Select the atomic models that
belong to the same element.

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

Explanation:

I hope this helped!

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- Zack Slocum

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Answer: 4.4 x 10^-7

Explanation:

0.000066 x 0.0000066 / 0.01

= 4.4 x 10^-7

Answer: d. Determining empirical formula

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Step 1: If percentage are given then we are taking total mass is 100 grams.  So, the mass of each element is equal to the percentage given.

Step 2 : convert given masses into moles

Step 3 : For the mole ratio, divide each value of moles by the smallest number of moles calculated

Step 4: The mole ratio gives the empirical formula of the compund.

Answer: The total pressure is 7 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_A+p_B+p_C...[/tex]

Given : [tex]p_{total}[/tex] =total pressure of gases = ?

[tex]p_{O_2}[/tex] = partial pressure of oxygen = 2 atm

[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 4 atm

[tex]p_{Ar}[/tex] = partial pressure of argon = 1 atm

putting in the values we get:

[tex]p_{total}=2atm+4atm+1atm[/tex]  

[tex]p_{total}=7atm[/tex]

Thus the total pressure is 7 atm

Answer:

temperature and rainfall

Explanation:

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.

Find more on molarity :

https://brainly.com/question/8732513

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Answer:

the first one

Explanation:

I looked it up:P

Only the 2nd answer is right, uncheck the 1st one

Explanation:

If we had a chemical formula like [tex]3(O_{2} )[/tex] the 3 means there are 3 molecules of [tex]O_{2}[/tex]. The 2 means there are two atoms of oxygen in each molecule.

Answer: The answers are:

When mixed, water and limestone give off heat.

On adding yellow turmeric to a transparent soap solution, the mixture turns bright red.

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:

B) 1S and 40

hope it helps

Answer:

b is ur answer

Explanation:

1S and 4O

Answer: True

Explanation:

Answer:

Coal

Explanation:

thx for points :D

Answer:

Coal

Explanation:

your welcome<3

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

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:

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

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:

Sodium chloride is the chemical name for salt and is composed of sodium and chlorine, so it is a compound.

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]

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:

nenrhj4rhty4bdwkwwa

Explanation:

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:

The most reactive compound is: Phenol  because the electron donating character of the alcohol group increases the rate of the reaction.

The least reactive compound is nitrobenzene

because the electron withdrawing character of the nitro group decreases the rate of the reaction.

Explanation:

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:

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:

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.