I have to make this question longer so im just typing thisgjfgjfjvndfi nufnvfjnvfjnjn vnfj
(Go all the way to the bottom)
Don't forget to drink water
Answer:
Ok, thank you
Explanation:
Solution remained colorless.
During the experimentation, the test tube was gently heated in a Bunsen burner flame for 60 seconds. What was the reason for this specific
procedure?
A)
Heating was done to initiate the combustion of the metal in water.
B)
Heating was done to confirm that no chemical reaction would take
place in acid
0
Heating was done to precipitate the chemical change in each test
tube containing water
D)
Heating helped released the hydrogen contained in water
molecules so students would have a positive H+ test.
Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated:
4KClO3 (s) ⟶ Δ3KClO4 (s) + KCl (s).
Use ΔHf ° and S° values to calculate ΔG_sys ° (which is ΔGrxn °) in kJ at 25°C for this reaction.
Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]\text{The chemical reaction is : } 4 KClO_{3(s)} \to 3 KClO_{4(s)} + KCl_{(s)}[/tex]
[tex]\text{To find} \ \Delta G^0_{rxn}\ \text{using the formula}: \\ \\ \Delta G^0_{rxn} = \sum n_p \times \Delta _f G^0 (Products) - \sum n_R \times \Delta _fG^0 ( Reactants) \\ \\ where; n_p = \text{no of moles of products } \ and; \\ \\ n_R = \text{no of moles of reactants }[/tex]
[tex]\implies G^0_{rxn} = 3 \times \Delta _fG^0 [KClO_4{(s)}] + \Delta_fG^0[KCl_{(s)}] - 4 \times \Delta _f G^0 [ KClO_3 (s) ][/tex]
[tex]\Delta _fG^0 \ values \ at \ 25^0 \ C (298 \ K) are\ given \ as:\\\\ \Delta _fG^0 [KClO_4(s)] = -303.09 \ kJ \\ \\ \Delta _fG^0 [KCl(s) ] = - 409.14 \ kJ \\ \\ \Delta_f G^0 [KClO_3_{(s)}] = -296.25 \ kJ \\ \\ replacing \ the \ above \ values \ into \ equation (1) ; then:\\ \\ \\\Delta G^0_{rxn} = 3 *(-303.09) + (-409.14) - 4*(-296.25) \ kJ \\ \\ = (-909.27 - 409.14 + 1185) \ kJ \\ \\ = -133.41 \ kJ \\ \\ \mathbf{\Delta G^0_{rxn} = -133.4 \ kJ }[/tex]
The standard free energy change of the reaction is -133 kJ.
From the reaction equation, we have; 4KClO3 ⇄ 3KClO4 (s) + KCl (s). The standard free energy of formation of each specie is given below;
ΔG°f KClO3 = -296.35 kJ
ΔG°f KClO4 = -303.09 kJ
ΔG°f KCl = -409.14 kJ
Hence;
ΔG°rxn = [3(-303.09)] + ( -409.14)] - [(4( -296.35))]
ΔG°rxn = (-909.27) + (-409.14) - (-1185.4)
ΔG°rxn = -133 kJ
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- Explain why the term greenhouse effect is used to describe the theory of global
warming.
Does the greenhouse effect affect life on Earth? If yes, explain how?
What are the possible effects of a buildup of greenhouse gases in our atmosphere?
Question 17 of 20
The reactants of a chemical equation have 1 Satom and 40 atoms. Which
set of atoms must also be found in the equation's products so that the
equation models the law of conservation of mass?
A. 4 S and 40
B. 1 S and 40
C. 1 S and 10
O D. 4 S and 10
Answer:
B) 1S and 40
hope it helps
Answer:
b is ur answer
Explanation:
1S and 4O
What is the other product for this reaction ? H3PO4 + Ca(OH)2 —> H20 + _________
*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆
Answer: h3po4 + ca(oh)2 = h2o + ca3(po4)2
Explanation:
I hope this helped!
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- Zack Slocum
*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆
A separatory funnel contains the two immiscible liquids water and toluene. Use the given densities to determine which layer is on top and which is on the bottom in the binary mixture.
Solvent Density (g/mL)
toluene 0.87
water 0.998
Drag and drop each label into the box to indicate the position of the liquid in the mixture.
Top layer
Bottom layer
Answer:
Top layer TolueneBottom layer WaterExplanation:
When two non-miscible liquids are put together, the one with the higher density will be on the bottom, while the one with the lower density will be on top.
