Answer:
The correct answer is MgCO₃.5H₂O.
Explanation:
Based on the given information,
The mass of hydrate of magnesium carbonate given is 15.67 grams, and the mass of anhydrous magnesium carbonate given is 7.58 grams.
Now the mass of water will be,
= 15.67 grams – 7.58 grams
= 8.09 grams
The molecular mass of magnesium carbonate is 84.32 grams, the moles of MgCO₃ can be determined by using the formula,
Number of moles = Mass/Molecular mass
Moles of MgCO₃ = 7.58/84.32
= 0.0898 moles
Similarly, the number of moles of water will be,
Moles of H₂O = 8.099/18.02
= 0.4489 moles
After getting the values, there is a need to choose the smaller value and use it to divide both the values to get the ratio,
MgCO₃ = 0.0898/0.0898 = 1
H₂O = 0.4489/0.0898 = 4.9 or 5
Thus, the molecular formula for the hydrate will be MgCO₃.5H₂O.
A 750-mL sample of hydrogen exerts a pressure of 822 torr at 325 K. What pressure does it exert if the temperature is raised to 475 K at constant volume
Answer:
1.20 × 10³ torr
Explanation:
Step 1: Given data
Initial pressure (P₁): 822 torrInitial temperature (T₁): 325 KFinal pressure (P₂): ?Final temperature (T₂): 475 KConstant volume: 750 mLStep 2: Calculate the final pressure of the gas
Considering the constant volume, if we assume the gas behaves ideally, we can calculate its final pressure using Gay-Lussac's law.
P₁/T₁ = P₂/T₂
P₂ = P₁ × T₂/T₁
P₂ = 822 torr × 475 K/325 K = 1.20 × 10³ torr
Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone
Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
Name one major function of the organ systems listed below.
1. Skeletal System
2. Muscular System
3. Digestive System
4. Respiratory System
5. Circulatory System
6. Nervous System
7. Urinary System
Plss help I’ll give u 35 points
The organ system has been defined as the collective working of the organs for performing a specified task. Each organ system has been assigned a major task to be performed.
The function of the following organ systems have been:
Skeletal system: It has been comprised of bones and muscle. The major function of skeletal system has to maintain the body structure, and allow the interaction of muscular system to nervous system.Muscular system: It has been dealing with the voluntary and involuntary muscles. The major function has been the operation of the skeletal system by following the commands from the nervous system.Digestive system: It has been comprised of esophagus, stomach, intestine. The major function has been the breakdown of food and absorption of nutrients.Respiratory system: The respiratory system has been comprised of lungs and trachea. It has a major function of release of harmful gases, as well as oxidation and reduction of the byproducts of glycolysis.Circulatory system: It has been majorly comprises heart. The major function has been purification of blood and transport to different parts of the body.Nervous system: It has been comprised of brain and neurons, the function has been transmission of signals to the stimuli and responsible for conscious and subconscious behavior.Urinary system: The major part has been kidneys and the nephrons. The function has been the filtration of the waste liquid and urination.For more information about organ system, refer to the link:
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I need help with my math but you can only choose one correct answer
Answer:
Explanation:
Decomposition is the answer.
If a compound breaks into two or more parts then it is called decomposition.
Give two examples of each of the following:
1.) Physical systems designed by people.
2.) Naturally occurring physical systems.
3.) Social systems that you are a part of.
4.) Naturally occurring social system.
g Given that ammonia (NH3) acts as a weak base in water with a Kb of 1.8 x 10-5 at 298.0 K, calculate the pH at 298.0 K of a solution prepared by mixing 100.0 mL of a 0.050 M aqueous solution of ammonia with 20.0 mL of a 1.00 M aqueous solution of nitric acid. The final volume of the solution is 120.0 mL.
Answer:
pH = 0.90
Explanation:
The NH3 reacts with nitric acid, HNO3 as follows:
NH3 + HNO3 → NH₄⁺ + NO₃⁻
Where 1 mole of NH3 reacts per mole of HNO3.
