Answer:
D) 2.01 x 10⁻⁴ .
Explanation:
pH = 2.11
[ H⁺ ] = [tex]10^{-2.11}[/tex]
Let the acid be HA
It will ionise as follows .
HA ⇄ H⁺ + A⁻
in equilibrium .30 [tex]10^{-2.11}[/tex] [tex]10^{-2.11}[/tex]
Acid ionisation constant Ka = [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]
= 2 x 10⁻⁴
Answer:
D) 2.01 x 10⁻⁴ is correct!
Explanation:
I got it in class!
Hope this Helps!! :))
If for a particular process, ΔH=308 kJmol and ΔS=439 Jmol K, in what temperature range will the process be spontaneous?
Answer:
The process will be spontaneous above 702 K.
Explanation:
Step 1: Given data
Standard enthalpy of the reaction (ΔH°): 308 kJ/molStandard entropy of the reaction (ΔS°): 439 J/mol.KStep 2: Calculate the temperature range in which the process will be spontaneous
The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
When ΔG° < 0,
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (308,000 J/mol)/(439 J/mol.K)
T > 702 K
The process will be spontaneous above 702 K.
A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure (p)will be
A.2/3p
B.3/5p
C.2/5p
D.1/3p
Answer this with reason
Answer:
B.3/5p
Explanation:
For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.
Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.
[tex]P_i=P_t_o_t_a_l*X_i[/tex]
Where:
[tex]P_i[/tex]=Partial pressure
[tex]P_t_o_t_a_l[/tex]=Total pressure
[tex]X_i[/tex]=mole fraction
With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:
moles of hydrogen gas
The molar mass of hydrogen gas ([tex]H_2[/tex]) is 2 g/mol, so:
[tex]6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2[/tex]
moles of oxygen gas
The molar mass of oxygen gas ([tex]O_2[/tex]) is 32 g/mol, so:
[tex]64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2[/tex]
Now, total moles are:
Total moles = 2 + 3 = 5
With this value, we can write the partial pressure expression for each gas:
[tex]P_H_2=\frac{3}{5}*P_t_o_t_a_l[/tex]
[tex]P_O_2=\frac{2}{5}*P_t_o_t_a_l[/tex]
So, the answer would be 3/5P.
I hope it helps!
If you have 2.4L of SO2 gas (at STP) how many moles of sulfur dioxide do you have?
Answer:
0.107 mole of SO2.
Explanation:
1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).
With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.
This can be obtained as follow:
22.4 L contains 1 mole of SO2.
Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.
Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.
plsssss help!!! Deep Space 1 was a spacecraft powered by an engine that gave off xenon particles to change velocity. It had a mass of 500 kg. Which expression can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N? A) 0.10 N 500 kg B) 500 kg · 0.10 N C) 500 kg 0.10 N D) 500 kg + 0.10 N
Answer:
Spacecraft’s acceleration (a) = 0.10 N / 500 kg
Explanation:
Given:
Mass of Spacecraft (M) = 500 Kg
Force generate by engine (F) = 0.10 N
Find:
Spacecraft’s acceleration (a)
Computation:
F = Ma
0.10 = 500 (a)
a = 0.10 / 500
Spacecraft’s acceleration (a) = 0.10 N / 500 kg
The expression which can be used to find the spacecraft’s acceleration if its engine created a net force of 0.10 N is 0.10 N/500Kg.
We know that force is the product of the mass a body and its acceleration. The result of motion is the action of an unbalanced force. We have the following information;
Mass of the spacecraft = 500 kg
Force on the engine = 0.10 N
From Newton's law;
F = ma
F = force
m = mass
a = acceleration
a = F/m
acceleration = 0.10 N/500Kg
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For a sample of stomach acid that is 2.02×10−2 M in HCl, how many moles of HCl are in 14.6 mL of the stomach acid?
Answer:
0.0002949 moles
Explanation:
Concentration = 2.02×10−2 M
Volume = 14.6 mL = 0.0146 L (Upon converting to litres)
Number of moles = ?