Meaning that in this problem's case toluene would be on the top layer and water in the bottom layer.
Definition of Acid, base and
salt
Answer:
nenrhj4rhty4bdwkwwa
Explanation:
HELP FAST 100 PTS Calculate the amount of heat needed to lower the temperature of 50.0g of ice from -40 °C to -100 °C.
Answer:
[tex]\Large \boxed{\sf -6000 \ J}[/tex]
Explanation:
Use formula
[tex]\displaystyle \sf Heat \ (J)=mass \ (kg) \times specific \ heat \ capacity \ (Jkg^{-1}\°C^{-1}) \times change \ in \ temperature \ (\°C)[/tex]
Specific heat capacity of ice is 2,000 J/(kg °C)
Substitute the values in formula and evaluate
[tex]\displaystyle \sf Heat \ (J)=0.05 \ kg \times 2000 \ Jkg^{-1}\°C^{-1} \times (-100\°C-(-40 \°C))[/tex]
[tex]Q=0.05 \times 2000 \times (-100-(-40)) =-6000[/tex]
A student uses a sample of KOH stock solution and dilutes it to a total of 120 mL. If the diluted solution is 0.60 M KOH and its original concentration was 2.25 M, what was the volume of the original sample? *
1.4 mL
89 mL
32 mL
5.5 mL
(DON'T POST LINKS PLEASE)
Answer:
5.5
Explanation:
i think so?????????
The diluted solution of volume 120 ml has the molarity of 0.60 M. Then, the volume of the original solution with a molarity of2.25 M is 32 ml.
What is molarity ?The molarity of a solution is the ratio of the number of moles of its solute particles to the volume of solution in liters.
To solve the given problem, we can use the formula for dilution:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
We are given that the diluted solution has a concentration of 0.60 M and a total volume of 120 mL. We are also given that the original concentration was 2.25 M, and we want to find the original volume.
Using the formula for dilution, we can write:
2.25 M x V1 = 0.60 M x 120 mL
Simplifying, we get:
V1 = (0.60 M x 120 mL) / 2.25 M
V1 = 32 mL
Therefore, the original volume of the KOH stock solution was 32 mL.
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Help this is for marks, who ever answers get brainliest
Answer:
Coal
Explanation:
thx for points :D
Answer:
Coal
Explanation:
your welcome<3
Your friend the archaeologist, who studies the ancient human past, has asked you to find a numerical date for some hominid artifacts found buried by ash from a volcanic eruption. To do this, you can use carbon-14 dating, which has a half-life of 5,730 years, or uranium-238, which has a half-life of 4.5 billion years. Based on their half-lives, which isotopic dating system will give your friend the most accurate numerical age
Answer:
I think it the carbon 14 which is a half life of 5730 because in 4.5billion years ago I don't think any human being existed on earth I'm not sure I hope that I helped you and good luck .
Carbon-14 dating will be used to find a numerical date for some hominid artifacts found buried by ash from a volcanic eruption.
What is carbon dating?
Radiocarbon dating also known as carbon dating or carbon-14 dating.
It is a method for establishing the age of an organic thing using the properties of radiocarbon, a radioactive carbon isotope.
Willard Libby created the approach in the late 1940s.
Thus, Carbon -14 dating will give your friend the most accurate numerical age.
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is chemical energy stored in coal, gas, and oil?
How many bromine atoms are present in 39.1 g of CH2Br2
Answer:
Answer:
2.77 × 10²³ Br Atoms
Solution:
Data Given:
Mass of CH₂Br₂ = 39.9 g
M.Mass of CH₂Br₂ = 173.83 g.mol⁻¹
Step 1: Calculate Moles of CH₂Br₂ as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 39.9 g ÷ 173.83 g.mol⁻¹
Moles = 0.23 mol
Step 2: Calculate number of CH₂Br₂ Molecules,
As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of CH₂Br₂ Molecules can be written as,
Moles = Number of CH₂Br₂ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of CH₂Br₂ Molecules,
Number of CH₂Br₂ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting value of moles,
Number of CH₂Br₂ Molecules = 0.23 mol × 6.022 × 10²³ Atoms.mol⁻¹
Number of CH₂Br₂ Molecules = 1.38 × 10²³ CH₂Br₂ Molecules
Step 3: Calculate Number of Br Atoms:
As,
1 Molecule of CH₂Br₂ contains = 2 Atoms of Br
So,
1.38 × 10²³ Molecules of CH₂Br₂ will contain = X Atoms of Br
Solving for X,
X = (1.38 × 10²³ CH₂Br₂ × 2 Br) ÷ 1 CH₂Br₂
X = 2.77 × 10²³ Br Atoms
Explanation:
Identify the techniques used in the work-up and characterization of benzoic acid. The analytical method used to confirm the structure and functional groups of the product NMR spectroscopy The technique used to separate the pure product from any excess reagent, impurities, and byproducts Recrystallization The quick, numeric analysis used to characterize the product and assess the purity Melting point.