The moles added of each reactant are:
Moles NH3:
100mL = 0.100L * (0.050mol / L) = 0.0050 moles NH3
Moles HNO3:
20mL = 0.0200L * (1.00mol / L) = 0.0200 moles HNO3
That means HNO3 is the excess reactant and, after the reaction, its moles are:
0.0200mol - 0.0050mol = 0.0150 moles HNO3
In a final volume of 120mL = 0.120L, the molar concentration of HNO3 is:
0.0150 moles HNO3 / 0.120L = 0.125M HNO3
As nitric acid is a strong acid, [HNO3] = [H+]
[H+] = 0.125M
pH = -log[H+]
pH = 0.90how many liters of hydrogen gas I needed to react with CS2 to produce 2.5 L of CH4 At STP
Answer:
9.8 L
Explanation:
The reaction that takes place is:
4H₂(g) + CS₂(g) → CH₄(g)+ 2H₂S(g)At STP, 1 mol of any gas occupies 22.4 L.
We calculate how many moles are there in 2.5 L of CH₄ at STP:
2.5 L ÷ 22.4 L/mol = 0.11 mol CH₄Then we convert CH₄ moles into H₂ moles, using the stoichiometric coefficients of the reaction:
0.11 mol CH₄ * [tex]\frac{4molH_2}{1molCH_4}[/tex] = 0.44 mol H₂Finally we calculate the volume that 0.44 moles of H₂ would occupy at STP:
0.44 mol * 22.4 L/mol = 9.8 LWhy do vinyl raincoats become brittle as they get old, even if they are not exposed to air or to any pollutants
Answer:
The substances made from polyvinyls are generally brittle and hard. This property prevails because the polymer strands present in the polyvinyls form a very regular and well-organized packing composition in the solid-state. The Van der Waals associations between the strands make the substance brittle. To make the substance soft and suitable to use as a raincoat, the incorporation of small molecules is done within the molten polymer before the hardening of the plastic.
These polymer strands act as plasticizers and prevent the strands of polymer from forming a hard and brittle conformation.
Correct order of the levels of classification in the ecosystem
Which of these half-reactions represents reduction?
I. Fe2+ → Fe3+
II. Cr2O72- → Cr3+
III. MnO4- → Mn2+
Answer: The half-reactions represents reduction are as follows.
[tex]Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}[/tex][tex]MnO^{-}_{4} \rightarrow Mn^{2+}[/tex]Explanation:
A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.
For example, the oxidation state of Cr in [tex]Cr_{2}O^{2-}_{7}[/tex] is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.
[tex]Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}[/tex]
Similarly, oxidation state of Mn in [tex]MnO^{-}_{4}[/tex] is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.
[tex]MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}[/tex]
Thus, we can conclude that half-reactions represents reduction are as follows.
[tex]Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}[/tex][tex]MnO^{-}_{4} \rightarrow Mn^{2+}[/tex]A 0.200 M solution of a week acid, HA, is 9.4% ionized. The molar concentration of H+ is 0.0188 M. the Acid-dissociation constant, Ka, for HA is...?
We are given:
Initial concentration of HA: 0.200 Molar
The acid is 9.4% ionized
Dissociation constant (α) = (Percent Ionized) / 100 = 0.094
Molar concentration of H+ = 0.0188
Let's Chill! (making the ICE box):
Reaction: HA ⇄ H⁺ + A⁻
Initial: 0.200M - -
Equilibrium: 0.200(1-α) 0.200α 0.200α
while we're here, let's confirm the given equilibrium concentration of H⁺ ions
from the table here, we can see that the equilibrium concentration of H⁺ ions is 0.200α, we know that α = 0.094
[H⁺] = 0.200α = 0.200 * 0.094 = 0.0188 M
which means that we're on the right track
We're basically scientists at this point (finding the dissociation constant):
Acid dissociation is nothing but the equilibrium constant, but for the dissociation of Acids
From the reaction above, we can write the equation of the acid dissociation constant:
Ka = [H⁺][A⁻] / [HA]
now, let's take the values from the 'equilibrium' row of the ice box the plug those in this equation
Ka = (0.200α)(0.200α) / [0.200(1-α)]
Ka = (0.200α)²/[0.200(1-α)]
plugging the value of α
Ka = (0.200*0.094)² / [0.200(0.906)]
Ka = (0.0188)² / 0.1812
Ka = 1.95 * 10⁻³
ins1502 assignment 03
Answer:
I think you forgot to post the question or picture
In the graphic,
X represents which element?
