.
Freezing point of solution = -12.266 oC.
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Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16 : 1.
The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution
Explanation:
The given equation is as follows.
[tex]PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^{-}[/tex]
s 2s
It is given that,
[tex]K_{sp} = [Pb^{2+}][Br^{-}]^{2} = 6.60 \times 10^{-6}[/tex]
Let the solubility of given ions be "s".
Since, KBr on dissociation will given bromine ions.
Hence, [tex]K_{sp} = [Pb^{2+}] \times ([Br^{-}])^{2}[/tex]
[tex]6.60 \times 10^{-6}[/tex] = [tex]s \times (2s)^{2}[/tex]
= [tex]1.18 \times 10^{-2}[/tex] M
Therefore, solubility of [tex][PbBr_{2}][/tex] is [tex]1.18 \times 10^{-2}[/tex] M in KBr.
Now, we will calculate the molar solubility of [tex]PbBr_{2}[/tex] in 0.5 M KBr solution as follows.
[tex]K_{sp} = (s) \times (2s + 0.5)^{2}[/tex]
[tex]6.60 \times 10^{-6}[/tex] = [tex]4s^{3} + 0.25s + 2s^{2}[/tex]
s = [tex]2.63 \times 10^{-5}[/tex]
Thus, we can conclude that molar solubility of [tex][PbBr_{2}][/tex] in 0.500 m KBr solution is [tex]2.63 \times 10^{-5}[/tex].
When preparing the diazonium salt, the solution is tested with potassium iodide-starch paper. a positive test is the immediate formation of a blue color. what is the ki-starch paper testing for ? explain?
Final answer:
Potassium iodide-starch paper is used to detect the presence of nitrogen dioxide in the preparation of diazonium salts by revealing a blue-black coloration upon reacting with iodine produced from the reaction between NO2 and potassium iodide.
Explanation:
The potassium iodide-starch paper is used to test for the presence of nitrogen dioxide (NO2), which is a byproduct of the reaction used to prepare diazonium salts. During the diazo coupling reaction, excess nitrous acid can decompose and produce nitrogen dioxide. This gas can then react with the potassium iodide (KI) in the starch paper to produce iodine (I2), which subsequently forms a blue-black complex with starch. The immediate formation of the blue color on the potassium iodide-starch paper is a positive test indicating the presence of nitrogen dioxide. It's important to monitor this because the presence of NO2 suggests that the diazonium salt solution might be unsafe due to the potential release of toxic gases.
The iodine-starch test is a well-known reaction in which iodine (I2), produced by the oxidation of iodide ions by NO2, interacts with starch to produce a characteristic blue-black color. This test provides a quick and sensitive method for detecting the presence of iodine, which, in this context, indirectly indicates the generation of nitrogen dioxide in the reaction mixture.
A solution has [oh−] = 2.3×10−2 m . use the ion product constant of water kw=[h3o+][oh−] to find the [h3o+] of the solution.
Answer:
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
Explanation:
Concentration of hydroxide ions = [tex][OH^-]=2.3\times 10^{-2}[/tex]
Concentration of hydroxide ions = [tex][H_3O^+]=?[/tex]
[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The ionic product of water is given as:
[tex]K_w=[H_3O^+][OH^-][/tex]
The value of ionic product of water, [tex]K_w=1\times 10^{-14}[/tex]
[tex]K_w=1\times 10^{-14}=[H_3O^+][OH^-][/tex]
[tex]1\times 10^{-14}=[H_3O^+]\times 2.3\times 10^{-2}M[/tex]
[tex][H_3O^+]=\frac{1\times 10^{-14}}{2.3\times 10^{-2}}=4.35\times 10^{-13} M[/tex]
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
The concentration of hydronium, ion [H₃O⁺] in the solution containing 2.3×10⁻² M concentration of hydroxide ion, [OH⁻] is 4.35×10⁻¹³ M
How to determine the concentration of hydronium, ion [H₃O⁺]?The following data were obtained from the question given above:
Concentration of hydroxide ion, [OH⁻] = 2.3×10⁻² MConcentration of hydronium, ion [H₃O⁺] = ?[H₃O⁺] × [OH⁻] = 10¯¹⁴
[H₃O⁺] × 2.3×10⁻² = 10¯¹⁴
Divide both side by 2.3×10⁻²
[H₃O⁺] = 10¯¹⁴ / 2.3×10⁻²
= 4.35×10⁻¹³ M
Thus, we can conclude that the concentration of hydronium ion, [H₃O⁺] in the solution is 4.35×10⁻¹³ M
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how many grams of salt are in 5 L of a solution with a concentration of 3 g/L
Which would be most useful if you were lost in the Sahara desert? A. Landsat satellite B. OSTM/Jason-2 satellite C. Global Positioning System D. topographic map of Africa
What solution has the lowest pH?
