Answer:
0.188 moles
Explanation:
moles=mass/molar mass
moles= 19.2/ (32+32+38)
moles= 0.188
Answer:
0.188 moles.
Explanation:
Molar mass of SO2F2
= 32 + 2*16 + 2 * 19
= 102 g.
So 19.2g = 19.2 / 102
= 0.188 moles.
43. Calculate the equilibrium constant at the temperature given. (a) O2 (g) + 2F2 (g) ⟶ 2F2 O(g) (T = 100 °C) (b) I2 (s) + Br2 (l) ⟶ 2IBr(g) (T = 0.0 °C) (c) 2LiOH(s) + CO2 (g) ⟶ Li2CO3 (s) + H2 O(g) (T = 575 °C) (d) N2 O3 (g) ⟶ NO(g) + NO2 (g) (T = −10.0 °C) (e) SnCl4 (l) ⟶ SnCl4 (g) (T = 200 °C)
Answer:
Explanation:
O₂(g) + 2F₂(g) ↔ 2F₂(g)
Stabdard ΔG values are
[tex]\Delta G_f[F_2O]=41.9kJ/mol =41900J/mol[/tex]
[tex]\Delta G_f[O_2]=0\\\\ \Delta G_f[F_2]=0[/tex]
[tex]\Delta G^0=\sum \Delta G^\circ (products)- \sum \Delta G ^\circ (reactants)[/tex]
[tex]\Delta G^\circ = [2 \times 41900]-0\\\\=83800J/mol[/tex]
Now,
[tex]\Delta G^\circ =-RTInK[/tex]
Given T = 100°C
= 100+ 273.15 = 373.15K
R = 8.314J/k / mol
so,
83800 = -8.314 * 373.15 * InK
InK = -27.0116
K = 1.858 * 10⁻¹²
Equilibrium constant = 1.858 * 10⁻¹²Oxygen in air can be removed using
Answer:Fractional distillation of air. About 78 per cent of the air is nitrogen and 21 per cent is oxygen. These two gases can be separated by fractional distillation of liquid air.
Explanation:
100 mL of a buffer that consists of 0.20 M NH3 and 0.20 M NH4Cl is titrated with 25 mL of 0.20 M HCl. Calculate the pH of the resulting solution given that the Kb for NH3 is 1.8 x 10-5.
Answer:
pH = 9.03
Explanation:
The equilibrium of the NH₄Cl / NH₃ buffer in water is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Initial moles of both NH₃ and NH₄⁺ are:
0.100L ₓ (0.20 mol / L) = 0.0200 moles
The NH₃ reacts with HCl producing NH₄⁺, thus:
NH₃ + HCl → NH₄⁺ + Cl⁻
That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺
Moles added of HCl were:
0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:
NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles
NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.
Using H-H equation for bases:
pOH = pKb + log [NH₄⁺] / [NH₃]
Where pKb is -log Kb = 4.745.
Replacing:
pOH = 4.745 + log 0.0250mol / 0.0150mol
pOH = 4.967
As pH = 14- pOH
pH = 9.03
Consider two solutions: one formed by adding 10.0 grams of glucose (C6H12O6) to 1.0 L of water
and another formed by adding 10.0 grams of sucrose (C12H22O11) to 1.0 L of water. Which of the
two solutes produces the lower reduction in the vapor pressure of the solvent? Explain fully.
Answer:
Explanation:
Lowering of vapour pressure of solvent is proportional to number of moles of solute dissolved in it per litre .
No of moles of glucose dissolved
= mass dissolved / molecular weight of glucose
= 10 / 180
= .055 moles.
No of moles of sucrose dissolved = 10 / 342
= .029 moles.
So reduction in vapour pressure will be lower of solution dissolving sucrose . It is so because , no of moles of solute dissolved in it is low.
Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.
Answer:
pH = 7.1581
Explanation:
The equilibrium of NaHSO₃ with Na₂SO₃ is:
HSO₃⁻ ⇄ SO₃²⁻ + H⁺
Where K of equilibrium is the Ka2: 6.5x10⁻⁸
HSO₃⁺ reacts with NaOH, thus:
HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺
As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:
HSO₃⁻: 0.252 moles
SO₃²⁻: 0.139 moles
Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:
0.0500L ₓ (1mol /L): 0.050 moles of NaOH.