Answer:
Explanation:
[tex]\text{From the list of the options given; we are to identify the suitable techniques} \\ \\ \text{for the characterization of benzoic acid.}[/tex]
[tex]\text{The analytical method used to confirm the structure and functional groups}\\ \\ \text{present in the product is} \ \ \mathbf{IR \ spectroscopy.}[/tex]
[tex]\text{The technique used to separate pure products from any excess reagents,} \\ \\ \text{impurities, and byproducts is}\ \ \mathbf{Recrystallization.}[/tex]
[tex]\text{The quick, numeric analysis done to characterize the product and assess the purity is}[/tex][tex]\mathbf{melting \ point.}[/tex]
Explain this method (Froth floatation method)..........
Answer:
froth flotation is a technique commonly used in the mining industry. In this technique, particles of interest are physically separated from a liquid phase as a result of differences in the ability of air bubbles to selectively adhere to the surface of the particles, based upon their hydrophobicity.
Explanation:
Froth floatation method is commonly used to concentrate sulphide ore such as galena (PbS), zinc blende (ZnS) etc. (ii) In this method, the metaalic ore particles which are perferentially wetted by oil can be separated from gangue. (iii) In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. (iv) A small quantity of sodium ethyl xanthate which act as a collector is also added. (v) A froth is generated by blowing air through this mixture. (vi) The collector molecules attach to the ore particles and make them water repellent. (vii) As a result, ore parrticles, wetted by the oil, rise to the surface along with the froth. (viii) The froth is skimmed off and dried to recover the concentration ore. (ix) The gangue particles that are preferentially wetted by water settle at the bottom.
is the sun the only star in our solar system
Answer:
Hey mate......
Explanation:
This is ur answer.......
The largest star, and indeed the only star in our solar system, is the sun. The sun is a bit under a million miles across. About 110 Earths put side by side would equal the size of the sun. The sun has 99.8 percent of the mass of our solar system.
Hope it helps!
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An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.0 wt%. If after 51 h the concentration of carbon is 0.35 wt% at a position 3.9 mm below the surface, determine the temperature at which the treatment was carried out. You will need to use data in the two tables below to solve this problem.
Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]C_o = 0.20 \ wt\% \\ \\ C_s = 1 \ wt\% \\ \\ t = 51 \ h \\ \\ x = 3.9 \ mm \\ \\ C_x = 0.35 \ wt\%[/tex]
[tex]\text{Using Fick's 2{nd} \ law \ of \ diffusion;} \\ \\ \dfrac{C_x- C_o}{C_s-C_o}= 1 - erf ( \dfrac{x}{2\sqrt{Dt}})[/tex]
[tex]\dfrac{0.35-0.20}{1-0.20}= 1 - erf ( \dfrac{x}{2\sqrt{Dt}})[/tex]
[tex]0.1875 = 1 - erf ( \dfrac{x}{2\sqrt{DT}}) \\ \\ erf ( \dfrac{x}{2\sqrt{DT}}) = 1 - 0.1875 \\ \\ erf ( \dfrac{x}{2\sqrt{DT}}) = 0.8125[/tex]
[tex]\text{To find the value of Z by Obtaining Data from Tabulation of Error Function}[/tex] [tex]\text{Table Values:}[/tex]
Z erf(z)
0.90 → 0.7970
0.95 → 0.8209
? → 0.8225
∴
[tex]\dfrac{z-0.90}{0.95-0.90}= \dfrac{0.8125-0.7970}{0.8209-0.7970}[/tex]
[tex]\dfrac{z-0.90}{0.05}= \dfrac{0.0155}{0.0239}[/tex]
[tex]z = 0.9324[/tex]
[tex]\text{To determine the diffusion coefficient;}[/tex]
[tex]erf (0.9324) = 0.8125 = erf (\dfrac{x}{2\sqrt{Dt}}) \\ \\[/tex]
[tex]\dfrac{x}{2 \sqrt{Dt}}= 0.9324 \\ \\ \dfrac{3.9 \times 10^{-3}}{2 \times \sqrt{D\times 51 \times 3600}} = 0.92324 \\ \\ \sqrt{D} = 4.88 \times 10^{-6} \\ \\ D = \sqrt{4.88 \times 10^{-6}} \\ \\ D = 2.38 \times 10^{-11} \ m^2 /s[/tex]
why iodide ions cannot be determined by Mohr method
Answer:
Exactly the same approach can be used for determination of bromides. Other halides and pseudo halides, like I- and SCN-, behave very similarly in the solution, but their precipitate tends to adsorb chromate anions making end point detection difficult.