1.0 M HNO2
1.0 M CH3COOH
1.0 M HCOOH
1.0 M HNO3
1.0 M HPO4-
A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate the freezing point of this solution. [kfp(benzene) = 5.12°c/m, kbp(benzene) = 2.53°c/m] (the boiling point and freezing point of benzene are 80.1°c and 5.5°c, respectively).
The new freezing point of the solution is -8.63 °C.
Using the ideal gas law, PV = nRT, we need to convert 4.00 L of gas at 27°C (which is 300 K) and 748 mmHg to moles of gas.
P = 748 mmHg / 760 mmHg atm = 0.984 atm
[tex]n = \frac{PV}{RT} = \frac{0.984 \, \text{atm} \times 4.00 \, \text{L}}{0.0821 \, \text{L atm / K mol} \times 300 \, \text{K}} = 0.160 \, \text{mol}[/tex]
Calculate molality:
Molality (m) = moles of solute / kg of solvent = 0.160 mol / 0.0580 kg = 2.759 m
Calculate freezing point depression (ΔTf):
ΔTf = i * Kf * m
For non-electrolytes, i = 1.
ΔTf = 1 * 5.12 °C/m * 2.759 m = 14.13 °C
Determine the new freezing point:
The freezing point of pure benzene is 5.5 °C.
New freezing point = 5.5 °C - 14.13 °C = -8.63 °C
Thus, the freezing point of the solution is -8.63 °C.
When hydrochloric acid is added to sodium hydrogen carbonate?
What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?
Write the balanced nuclear equation for β− decay of sodium−26. include both the mass numbers and the atomic numbers with each nuclear symbol. use the sup-subscript button in the answer palette to enter these numbers correctly. greek letters can be accessed in the drop-down menu that says -select−.
How much 0.100 m hcl is required to completely neutralize 20.0 ml of 0.250 m naoh?
To neutralize 20.0 mL of 0.250 M NaOH, you will need 50.0 mL of 0.100 M HCl.
Explanation:To determine how much 0.100 M HCl is required to neutralize 20.0 mL of 0.250 M NaOH, we can use the balanced equation for the reaction between HCl and NaOH: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq). The mole ratio between HCl and NaOH is 1:1, meaning that 1 mole of HCl reacts with 1 mole of NaOH. We can use this ratio to calculate the amount of HCl needed.
First, find the number of moles of NaOH:
0.250 M NaOH x 0.0200 L = 0.005 moles NaOH
Since the mole ratio between HCl and NaOH is 1:1, we need 0.005 moles of HCl to neutralize the NaOH.
Now, calculate the volume of 0.100 M HCl needed to contain 0.005 moles:
0.005 mol HCl / 0.100 mol/L = 0.050 L = 50.0 mL
Therefore, 50.0 mL of 0.100 M HCl is required to completely neutralize 20.0 mL of 0.250 M NaOH.
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What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?
The chemical reaction is given as:
[tex]2NaDP^{+} +2H_{2}O\rightarrow 2NaDPH+2H^{+}+O_{2}[/tex]
Here, oxygen is oxidised and [tex]NaDP^{+}[/tex] is reduced. Thus, redox reaction occurs.
For cell reaction, [tex]\Delta G^{o} = -nFE^{o}_{cell}[/tex] (2)
where, [tex]\Delta G^{o} [/tex] = standard state free energy
n= number of electrons
F= Faraday constant ([tex]96485.33 C/mol[/tex])
[tex]E^{o}_{cell}[/tex] = cell potential
Substitute the value of number of electrons i.e. 2, Faraday constant and cell potential in the formula to determine the value of [tex]\Delta G^{o} [/tex].
Now, calculate the value of cell potential
[tex]E^{o}_{cell} = E^{o}_{cathode}- E^{o}_{anode}[/tex] (1)
[tex]E^{o}_{cathode}[/tex] = [tex]-0.324 V[/tex] (standard reduction potential of [tex]NaDP^{+}[/tex])
[tex]E^{o}_{anode}[/tex] = [tex]1.23 V[/tex] (standard reduction potential of [tex]O_{2}[/tex])
Put the above values in formula (1), we get:
[tex]E^{o}_{cell} = -0.324 V-1.23 V[/tex]
= [tex]-1.554 V[/tex]
Now, substitute above value in formula (2)
[tex]\Delta G^{o} = -2\times 96485.33 C/mol \times(-1.554 V) [/tex]
= [tex]299876.40564 CV/mol[/tex]
Since, one coulomb volt is equal to one joule.
Thus, value of [tex]\Delta G^{o}[/tex] is equal to [tex]299876.40564 J/mol[/tex] or [tex]299.87640564 kJ/mol[/tex]
a student reacts 0.600 g of lead (ii) nitrate with 0.850 g of potassium iodide
write and balance the chemical equation.
how many miles of Pb(No3)2 were used?
how many miles of KI were used?
how many miles of PbI2 would form based on the moles of Pb(No3)2 used?
how many miles of PbI2 would form based on the moles of KI used?
which is the limiting reactant?
what is the theoretical yeild of PbI2 in grams?
if the student obtained 0.475 grams of PbI2 product after conducting it by filtration, what is the percent yeild of PbI2 obtained ?