Thus, final moles of both compounds are:
HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles
SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles
Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:
pH = pka + log [SO₃²⁻] / [HSO₃⁻]
Where pKa is - log Ka = 7.187
Replacing:
pH = 7.187 + log [0.189] / [0.202]
pH = 7.1581A solution is made by mixing 10.2 grams of CaCl2 in 250 grams of water. What is the molality of the
solution?
Answer:
The molality of this solution is 0.368 molal
Explanation:
Step 1: Data given
Mass of CaCl2 = 10.2 grams
Molar mass of CaCl2 = 110.98 g/mol
Mass of water = 250 grams = 0.250 kg
Step 2: Calculate the number of moles CaCl2
Moles CaCl2 = mass CaCl2 / molar mass CaCl2
Moles CaCl2 = 10.2 grams / 110.98 g/mol
Moles CaCl2 = 0.0919 moles
Step 3: Calculate the molality of the solution
Molality solution = moles CaCl2 / mass water
Molality solution = 0.0919 moles / 0.250 kg
Molality solution = 0.368 mol / kg = 0.368 molal
The molality of this solution is 0.368 molal
Answer:
0.367 m
Explanation:
We have to start with the molality equation:
[tex]m=\frac{mol~of~solute}{Kg~of~solvent}[/tex]
If we want to calculate the molality we need the moles of the solute and the Kg of the solvent. In this case the solute is [tex]CaCl_2[/tex] and the solvent is [tex]H_2O[/tex].
Lets start with the the moles of solute. For this calculation we need the molar mass of [tex]CaCl_2[/tex] ([tex]111~g~CaCl_2[/tex]) and the mass, so:
[tex]10.2~g~CaCl_2\frac{1~mol~CaCl_2}{111~g~CaCl_2}[/tex]
[tex]0.0918~mol~CaCl_2[/tex]
Now, we can calculate the Kg of solvent, if we do the conversion from grams to Kg:
[tex]250~g~CaCl_2\frac{1~Kg}{1000~g}[/tex]
[tex]0.25~Kg[/tex]
With these values we can calculate the "molality":
[tex]m=\frac{0.0918~mol~CaCl_2}{0.25~Kg}[/tex]
[tex]m=0.3672[/tex]
I hope it helps!
An unknown substance has a mass of 15.5 g . When the substance absorbs 1.395×102 J of heat, the temperature of the substance is raised from 25.0 ∘C to 45.0 ∘C . What is the most likely identity of the substance?
Answer:
iron (Fe)
Explanation:
Use q = mcΔT.
q = 1.396 x 10^2 J
m = 15.5g
ΔT = (45.0 - 25.0) = 20 degrees celcius
1.396 x 10^2 = 15.5 x c x 20
c = .450 J/g x degrees celcius
This is closest to the specific heat of iron/steel.
516 mL of a 3.82 M sodium sulfate (Na2S04) solution is diluted with 0.875 L of water. What is the new concentration in molarity?
Answer
Molarity = [tex]1.4mol/L[/tex]
Explanation:
Molarity provides the number of moles of solute per liter of solution (moles/Liter). It is a means by which concentration of solution is measured.
SEE THE ATTACHMENT BELOW FOR STEP BY STEP SOLUTION.
Answer:
1.42 M
Explanation:
In this case have a dilution problem, therefore we need to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
What values we have?
[tex]C_1=3.82M[/tex]
[tex]V_1=526mL(0.516L)[/tex]
[tex]C_2=?[/tex]
[tex]V_2=?[/tex]
Now, we can calculate [tex]V_2[/tex] if we add the volumes, so:
[tex]0.516~L+~0.875~L=1.391~L[/tex]
So, [tex]V_2=1.391~L[/tex]
We can plug the values in the equation:
[tex]3.82~M*0.516~L=C_2*1.391~L[/tex]
[tex]C_2=\frac{3.82~M*0.516~L}{1.391~L}[/tex]
[tex]C_2=1.42~M[/tex]
I hope it helps!