40
20
40x
20
Calcium
Ca
40
Zirconium
60
Neodymium
Nd
Zr
1. What is the oxidation state of iron in FeC2O4
Answer:
+2
Explanation:
I am first with correct answer
Answer:
the oxidation state of iron in FeC2O4 is +2
Calculate the volume of 0.07216 M AgNO3 needed to react exactly with 0.3572 g of pure Na2CO3 to produce solid Ag2CO3.
Answer:
93.4 mL
Explanation:
Let's state the reaction:
2AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2NaNO₃
We determine the moles of sodium carbonate:
0.3572 g . 1mol / 105.98g = 3.37×10⁻³ moles
Ratio is 1:2. We say:
1 mol of sodium carbonate react to 2 moles of silver nitrate
Then, our 3.37×10⁻³ moles of carbonate may react to: 3.37×10⁻³ . 2
= 6.74×10⁻³ moles
If we convert to mmoles → 6.74×10⁻³ mol . 1000 mmol / mol = 6.74 mmol
Molarity is mol/L but we can use mmol /mL
6.74 mol / volume in mL = 0.07216 M
6.74 mol / 0.07216 M = volume in mL → 93.4 mL
Silver nitrate, AgNO3, reacts with ferric chloride, FeCl3, to give silver chloride, AgCl, and ferric nitrate, Fe(NO3)3. In a particular experiment, it was planned to mix a solution containing 25.0 g of AgNO3 with another solution containing 45.0 grams of FeCl3. Write the chemical equation for the reaction.
Answer:
[tex]\mathbf{FeCl_3 + AgNO_3 \to 3 AgCl+Fe(NO_3)_3}[/tex]
Explanation:
From the given question,
The chemical equation is:
[tex]\mathbf{FeCl_3 + AgNO_3 \to 3 AgCl+Fe(NO_3)_3}[/tex]
We are given that:
mass of [tex]FeCl_3[/tex] = 45 g
number of moles of [tex]FeCl_3[/tex] = 45 g/162.2 g/mol
= 0.28 mol
mass of [tex]AgNO_3[/tex] = 25g
number of moles of [tex]AgNO_3[/tex] = 25/169.87
= 0.147 mol
From the given equation 1 mole of [tex]FeCl_3[/tex] is required to make 3 mole of [tex]AgNO_3[/tex]
0.28 mole of [tex]FeCl_3[/tex] = 0.28 × 3 = 0.84 mol of [tex]AgNO_3[/tex]
Here [tex]AgNO_3[/tex] is the limiting reagent.
Thus,
3 mole of [tex]AgNO_3[/tex] = 3 moles of AgCl
0.147 mole of [tex]AgNO_3[/tex] = 0.147 mole of AgCl produced.
H
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C C
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CH H
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C C
1
NO ,
2
BY
Answer:
hakdog
Explanation:
pangitag lainbbbsjsbshsisbsjsbshxx
compare and contrast the three types oflevers.
Answer:
The difference between the three classes depends on where the force is, where the fulcrum is and where the load is. In a first class lever, the fulcrum is located between the input force and output force. In a second class lever, the output force is between the fulcrum and the input force. write the class of lever.