These variables are related by the fllowing equation;
Concentration = Number of moles / Volume
Number of moles = Concentration * Volume
Number of moles = 2.02×10−2 * 0.0146 = 0.0002949 moles
what is the colour before and after when bromine reacts with chlorine ??
ANSWER
I need great answers
EXPLANATION
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If I make a solution by adding 83 grams of sodium hydroxide to 750 mL of water. a. What is the molality of sodium hydroxide in this solution
Answer:
2.77 mol/kg
Explanation:
Molality is a sort of concentration that indicates the moles of solute in 1kg of solvent. In this case our solvent is water and, if we consider water's density as 1g/mL, we determine that the mass of solvent is 750 g.
We convert the mass to kg → 750 g . 1kg /1000g = 0.750 kg
Our solute is the NaOH → 83 g.
We convert the mass to moles → 83 g . 1mol /40g = 2.075 mol
Molality (mol/kg) = 2.075 mol / 0.75kg = 2.77 m
Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that could be present in hard water. Assume fluoride is the only anion present that will precipitate calcium ion.
Answer:
[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:
[tex]Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)[/tex]
Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
[tex][F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\[/tex]
[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]
Best regards.
During which phase do the centromeres split, allowing the two linked chromatids to separate?
Answer:
Anaphase
Explanation:
The centromere splits during the anaphase of the cell division. Thus, allowing the two linked chromatids to separate.
A typical chromosome is made up of two sister chromatids joined together by a structure known as the centromere. During cell division - at the metaphase stage - the chromosomes align at the equator of the cell, forming the metaphase plate. The spindle from the opposing ends of the cell engages each chromosome at the kinetochore of the centromere.
At the anaphase stage, the centromere splits, leading to the separation of the sister chromatids of each chromosome. The sister chromatids of the same then start migrating in the opposite direction as a result of the shortening of the spindle fiber.
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, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC
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
If the heat released during condensation goes only to warming the iron block, what is the final temperature (in ∘C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol and a heat capacity for iron of 0.449 J⋅g−1⋅∘C−1.)
Answer:
[tex]91°C[/tex]
Explanation:
CHECK THE COMPLETE QUESTION BELOW;
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)
Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree
From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.
Then number of moles =given mass/ molecular mass
Molecular mass of water= 18 g/mol
Given mass= 0.95 g
( 0.95 g/18 g/mol)
= 0.053 moles
Then Heat evolved during condensation = moles of water x Latent heat of vaporization
Q= heat absorbed or released
H=enthalpy of vaporization for water
n= number of moles
Q=nΔH
Q = 0.053 moles x 44.0 kJ/mol
= 2.322 Kj
=2322J
We can now calculate Heat gained by Iron block
Q = mCΔT
m = mass of substance
c = specific heat capacity
=change in temperature
m = 75 g
c = 0.450 J/g/°C
If we substitute into the above formula we have
Q= 75 x 0.450 x ΔT
2322 = 75 x 0.450 x ΔT
ΔT = 68.8°C
Since we know the difference in temperature, we can calculate the final temperature
ΔT = T2 - T1
T1= Initial temperature = 22°C
T2= final temperature
ΔT= change in temperature
T2 = T1+ ΔT
= 68.8 + 22
= 90.8 °C
=91°C
Therefore, final temperature is [tex]91°C[/tex]
The final temperature of the iron block is 91∘C.
Given that;
Heat lost during condensation of the water = Heat gained by iron block
Mass of water(mw) = 0.95 g
Latent heat of vaporization = Latent heat of condensation(L) = 44.0 kJ/mol
Mass of iron(mi) = 75.0 g
Initial temperature of iron(T1) = 22∘C
Final temperature of iron(T2) = ?
Heat capacity of iron(ci) = 0.449 J⋅g−1⋅∘C−1
So;
mwL = mici(T2 - T1)
Substituting values;
(0.95g/18g/mol) × 44.0 × 10^3(J/mol) = 75.0(g) × 0.449 J⋅g−1⋅∘C−1 (T2 - 22∘C)
2322.2 = 33.7T2 - 741.4
2322.2 + 741.4 = 37.4T2
T2 = (2322.2 + 741.4)/ 33.7
T2 =91∘C
Missing parts;
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)
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A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
Answer:
The volume of the container is 59.112 L
Explanation:
Given that,
Number of moles of Oxygen, n = 3
Temperature of the gas, T = 300 K
Pressure of the gas, P = 1.25 atm
We need to find the volume of the container. For a gas, we know that,
PV = nRT
V is volume
R is gas constant, R = 0.0821 atm-L/mol-K
So,
[tex]V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L[/tex]
So, the volume of the container is 59.112 L
Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.