Given that a 15.00 g milk chocolate bar contains 9.500 g of sugar, calculate the percentage of sugar present
in 15.00 g of milk chocolate bar keeping in mind that the answer should have four significant figures (two
decimal places).
Answer:
63.33%
Explanation:
To find the percentage of sugar in the milk chocolate just divide 15 by 9.5. This will give you 0.63333 (recurring). To find the percentage, times it by 100, which will give you 63.33%.
Hope this helped :)
The percentage of sugar present in the milk would be 63.33 %.
What are significant figures?In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.
As given in the problem a 15.00 g milk chocolate bar contains 9.500 grams of sugar, and we have calculated the percentage of sugar present,
The percentage of sugar present in the milk = 9.500 ×100/15
= 63.33 %
Thus, the percentage of sugar present in the milk would be 63.33 %.
To learn more about significant figures here, refer to the link;
brainly.com/question/14359464
#SPJ2
need help asappp please
Answer:
C.POLARExplanation:
According to research :
Methanol is a polar molecule:
The alcohol (-OH) group dominates the molecule making it definitely polar.
When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to effuse as the same amount of CO2 under the same conditions of temperature and pressure. What is the molar mass of the phosphorus chloride
Answer:
molar mass of the phosphorus chloride = 138.06 g/mol
Explanation:
mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.
mass = 3.51 g
lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec
From this we can say that
rate of effusion of CO2 = 3.51/1 = 3.51 g/s
rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s
From graham's equation of effusion,
[tex]\frac{Rc}{Rp}[/tex] = [tex]\sqrt{\frac{Mp\\}{Mc} }[/tex]
Rc = rate of effusion of CO2 = 3.51 g/s
Rp = rate of effusion of phosphorus chloride = 1.98 g/s
Mc = molar mass of CO2 = 44.01 g/mol
Mp = molar mass of the phosphorus chloride = ?
Imputing values into the equation, we have
[tex]\frac{3.51}{1.98}[/tex] = [tex]\sqrt{\frac{Mp\\}{44.01} }[/tex]
1.77 = [tex]\frac{\sqrt{Mp} }{6.64}[/tex]
11.75 = [tex]\sqrt{Mp}[/tex]
Mp = [tex]11.75^{2}[/tex]
Mp = molar mass of the phosphorus chloride = 138.06 g/mol
There is a pattern to the phases of the moon. The time between one full moon to another is about a;new b;waxing crescent c;1st quarter d;full
Answer: a. new
Explanation:
The time between one full moon to another is new moon. The lunation is the average time taken for the formation of new moon to the next. The new moon is the initial visible crescent of the Moon after being conjugated with the Sun. Per lunar cycle can be assigned with a lunation number for the purpose of identification.
How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Answer:
Grams of mercury= 0.06 g of Hg
Note: The question is incomplete. The complete question is as follows:
A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Explanation:
Since one fluorescent light bulb contains 4 mg of mercury,
15 such bulbs will contain 15 * 4 mg of mercury = 60 mg
1 mg = 0.001 g
Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.
Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.
Total amount of mercury in 15 compact fluorescent light bulbs is 0.06 gram of mercury.
Compact fluorescent light bulbs and mercury:What information do we have?
Number of compact fluorescent light bulbs = 15 bulb
Amount of mercury in each bulb = 4 mg
Total amount of mercury = Number of compact fluorescent light bulbs × Amount of mercury in each bulb
Total amount of mercury = 15 × 4
Total amount of mercury = 60 mg
Total amount of mercury = 60 / 1000
Total amount of mercury = 0.06 gram of mercury
Find out more information about 'Mercury'.
https://brainly.com/question/2279846?referrer=searchResults
What volume does 49.0g of KBr need to be dissolved in to make a 8.48% (w/v) solution?