Explanation:
Calculate the pOH in a solution with [H+] =9.8 • 10-3 M
The pOH of the solution containing 9.8×10⁻³ M of hydrogen ion concentration, [H⁺] is 11.99
How do i determine the pOH of the solution?First, we shall determine the hydroxide ion concentration [OH⁻] in the solution. Details below:
hydrogen ion concentration, [H⁺] = 9.8×10⁻³ MHydroxide ion concentration [OH⁻] =?[H⁺] × [OH⁻] = 10¯¹⁴
9.8×10⁻³ × [OH⁻] = 10¯¹⁴
Divide both side by 9.8×10⁻³
[OH⁻] = 10¯¹⁴ / 9.8×10⁻³
[OH⁻] = 1.02×10⁻¹² M
Finally, we shall determine the pOH of the solution. Details below:
Hydroxide ion concentration [OH⁻] = 1.02×10⁻¹² MpOH of solution =?pOH = -Log [OH⁻]
pOH = -Log 1.02×10⁻¹²
pOH = 11.99
Thus, from the above calculation, we can conclude that the pOH of the solution is 11.99
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the solubility of alcohol decrease with increasing numbers of carbon atom explain
Explanation:
bdjdiififjc mfkxkdkd has also had a good time in the
Please type out all of your calculations for this dilution equation: In your vitamin C experiment, it calls for a 5% concentration of iodine. However, your 7 fluid oz. tincture of iodine contains 70% iodine. How would you dilute this
Answer:
The original7 fluid oz. 70% tincture of iodine is diluted to 98 fluid oz. to obtain a 5% solution.
Explanation:
Dilution is a technique employed in experimental sciences such as chemistry and biochemistry as well as in medicine to obtain a less concentrated solution from a more concentrated one.
The dilution formula is the most important formula required in dilution. The dilution formula is given as: C1V1 = C2V2
Where C1 is the initial concentration of the stock solution;
V1 is the volume of the stock solution required;
C2 is the final concentration of the diluted solution to be prepared;
V2 is the final volume of the diluted solution.
Using the dilution formula to determine the answer to the above question:
C1 = 70% = 0.7; V1 = 7 fluid oz.; C2 = 5% = 0.05; V2 = ?
To determine V2, it is made subject of the dilution formula:
V2 = C1V1/C2
V2 = (0.7 × 7) / 0.05
V2 = 98 fluid oz.
Therefore, the original7 fluid oz. 70% tincture of iodine is diluted to 98 fluid oz. to obtain a 5% solution.
To which family of organic compounds does CH3COCH2CH2CH3 belong?
Answer:
I think:
CH3COCH2CH2CH3 belongs to ketone because it has a carbonyl group C=O in its structure.
need it asap please!!
If a 30.0 mL of a 0.20 mol/L HOCN is titrated with 0.250 mol/L KOH,
What is the pH of the solution if 10.0 mL KOH is added?
Answer:
[tex]1 \: l \: contains \: 0.20 \: moles \: of \: HOCN \\ 0.03 \: l \: will \: contain \: (0.03 \times 0.20) \: moles \\ = 0.006 \: moles \: of \: HOCN \\ 1 \: l \: contains \: 0.250 \: moles \: of \: potassium \: hydroxide \\ 0.01 \: l \: will \: contain \: (0.01 \times 0.250) \\ = 0.0025 \: moles \: of \: potassium \: hydroxide \\ KOH + HOCN → KCN + H _{2}O \\ k _{b} = \frac{[K {}^{ + } ][OH {}^{ - } ]}{[KOH]} \\ moles \: of \: OH {}^{ - } \: remaining = 0.006 - 0.0025 \\ = 0.0035 \: moles \\ total \: volume = 0.03 + 0.01 \\ = 0.04 \: litres \\ 0.04 \: l \: contains \: 0.0035 \: moles \\ 1 \: l \: will \: contain \: ( \frac{1 \times 0.0035}{0.04} ) \\ = 0.0875 \: mol \: per \: litre \\ [H {}^{ + } ] = \frac{kw}{[OH {}^{ - } ]} \\ = \frac{1 \times {10}^{ - 14} }{0.0875} \\ = 1.143 \times {10}^{ - 13} \\ pH = - log(1.143 \times {10}^{ - 13} ) \\ pH = 12.94[/tex]
Bromine is a red-brown liquid with a density of 3.10 g/mL. A sample of bromine weighing 88.5 g occupies what volume
Answer:
28.55 mL or 0.02855 L
Explanation:
Applying,
Density(D) = mass(m)/Volume(V)
D = m/V............... Equation 1
Where D = Density of the Bromine, m = mass of the bromine, V = volume of the bromine
make V the subeject of the equation
V = m/D............. Equation 2
From the question,
Given: m = 88.5 g, D = 3.10 g/mL
Substitute these values into equation 2
V = 88.5/3.10
V = 28.55 mL
Hence the volume occuppied by the Bromine gas is 28.55 mL or 0.02855 L
55
Tam pours salt water into one beaker and then pours acid into another beaker.