How many times greater is the rate of effusion of molecular fluorine than that of molecular bromine at the same temperature and pressure
A 3.00-l flask is filled with gaseous ammonia, nh3. the gas pressure measured at 15.0 ∘c is 2.15 atm . assuming ideal gas behavior, how many grams of ammonia are in the flask? express your answer to three significant figures and include the appropriate units.
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] = 6.3×10−4m , and [hno2] = 0.21m ? be sure to express your answer in units of kj in the box below. answers without units will not be given credit.
By applying the modified Gibbs Free Energy formula with given values for equilibrium constants, atom concentrations, and other parameters, we find ΔG for nitrous acid at 25°C to be around 27.94 KJ/mol.
Explanation:In chemistry, the Gibbs free energy (ΔG) is calculated using the equation ΔG = -RTlnK, where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. However, since we're given Ka, the equation must be adapted. Therefore, we use ΔG = -RTlnKa + RTlnQ, where Q is the reaction quotient given by [NO₂⁻][H⁺] / [HNO₂].
Inserting the given values, such as Ka = 4.5×10⁻⁴, R (in appropriate units) as 0.0083145 KJ/(mol.K), T as 298.15K (25°C in Kelvin), and Q = ([NO₂⁻][H⁺]) / [HNO₂] = (6.3×10⁻⁴ × 5.9×10−2) / 0.21, we can now solve for ΔG. Doing the math, we find that ΔG ≈ 27.94 KJ/mol.
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Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. CH3OH + O2 → CH2O + H2O
Answer:
look at the screenshot below
Explanation:
Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen. determine the empirical formula for a compound that is 70.79 carbon, 8.91 hydrogen, 4.59 nitrogen, and 15.72 oxygen. c18h27no2 c18h27no3 c17h27no3 c17h26no3
The empirical formula of the compound is C₁₈H₂₇NO₃
From the question given above, the following data were obtained:
Carbon (C) = 70.79%
Hydrogen (H) = 8.91%
Nitrogen (N) = 4.59%
Oxygen (O) = 15.72%
Empirical formula =?The empirical formula of the compound can be obtained as follow:
C = 70.79%
H = 8.91%
N = 4.59%
O = 15.72%
Divide by their molar massC = 70.79 / 12 = 5.899
H = 8.91 / 1 = 8.91
N = 4.59 / 14 = 0.328
O = 15.72 / 16 = 0.9825
Divide by the smallestC = 5.899 / 0.328 = 18
H = 8.91 / 0.328 = 27
N = 0.328 / 0.328 = 1
O = 0.9825 / 0.328 = 3
Therefore, the empirical formula of the compound is C₁₈H₂₇NO₃
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Calculate the volume in liters of a ×4.3910−5/moll mercury(ii) iodide solution that contains 500.mg of mercury(ii) iodide hgi2 . round your answer to 3 significant digits.
Final answer:
To calculate the volume of the mercury(II) iodide solution in liters, convert the mass of the solute to moles and use the Molarity equation. The volume is approximately 25.1 L.
Explanation:
To calculate the volume in liters of the mercury(II) iodide solution, we need to convert the mass of the solute (HgI2) to moles, using the molar mass of HgI2. Then, we can use the equation Molarity = moles of solute / volume of solution in liters to find the volume.
First, calculate the moles of HgI2:
Moles = mass / molar mass = 500 mg / (454.39 g/mol) = 1.101 x 10^-3 mol
Next, rearrange the equation to solve for volume:
Volume = moles of solute / Molarity = 1.101 x 10^-3 mol / (4.3910^-5 M) = 25.07 L
Rounding to three significant digits, the volume of the solution is approximately 25.1 L.
N aqueous solution of sodium hydroxide is standardized by titration with a 0.154 m solution of hydrochloric acid. if 17.5 ml of base are required to neutralize 17.6 ml of the acid, what is the molarity of the sodium hydroxide solution?
Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.
Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
The anode is positive and the cathode is negative.
Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
a. treatment of compound a with m−chloroperoxybenzoic acid (mcpba) gave compound b, which on reduction with lithium aluminum hydride gave compound
c. oxidation of compound c with chromic acid gave compound d (c9h14o). identify compounds a through d in this sequence, and do not indicate stereochemistry.
Notice that " s o 4 " appears in two different places in this chemical equation. s o 2− 4 is a polyatomic ion called "sulfate." what number should be placed in front of cas o 4 to give the same total number of sulfate ions on each side of the equation? ?cas o 4 +alc l 3 →cac l 2 +a l 2 (s o 4 ) 3 express your answer numerically as an integer.