Hope it helps!
Mohr method is the titration method, used for the determination of the chloride ion concentration in solution. It can not be used for iodine ions, as the alkaline titration is not suitable for iodine.
What is titration?Titration is the analytical process used for the determination of the concentration of a compound, by reacting it with titrant.
The mohr method is the titration method, that determines the concentration of the cations in the solution by titrating them with the salt. The salt formed precipitate in the reaction and the end point is determined.
The limitation to the use of mohr method, is its inability for the analysis of the chlorides of cations, as the reaction is carried out at high pH in alkaline conditions.
The iodide slats are suitably precipitated in the alkaline solution, and thus limits the use of the mohr method.
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What is a similarity between a fish embryo and a human embryo in the late stages of development?
They both have tails
They both have gills.
They both have spines.
They both have arms.
Answer:In the LATER stages of development they would have a spine
Explanation:
Please help me as soon as possible!!! How many grams of H3PO4 are in a 4 L sample of a 17 M solution?
Answer:
about 6664 grams of hydro phosphoric acid are in a sample of 4 L of 17 M solution
Explanation:
grams of H3PO4 = 4 L | 17 mol | 98 grams H3PO4
| 1 L | 1 mol
Solid iron is mixed with a solution of copper (I) nitrate to form iron (III) nitrate solution and metal copper. what is the balanced equation?
1°C if the pressure is 149.3 kPa? How many moles of hydrogen sulfide H2S are contained in a 327.3 mL bulb at 48.
Answer:
Number of moles of H₂S gas = 0.0183 moles
Note: The question is incomplete. The complete question is given below:
How many moles of hydrogen sulfide H2S are contained in a 327.3 mL bulb at 48.1°C if the pressure is 149.3 kPa?
Explanation:
The following values are given in the question :
Volume of H₂S gas = 327.3 M = 0.3273 L
Temperature of gas = 48.1°C = (273.15 + 48.1) K = 321.25 K
Pressure of gas = 149.3 kPa
1 kPa = 0.00987 atm; 149.3 kPa = 149.3 × 0.00987 atm = 1.474 atm
Molar gas constant, R = 0.0821 liter·atm/mol·K.
number of moles, n = ?
Using PV = nRT
n = PV/RT
n = (0.3273 × 1.474)/(0.0821 × 321.25)
n = 0.0183 moles
Therefore, number of moles of H₂S gas = 0.0183 moles
g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thin film gate of 500 nm in 50 minutes. What is the original field oxide thickness d0 (in nm)?
Answer:
Explanation:
From the information given:
oxidation of oxidized solution takes place at 950° C
For wet oxidation:
The linear and parabolic coefficient can be computed as:
[tex]\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big][/tex]
Using [tex]D_o[/tex] and [tex]E_a[/tex] values obtained from the graph:
Thus;
[tex]\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr[/tex]
[tex]B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr[/tex]
So, the initial time required to grow oxidation is expressed as:
[tex]t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)[/tex]
[tex]where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719[/tex]
∴
[tex]2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)[/tex]
[tex]2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr[/tex]
NOW;
[tex]1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453[/tex]
[tex]d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}[/tex]
[tex]d_o =0.02609 \ OR \ -0.0939[/tex]
Thus; since we will consider the positive sign, the initial thickness [tex]d_o[/tex] is ;
≅ 0.261 μm
Use the following balanced reaction to solve:
P4 (s) + 6H2 (g) → 4PH3 (g)
How many grams of phosphorus trihydride will be formed by reacting 60 L of Hydrogen gas with an excess of P4?