Answer:
The pH of the solution is 9.06.
Explanation:
The reaction of the dissociation of NH₃ in water is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq) (1)
[NH₃] - x [NH₄⁺] + x x
The concentration of NH₃ and NH₄⁺ is:
[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]
[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]
From equation (1) we have:
[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]
[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]
[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]
By solving the above equation for x we have:
x = 1.15x10⁻⁵ = [OH⁻]
The pH of the solution is:
[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]
[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]
Therefore, the pH of the solution is 9.06.
I hope it helps you!
How many kg/hr of steam are produced by a 50HP boiler?
Answer:
Explanation:
50 HP = 50 x 746 watt
= 37300 watt
= 37300 J /s
heat produced in one hour = 60 x 60 x 37300 J
= 134280 x 10³ J
latent heat of vaporization = 2260 x 10³ J / kg .
for evaporation of water of mass 1 kg , 2260 x 10³ J of heat is required .
kg of water being evaporated by boiler per hour
= 134280 x 10³ / 2260 x 10³
= 59.41 kg
rate of production of steam
= 59.41 kg / hr .
If 11.2 g of naphthalene, C10H8, is dissolved in 107.8 g of chloroform, CHCl3, what is the molality of the solution
Answer:
CHC12
Explanation:
i am not really sure i am onna do a quick research 4 u tho
Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
A particular reaction at constant pressure is spontaneous at 390K. The enthalpy change for this reaction is +23.7kJ. What can you conclude about the sign and magnitude of ΔS for this reaction?a. smallb. largec. + smalld. + largee. 0.0
Answer:
+ small
Explanation:
The entropy is obtained from;
∆S= ∆H/T
Where;
∆S= entropy of the system
∆H= enthalpy if the system = +23.7 KJ
T= absolute temperature of the system = 390 K
∆S= 23.7 ×10^3/390 = 60.8 JK^-
There is a small positive change in entropy.
For an ideal gas condition, what is the mass (g) of N2 if the pressure is 2.0 atm, the volume is 25 mL and the temperature is 290 Kelvin.
Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
clacium hydroxide is slightly soluable in water about 1 gram will dissolve in 1 liter what are the spectator ions in the reaction ions in the reaction of such a dilute solution of calcium hydroxide with hydrochloric acid
Answer:
Ca²⁺ and Cl⁻
Explanation:
In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.
In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:
Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻
The ions that react are H⁺ and OH⁻ (Acid and base producing water)
And the ions that are not reacting, spectator ions, are:
Ca²⁺ and Cl⁻Which part of an atom is mostly empty space?
A. nucleus
B. proton cloud
C. electron cloud
D. neutron
Answer:
C. Electron cloud
the electron is around 1/2000 times the size of the proton.
If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row
The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Calculate the value of Ksp for silver carbonate from this data.
Answer:
2.3 × 10⁻⁹
Explanation:
Step 1: Write the reaction for the solution of calcium oxalate
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
Step 2: Make an ICE chart
We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹
Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . What mass of ammonium phosphate is produced by the reaction of of ammonia?
Answer:
The mass of ammonium phosphate produced is 14.3g
Explanation:
Full question contains: "What mass of ammonium phosphate is produced by the reaction of 4.9g of ammonia"
Ammonium phosphate ((NH₄)₃PO₄) can be produced by the reaction of phosphoric acid (H₃PO₄) with ammonia (NH₃) as follows:
H₃PO₄ + 3NH₃ → (NH₄)₃PO₄
Where 1 mole of phosphoric acid reacts with 3 moles of ammonia producing 1 mole of ammonium phosphate.