Answer:
578mL is the volume you need to dissolve the mass of KBr
Explanation:
The percent by mass volume, % (w/v) is defined as the ratio between mass of solute in grams and volume of the solution in mililiters times 100. The formula is:
% (w/v) = mass solute (g) / volume solution (mL) × 100
As you want a solution 8.48% and the solute is 49.0g of KBr:
8.48% = 49.0 (g) / volume solution (mL) × 100
Volume solution = 578mL is the volume you need to dissolve the mass of KBr
How many moles of hydrogen atoms are there in 120 g of C6H12O6
Answer: 8.0 moles
Explanation:
0.6661 moles×12 H≈8.0 moles
120 g of C₆H₁₂O₆ contain 8 moles of Hydrogen atoms
To obtain the answer to the question given above, we'll begin by calculating the mass of 1 mole of C₆H₁₂O₆
1 mole of C₆H₁₂O₆ = (12×6) + (12×1) + (16×6)
= 72 + 12 + 96
= 180 gFrom the above,
We can see that 180 g of C₆H₁₂O₆ contains 12 moles of Hydrogen.
Finally, we shall determine the number of mole of Hydrogen atoms in 120 g of C₆H₁₂O₆. This can be obtained as follow:
180 g of C₆H₁₂O₆ contains 12 moles of Hydrogen.
Therefore,
120 g of C₆H₁₂O₆ will contain = [tex]\frac{120 * 12}{180}[/tex] = 8 moles of Hydrogen.
Thus, 120 g of C₆H₁₂O₆ contain 8 moles of Hydrogen atoms
Learn more: https://brainly.com/question/24707125
Select the correct answer from each drop-down menu.
Describe an oxidation-reduction reaction.
In an oxidation-reduction reaction, oxidation is what happens when a reactant___ . Reduction occurs when a reactant ___ in the reaction.
Explanation:
First off, we have to understand what we means by oxidation and reduction.
Oxidation is the loss of electrons during a reaction by a molecule, atom or ion. Oxidation occurs when the oxidation state of a molecule, atom or ion is increased.
Reduction on the other hand is the opposite of oxidation. It is the gain of electrons during a reaction by a molecule, atom or ion. Reduction occurs when the oxidation state of a molecule, atom or ion decreases.
Back to the question;
oxidation is what happens when a reactant loses electrons.
Reduction occurs when a reactant gains electrons in the reaction.
Which is a substance that could be found in air , water , or soil that is harmful to humans or animals?
Answer: Pollutant
Explanation:
Answer:
Explanation:
The substance that could be found in air, water or soil that is harmful to humans or animals is called pollutant.
A chemistry student needs of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of ethanolamine is . Calculate the mass of ethanolamine the student should weigh out. Round your answer to significant digits.
Answer:
mass of ethanolamine 61.2 g
Explanation:
Data:
V (Volume) = 60 mL
ρ (Density)= 1.02 [tex]\frac{g}{cm^{3} }[/tex] = 1.02 [tex]\frac{g}{mL}[/tex]
With the given data and using the density formula:
ρ = [tex]\frac{mass}{volume}[/tex]
It is possible to calculate the grams of ethanolamine that the student should weigh out. The mass in the formula is cleared and the data is replaced.
mass = ρ * volume
mass = 1.02 [tex]\frac{g}{mL}[/tex] * 60 mL = 61.2 g of ethanolamine
Suppose of potassium iodide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it.
The given question is incomplete, the complete question is:
Suppose 1.27 g of potassium iodide is dissolved in 100. mL of a 44.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it. Round your answer to 3 significant digits.
Answer:
The correct answer is 0.0325 M.
Explanation:
The mass of potassium iodide or KI mentioned in the question is 1.27 grams, the molar mass of KI is 166 g/mol. The formula for determining the no of moles of the substance is mass/molar mass. Thus, the moles of KI in 1.27 grams will be,
= 1.27g / 166 g/mol = 0.00765 moles.
KI = K⁺ + I⁻
Therefore, the moles of KI will be equivalent to moles of iodide anion, that is, 0.00765 moles.
The moles of silver nitrate or AgNo3 in the solution can be determined by using the formula, molarity (M) * volume in liters. The molarity of silver nitrate given in the question is 44 mM and the volume used is 100 ml or 100/1000 L. Now putting the values we get,
= (44 M/1000) * (100 L/1000) = 0.0044 moles
The moles of silver nitrate is equivalent to moles of silver ion, which is further equivalent to the moles of iodide ion that has taken part in precipitation = 0.0044 moles.
The moles left of I⁻ in the solution will be,
0.00765 - 0.0044 = 0.00325
Now, the final molarity of iodide ion in the solution will be,
= moles/volume in liters
= 0.00325 moles / 0.100 L = 0.0325 M
Interpret and describe an example of a chemical formula. Summarize the two parts a chemical formula must have.