Tam realises that the liquids are not labelled.
What should Tam do?
A
Smell each liquid in order to identify them.
В.
Guess which beaker contains each liquid.
С
Mix the liquids together to find out what they are.
D
Safely throw away the liquids and get new ones.
3
PM
2021
Answer:
correct answer i C i think so
Answer:
Safely throw away the liquids and get new ones.
Explanation:
Not A because
you shouldn't smell them cause it could be toxic and hurt you
Not B
you should never guess which chemical are which cause if your not correct and use the wrong one you could have a chemical reaction
Not C
you cant mix random liquids it could make a chemical reaction to exploded ore worse create a hurtful gas
5.
When 0.421 g of BaCl2:2H20 and 0.722g of Na3PO4.12H2O mix with water
forming 500 mL of solution, how many grams of Ba3(PO4)2 precipitate?
Explanation:
2H2O and 0.722g of Na3PO4.12H20 mix with water forming 500 mL of solution, how many grams of Ba3(. ... Forming 500 ML Of Solution, How Many Grams Of Ba3(PO4)2 Precipitate?
When 0.421 g of BaCl₂ · 2 H2O and 0.722g of Na₃PO₄ · 12 H₂O mix with water forming 500 mL of solution then 0.003 mol/L of Ba₃(PO4)₂ precipitate.
What is Stoichiometry ?Stoichiometry is an important concept which helps us use the balanced chemical equation to measures the amount of reactants and products.
What is Molarity ?Molarity of solution is the number of moles of solute per liter of solution. Molarity is also called Molar Concentration.
Molarity (M) = [tex]\frac{\text{number of moles}}{\text{volume of solution (in liters)}}[/tex]
How to find the number of moles ?Number of moles = [tex]\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
First we have to write the balanced chemical equation
2Na₃PO₄ · 12 H₂O + 3BaCl₂ · 2 H2O → Ba₃(PO4)₂ + 6 NaCl + 30 H₂O
Here 3 moles of BaCl₂ · 2 H2O produce 1 mole of Ba₃(PO4)₂
So,
Number of moles of BaCl₂ · 2 H2O = [tex]\frac{\text{Mass of}\ BaCl_2 . 2H_2O}{\text{Molar mass of}\ BaCl_2. 2H_2O}[/tex]
= [tex]\frac{0.421}{244.26}[/tex]
= 0.017
Number of moles of 2Na₃PO₄ · 12 H₂O = [tex]\frac{\text{Mass of}\ Na_3PO_4. 12H_2O}{\text{Molar mass of}\ Na_3PO_4 . 12H_2O}[/tex]
= [tex]\frac{0.722}{380.12}[/tex]
= 0.002
Therefore, Na₃PO₄ · 12 H₂O is the limiting reagent.
Now, put the value in above formula to find the molarity
Molarity (M) = [tex]\frac{\text{number of moles}}{\text{volume of solution (in liters)}}[/tex]
= [tex]\frac{0.0017}{0.5}[/tex]
= 0.003 mol/L
Thus, from above conclusion we can say that 0.003 mol/L of Ba₃(PO4)₂ precipitate.
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A potted plant is placed under a grow lamp, which provides 6,400 J of energy to the plant and the soil over the course of an hour. The specific heat capacity of the soil is about 0.840 J/g°C and the temperature goes up by 9.25°C of soil. How many grams of soil are there?
Answer:
823.7g
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
Using the information given in this question as follows:
Q = 6,400 J
m = ?
c of soil = 0.840 J/g°C
∆T = 9.25°C
Using Q = mc∆T
m = Q ÷ c∆T
m = 6,400 ÷ (0.840 × 9.25)
m = 6400 ÷ 7.77
m = 823.7g
Answer:
There are 823.68 grams.