To know how many grams of ammonium phosphate we need to find moles of ammonia that react, and, with the chemical equation we can find moles of ammonium phosphate and its mass as follows:
Moles ammonia (Molar mass: 17.031g/mol):
4.9g × (1mol / 17.031g) = 0.288 moles of ammonia you have in 4.9g
Moles of ammonium phosphate (149.09g/mol) and its mass:
As 0.288 moles of NH₃ are reacting and 3 moles of ammonia produce 1 mole of ammonium phosphate, moles produced are:
Moles (NH₄)₃PO₄:
0.288 moles NH₃ ₓ (1 mol (NH₄)₃PO₄ / 3 mol NH₃) = 0.0959 moles (NH₄)₃PO₄
These moles are, in grams:
0.0959 moles (NH₄)₃PO₄ ₓ (149.09g / mol) = 14.3g ammonium phosphate.
The mass of ammonium phosphate produced is 14.3gAutomotive air bags inflate when sodium azide decomposes explosively to its constituent elements: 2NaN3 (s) → 2Na (s) + 3N2 (g) How many grams of sodium azide are required to produce 30.5 g of nitroge
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Explanation:
The standard free energy change for a reaction can be calculated using the equation ΔG∘′=−nFΔE∘′ ΔG∘′=−nFΔE∘′ where nn is the number of electrons transferred, FF is Faraday's constant, 96.5 kJ·mol−1·V−1, and ΔE∘′ΔE∘′ is the difference in reduction potential. For each of the given reactions, determine the number of electrons transferred (n)(n) and calculate standard free energy (ΔG∘′)(ΔG∘′) . Consider the half-reactions and overall reaction for reaction 1. half-reactions:fumarate 2−+2H+CoQH2↽−−⇀succinate−↽−−⇀CoQ+2H+ half-reactions:fumarate−+2H+↽−−⇀succinate2−CoQH2↽−−⇀CoQ+2H+ overall reaction:fumarate2−+CoQH2↽−−⇀succinate2−+CoQΔE∘′=−0.009 V
Answer:
ΔG°′ = 1.737 KJ/mol
Explanation:
The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.
The overall equation of reaction is as follows:
fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ; ΔE∘′=−0.009 V
Using the equation for standard free energy change; ΔG°′ = −nFΔE°′
where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V
ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V
ΔG°′ = 1.737 KJ/mol
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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Zinc is used as a coating for steel to protect the steel from environmental corrosion. If a piece of steel is submerged in an electrolysis bath for 24 minutes with a current of 6.5 Amps, how many grams of zinc will be plated out? The molecular weight of Zn is 65.38, and Zn+2 + 2e– → Zn. Question 7 options: A) 3.17 g of Zn B) 1.09 g of Zn C) 6.34 g of Zn D) 12.68 g of Zn
Answer:
A) 3.17 g of Zn
Explanation:
Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.
Zn⁺²(aq) + 2e⁻ → Zn(s)
We can establish the following relations:
1 min = 60 s1 A = 1 C/sThe charge of 1 mole of electrons is 96,468 C (Faraday's constant).When 2 moles of electrons circulate, 1 mole of Zn is deposited.The molar mass of Zn is 65.38 g/molThe mass of Zn deposited under these conditions is:
[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]
Answer:
A.) 3.17
Explanation:
I got it right in class!
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Which gas will have the most collisions between its particles?
Answer:
The gas is Methane at 340K
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:
Chelsi has talked to her artist friends about how much money they earn each year from working in the arts. She gathers these values from seven people: [$1,500; $6,700; $2,200; $8,100; $50,500; $12,000; $2,200].
What is the median of this data set?
Answer:
The median would be 6700
Explanation:
Arrange data values from lowest to highest value
The median is the data value in the middle of the set
.
Ordering a data set x1 ≤ x2 ≤ x3 ≤ ... ≤ xn from lowest to highest value, the median x˜ is the data point separating the upper half of the data values from the lower half.
If the size of the data set n is odd the median is the value at position p where
Formula for the median
p=n+12
x˜=xp
If n is even the median is the average of the values at positions p and p + 1 where
p=n2
x˜=xp+xp+12
If there are 2 data values in the middle the median is the mean of those 2 values.
