When we report the concentration of a solution we are using the moles of the entire compound.

a. True
b. False

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

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Answer: h3po4 + ca(oh)2 = h2o + ca3(po4)2

Explanation:

I hope this helped!

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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.

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.

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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Answer:In the LATER stages of development they would have a spine

Explanation:

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]

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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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

Answer:

Hot spot is like someone who has data or min on there phone turn on their phone wifi so you can use it pretty much

Answer:

a hot spot is a form of wifi that u can use anywhere at anytime. it connects to near satalites or wifi towers. it allows you to use devices r games without the need for wifi.

Explanation:

Answer:

nenrhj4rhty4bdwkwwa

Explanation:

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]

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.

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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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]

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]

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.

Answer:

2-propanol

Explanation:

From the given information

2 drops of an unknown sample were said to be placed in a test tube followed by the addition of 2 ml of ethanol then gentle mixing. They then initiate a further addition of  2 drops of potassium permanganate reagent (KMNO₄) to the test tube and mixed the contents of the test tube thoroughly. After adding 2 drops of potassium permanganate reagent, the reagent oxidizes the secondary alcohols(2-propanol) to ketone(i.e acetone) and no further reaction will take place since there are no reactive C-H bonds left. The diagram attached below shows how the reaction proceeds.

Answer:

46 g

Explanation:

First we convert 1.2 x 10²⁴ atoms of sodium into moles, using Avogadro's number:

Then we convert 2.0 moles of sodium into grams, using sodium's molar mass:

Thus, there are 46 grams in 1.2x10²⁴ atoms of sodium.

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

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

Answer:

NaCl

Explanation:

If you follow traditional rounding, Na atomic mass is 23 and Cl atomic mass 35, 23 +35 = 58.

I was taught that Cl is one of the elements you round to 35.5, but this works if you’re not in an advanced chem class.

Answer:

Explanation:

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.

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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Answer:

Coal

Explanation:

thx for points :D

Answer:

Coal

Explanation:

your welcome<3

Answer:

The answer is "Option C".

Explanation:

Given equation:

[tex]2Na_20_2 (s)+S(s)+2H_2O \longrightarrow 4NaOH(aq)+SO_2(aq)[/tex]

[tex]\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}[/tex]

[tex]\to Na_2O_2 \ Mass = 7.8 \ g\\\\ \to Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}[/tex]

[tex]Na_2O_2[/tex] Has been the reactant which is limited since the two experiments are equal to[tex]Na_2O_2[/tex] for relationship between stress amounts.

[tex]Na_2O_2, n =\frac{Mass of Na_2O_2}{Molar mass of Na_2 O_2}=\frac{7.8 \ g}{78 \frac{g}{mol}} =0.1 \ mol \\\\q=\Delta H^{\circ}_{rxn} \times n = \frac{ -610 \ kJ}{ 2 \ mol \ Na_2 O_2} \times 0.1 \ mol \ Na_2O_2= 30.5 \ KJ\\\\[/tex]

Limiting reactant =[tex]Na_2O_2[/tex]

[tex]q=30.5 \ kJ \approx 30 \ kJ[/tex]