Answer: 60.7 g of [tex]PH_3[/tex] will be formed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}[/tex]
[tex]\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles[/tex]
The balanced chemical reaction is
[tex]P_4(s)+6H_2(g)\rightarrow 4PH_3(g)[/tex]
[tex]H_2[/tex] is the limiting reagent as it limits the formation of product and [tex]P_4[/tex] is the excess reagent.
According to stoichiometry :
6 moles of [tex]H_2[/tex] produce = 4 moles of [tex]PH_3[/tex]
Thus 2.68 moles of [tex]H_2[/tex] will produce=[tex]\frac{4}{6}\times 2.68=1.79moles[/tex] of [tex]PH_3[/tex]
Mass of [tex]PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g[/tex]
Thus 60.7 g of [tex]PH_3[/tex] will be formed by reactiong 60 L of hydrogen gas with an excess of [tex]P_4[/tex]
I need help fast pls someone
Answer:
I would say A. I'm no expert, but it can't be C obviously, and I think wind would hit all of it, wearing off the top as well like the great pyramids. B would be my next choice, but A i think would be best.
When 577 J of energy is added to 32.3 g of aluminum at 17.4ºC, the temperature increases to 46.6ºC. What is the specific heat of aluminum?
Answer:
0.612 J/g°C
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (Joules)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C) i.e. final - initial temperature
According to the information provided in this question;
Q = 577 J
m = 32.3 g
c = ?
Final temp = 46.6ºC, initial temp. = 17.4°C
∆T = (46.6 - 17.4) = 29.2°C
Using Q = m × c × ∆T
c = Q ÷ m.∆T
c = 577 ÷ (32.3 × 29.2)
c = 577 ÷ 943.16
c = 0.6117
c = 0.612 J/g°C
How many Calories are in 5,926 joules
How many grams are in 1.2 x 1024 atoms of sodium?
Answer:
46 g
Explanation:
First we convert 1.2 x 10²⁴ atoms of sodium into moles, using Avogadro's number:
1.2x10²⁴ atoms ÷ 6.023x10²³ atoms/mol = 2.0 molThen we convert 2.0 moles of sodium into grams, using sodium's molar mass:
2.0 mol Na * 23 g/mol = 46 gThus, there are 46 grams in 1.2x10²⁴ atoms of sodium.
2.0M Propionic Acid HC3H5O2 Dissolves In Distilled Water. If It Has A Ka Of 1.3*10-5, What Is The Final PH?
The base ethylamine C2H5NH2 has a Kb of 5.6*10-4. What will the pH be in .53 molar solution?
Answer: first you have to calculate the amount ionized. We will say it is x mol / L
then % ionization = (amount ionized / initial concentration) * 100%
x can be calculated using an ice chart
HC3H5O2 -----> H+ + C3H5O2-
Initial HC3H5O2 = 0.250
change = -x
equilibrium = 0.250 - x
initial H+ = 0
change = +x
equilibrium = x
C3H5O2- initial = 0
change = +x
equilibrium = x
Ka = [H=][C3H5O2-] / HC3H5O2]
1.3 * 10 ^ -5 = [x][x] / (0.250 - x)
So 1.3 * 10 ^ -5 * (0.250 - x) = x ^ 2
3.25 * 10^ -6 - (1.3 * 10^-5)x = x^2 now this is a quadratic equation and you have to rearrange it and solve for x
x^2 + 1.3 * 10^-5)x - 3.25 * 10^ -6 = 0
use the equation x = {-b (+ or -)[b^2 - 4.a.c] ^ 1/2} / 2a
you should get x = 1.80 * 10 ^ -3 or x = -1.80* 10^-3
but x can not be negative..
so x = 1.80 * 10 ^ -3
so percent ionization = (1.80 * 10 ^ -3 / 0.250) * 100%
=0.72 %
the other way which is more easier is
assuming that x is very small and therefore 0.250 - x is approximately equals to 0.250
then 1.3 * 10^-5 = x^2 / 0.250
so x^2 = 1.3 * 10^-5 * 0.250
x = 1.80 * 10 ^-3
then percent ionization is = (1.80 * 10 ^ -3 / 0.250) * 100%
=0.72 %
if the percent ionization is > 5 % you can not do that approximation. in such a case you have to solve the quadratic equation. that is why I showed both methods.
now you can do the parts b and c
b answer : percent ionization = 1.27 %
c answer : 2.54%
good luck