Explanation:
Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.
Between heat and temperature there is a relationship of direct proportionality. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body. In summary, the amount of heat Q that receives or transmits a mass m of a substance with specific heat C to raise its temperature from T1 to T2 is given by the formula:
Q= C*m* (T2- T1) = C*m* ΔT
In this case:
Q= 6400 JC= 0.840 [tex]\frac{J}{g*C}[/tex]m= ?ΔT= 9.25 CReplacing:
6400J= 0.840 [tex]\frac{J}{g*C}[/tex] *m* 9.25 C
Solving:
[tex]m=\frac{6400 J}{0.840 \frac{J}{g*C} *9.25 C}[/tex]
m=823.68 grams
There are 823.68 grams.
Brainiest and 10 points
Which has MORE energy?
A. Radio
B. Infrared
Answer:
B. infrared
hope this helps
have a good day :)
Explanation:
what is the shape of the TeF5 anion
Answer:
There are four atoms and two lone pairs around the central atom, which corresponds to AX4E2 or square planar. The shape of XeF4 is square planar.
Explanation:
I hope it's help u
A student wishes to determine the chloride ion concentration in a water sample at 25 °C using a galvanic cell constructed with a graphite electrode and a half-cell of AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq) E°red = 0.2223 V And a copper electrode with 0.500 M Cu²⁺ as the second half cell Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red= 0.337 V The measured cell potential when the water sample was placed into the silver side of the cell was 0.0925 V. What is the standard cell potential for this cell in V?
The question is incomplete, the complete question is;
And a copper electrode with 0.500 M Cu²⁺ as the second half cell
Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red= 0.337 V
The measured cell potential when the water sample was placed into the silver side of the cell was 0.0925 V.
A- What is the standard cell potential for this cell in V?
B- What is the value of the standard free energy (in kJ) for this reaction?
C- Write the balanced equation for the overall reaction in acidic solution?
D- And the measured cell potential is 0.0925, what is the concentration of chloride ions in the solution?
Answer:
See Explanation
Explanation:
a) E°cell = E°cathode - E°anode
E°cell = 0.337 V - 0.2223 V
E°cell = 0.1147 V
b) ΔG°cell = −nFE°cell
Where n=2 and F = 96500C
ΔG°cell =-(2 * 96500 * 0.1147 )
ΔG°cell =-22,137.1 J or -22.1371 KJ
c) 2Ag(s) + 2Cl⁻(aq) + Cu²⁺(aq) -----> 2AgCl(s) + Cu(s)
d) From Nernst Equation;
E= E°cell - 0.0592/n log Q
0.0925 = 0.1147 - 0.0592/2 log 1/[0.500] [Cl⁻]^2
0.0925 - 0.1147 = - 0.0592/2 log 1/[0.500] [Cl⁻]^2
-0.0222 = -0.0296 log 1/[0.500] [Cl⁻]^2
-0.0222/-0.0296 = log 1/[0.500] [Cl⁻]^2
0.75 = log 1/[0.500] [Cl⁻]^2
Antilog (0.75) = 1/[0.500] [Cl⁻]^2
5.6234 * 0.500 = [Cl⁻]^2
[Cl⁻] = √2.8117
[Cl⁻] = 1.68 M
A galvanic cell is a voltaic cell that produces electrical energy from the oxidation-reduction process. The standard cell potential of the cell is 0.1147 V.
What is standard cell potential?The standard cell potential is a difference between the electrode potential of the cathode and the anode of the cell.
The balanced overall reaction of the cell is given as,
[tex]\rm 2Ag(s) + 2Cl^{-}(aq) + Cu^{2+}(aq) \rightarrow 2AgCl(s) + Cu(s)[/tex]
The standard cell potential is calculated as:
[tex]\begin{aligned} \rm E^{\circ}cell &= \rm E^{\circ}cathode - E^{\circ} anode\\\\&= 0.337 - 0.2223 \\\\&= 0.1147\;\rm V\end{aligned}[/tex]
Therefore, 0.1147 V is the standard cell potential for the cell.
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