Answer and Explanation:
A chemical formula is composed by chemical symbols (in letters, which indicate the chemical elements) and subscripts (numbers). For example, let see the chemical formula for carbon dioxide:
CO₂
The letters C and O are the chemical symbols for the elements carbon (C) and oxygen (O). The number 2 in subscript indicates that there are 2 atoms of O per molecule. The subscript 1 is generally not indicated in chemical formulae, so it is assumed that the number of atoms of C is 1.
Summarizing, the chemical formula CO₂ indicates us that the molecule if formed by 1 atom of the element carbon (C) and 2 atoms of the element oxygen (O).
If you are diving and playing in the waves near a beach and suddenly find that you are being
pulled out into the ocean at a relatively quick pace, what current are you most likely trapped in?
Answer: Wind current
Explanation:
The wind current is the main driving force along with the water currents when one reaches to the seashore. The wind currents above surface of water generates due to the evaporation of water.
The strong wind currents can pull any animal or human walking or running near the sea shore. These currents bring the animal or human towards the sea.
A student wishes to prepare 65.0 mL of 0.875 M HCl from 12.0 M HCl.
What volume of the
12.0 M HCl should he/she start with?
A student should start with 4.74 mL of the 12.0 M HCl to prepare 65.0 mL of 0.875 M HCl from 12.0 M HCl.
What is Molarity ?Molarity (M) is the amount of a substance in a certain volume of solution.
Molarity is defined as the moles of a solute per liters of a solution.
Molarity is also known as the molar concentration of a solution.
Formula used for the given question ;
M₁V₁ = M₂V₂
M₁ = 12.0 M HCl.V₁ = ?M₂ = 0.875 M HCV₂ = 65.0 mLM₁V₁ = M₂V₂
V₁ = 0.875M x 65.0 mL / 12.0 M HCl.
= 4.74 ml
Therefore , A student should start with 4.74 mL of the 12.0 M HCl to prepare 65.0 mL of 0.875 M HCl from 12.0 M HCl.
Learn more about Molarity here ;
https://brainly.in/question/11172828
#SPJ2
Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.50 M HC2H3O2, 0.10 M NaC2H3O2 Which of the above buffers has the highest buffer capacity? [ Select ] Which of the above buffers has the lowest buffer capacity? [ Select ]
From the data provided;
Buffer B has the highest buffer capacityBuffer C has the lowest buffer capacityWhat is a buffer?A buffer is a solution which resists changes to its pH when a small quantity of acid or base is added to it.
The buffer capacity is determined using the Henderson-Hasselbach equation:
pH = pKa + log ([CH3COO-]/[CH3COOH])For Buffer A: pH = 4.74 + log(0.10/010) = 4.74
Buffer B: pH = 4.74 + log(0.30/030) = 4.74
For Buffer C: pH = 4.74 + log (0.10/0.50) = 4.04
Buffer A and Buffer B has same pH value as the pKa of acetic acid.
However, Buffer B has higher concentration of the components compared to buffer A, therefore, Buffer B has the highest buffer capacity.
The pH of buffer C is lower than pKa of acetic acid. Hence, buffer C has the lowest buffer capacity.
Learn more about buffers at: https://brainly.com/question/22390063
Is electricity an essential property of matter ?
Answer: I think so.. (sorry if i am wrong)
Explanation: Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. A physical property is a characteristic of matter that is not associated with a change in its chemical composition.
An increase in the temperature of reactant causes an increase in the rate of reaction.
Which of the following is the best explanation?
a)the concentration of reactant increases
b) the activation energy decreases
c) the collision frequency increases
d) the fraction of collision with total kinetic energy larger than activation energy
increases
Answer:
d) The fraction of collision with total kinetic energy larger than activation energy increases.
Explanation:
Hello,
In this case, kinetic models explain how the rate of a chemical reaction is affected by several factors. In such a way, specifically for temperature, when it increases, the average velocity of the particles is also increased, for that reason, the collision frequency increases since the molecules are more likely to collide as they move faster and encounter to each other.
Nonetheless, it is the minor reason because the main reason is that the effective collisions increase when the temperature is increased, and they are related with the fraction of collision with total kinetic energy that turns out larger than the activation energy, therefore, answer is d).
Best regards.
Which phrase best describes heat?
A. the energy that an object has as a result of its temperature
B
the average translational kinetic energy of the particles in an object
C
the energy transferred between objects at different temperatures
D
the total amount of energy possessed by the particles in an object
Answer: A
Explanation: Heat is the temperature of an object.
Answer:
Option C
the energy transferred between objects at different temperature.
An electrochemical cell contains a standard hydrogen electrode and a cathode consisting of a metallic chromium electrode, Cr(s), in contact with a 1.00 M chromium solution, Cr3 (aq). The voltage produced by this cell was measured at 25oC. Which statements describe the results of this measurement, assuming the conditions are ideal? The cell voltage with the appropriate sign equals
I. the cell potential.
II. the electromotive force.
III. the standard cell potential.
IV. the standard reduction potential for Cr3+/Cr.
a. I only
b. l and Il
c. I, II and III
d. I, II, II, and IV
e. ll only
Answer:
c. I, II and III
Explanation:
The cell is as follows
Cr / Cr⁺²(1M) // H⁺ ( 1 M ) / H₂
Standard reduction potential of hydrogen cell is zero . Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell
Cell potential = Ecathode - Eanode
= 0 - ( - E)
= E
E is cell potential and also standard cell potential or emf of the cell .
Standard reduction potential that is for Cr3+/Cr. is - E .
Hence statement I , II , III are right . IV th statement is wrong because of sign
Option c is correct.
The branch of chemistry which deals with electricity is called electrochemistry.
The correct answer is C
The cell representation is as follows:-
[tex]Cr / Cr^{2+}(1M) // H^+ ( 1 M ) / H_2[/tex]
The standard reduction potential of hydrogen cells is zero. Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell because it will behave as a cathode.
The formula of the electric cell is as follows :-[tex]Cell potential = E_{cathode} - E_{anode[/tex]
After putting the value the cell potential will be:-
Cell potential = 0 - ( - E)
The cell potential will be= E
The standard reduction potential that is for [tex]Cr^{3+}/Cr = - E .[/tex]
Hence, the correct option is C that is I, II, and IV.
For more information, refer to the link:-
https://brainly.com/question/25026730
How does the disturbance travel through the coil when you raise your arm up and down?
Answer:
What happens to the wavelength when you increase the rate of vibration (how fast your hand moved back and forth). After increasing the rate of vibration, the wavelength decreased.\
PLEASE HELP I HAVE LIMITED TIME!!
How many particles are in 0.500 moles of
N.05?
Answer:
3.01×10²³ particles.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles. This implies that 1 mole of N2O5 also contains 6.02×10²³ particles.
Now, if 1 mole of N2O5 contains 6.02×10²³ particles ,
Then 0.5 mole of N2O5 will contain = 0.5 × 6.02×10²³ = 3.01×10²³ particles.
Therefore, 0.5 mole of N2O5 contains 3.01×10²³ particles.
An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:
Answer:
101
Explanation:
Provided that
[tex]S_1 = S_2 = same\ V_{max}[/tex]
And,
[tex]S_1\ k_M = 2.0mM\\S_2\ k_M = 20mM[/tex]
Now we expect the same
{S} (0.1mM)
This determines that [tex]S_1[/tex] generates a higher rate of product formation as compared to the [tex]S_2[/tex]
So we can easily calculate the [tex]V_{max}[/tex] for either of [tex]S_1[/tex] or [tex]S_2[/tex] as we know that Tube 1 is [tex]S_2[/tex] and tube 2 is [tex]S_1[/tex]
As we know that
[tex]V_0 = V_{max}\ {S} / (K_M + {S})[/tex]
As the rates do not include any kind of units so we do not consider the units for [tex]V_{max}[/tex]
Now the calculation is
[tex]0.5 = V_{max} (0.1\ mM) / (20\ mM + 0.1\ mM)[/tex]
[tex]V_{max} = 0.5 (20.1\ mM) / 0.1\ mM[/tex]
= 100.5
≈ 101