Could anyone provide me with a run down on Electrolysis?​

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

Answer:

Explanation:

i will show in details how 2 do the 1st Q n u can do the rest by following the way how it is done

3. given C^2H^5OH + CH^3CO^2H → CH^3CO^2C^2H^5 + H^2O

molar ratio of ethanol, ethanoic acid and ethyl ethanoate is 1 : 1 : 1

so mass = no. of moles * molecular mass

for same no. of moles, mass / molecular mass is the same

molecular mass of CH^3CO^2C^2H^5 = 12+1*3+12+16*2+12*2+5*1 = 88

molecular mass of C^2H^5OH = 12*2+1*5+16+1 = 46

molecular mass of CH^3CO^2H = 12+1*3+12+16*2+1 = 60

1 g of ethyl ethanoate = 1/88 mole

it requires 1/88*46 = 0.5227 g of ethanol; and

1/88*60 = 0.6818 g of ethanoic acid

to react together to form 1 g of ethyl ethanoate

Metal containers are not used for storing acid because most of the time acid reacts with almost every metal and produces salts or oxides which alters the acid characteristics making it useless....

Explanation:

2C +O2 = 2CO

this will be the answer

Answer:

5 kg of feather

Explanation:

0Answer: 0.6094

Explanation:

no of moles = mass / molar mass = 100/164.088= 0.6094 mole

Answer:

0.609 moles

Explanation:

mass in g ÷ atomic mass = moles

Ca(NO₃)₂ = 1 Ca 40.078 amu

2 N 28.0134 amu

+ 6 O 95.994 amu

____________________

164.0854 amu

100 g ÷ 164.0854 amu = 0.609 moles

three significant digits

Answer:

He should ask an adult if they know what the solution is. if they dont, put it back and find a different cleaning solution to use

Answer:

Swimming pool is an example of solution

Explanation:

It has water (solvent) and dissolve ion (solute)

Answer:

it is a solution

Explanation:

the combining of (water and Cl and other pool chemicals) make it a solution.

Answer:

Increased surface area - finely divided solute

like dissolves like - matching polarity

temperature - rate proportional to kinetic energy

stirring spreads - solute throughout solution

Answer:

18,9g de hexano son necesarios

Explanation:

Basados en la ley de Raoult, la presión de vapor ejercida por una solución es:

P = P°*Xsolvente

Donde P es la presión de la solución deseada = 700mmHg, P° la presión de vapor de vapor del benceno = 1atm = 760mmHg y X es la fracción molar del solvente (Benceno).

Reemplazando:

700mm Hg = 760mmHg * X(Benceno)

0.9211 = X(Benceno)

La fracción molar de benceno se define como:

X = Moles benceno / Moles benceno + Moles hexano

Moles benceno -Masa molar: 78g/mol-

200g * (1mol/78g) = 2.5641 moles benceno

X = Moles benceno / Moles benceno + Moles hexano

0.9211 = 2.5641 moles benceno / 2.5641 moles benceno + Moles hexano

0.9211moles Hexano + 2.3618 = 2.5641

0.9211*Moles Hexano = 0.2023

Moles hexano = 0.2023/0.9211 = 0.2196 moles hexano.

Masa Hexano -Masa molar: 86g/mol-

0.2196 moles hexano * (86g/mol) =

La cantidad de hexano (C6H14) que se debe agregar a 200 g de benceno para permitir que la presión de vapor se convierta en 700 mm de Hg sería:

18.9 g

Usando la ley de Raoult, la presión de vapor que libera una solución se encuentra por:

P = P ° × X disolvente  

Donde (P) denota la presión de la solución deseada

P ° denota la presión de vapor del benceno y

X denota la fracción molar del disolvente (benceno).

En el caso dado,

(P) = 700 mmHg,

P° = 1atm = 760mmHg

Fracción molar de benceno = 0,9211           (∵ 700/760)

La fracción molar de benceno (X)  = Moles de benceno/Moles de benceno + Moles de hexano

Moles de benceno - Masa molar: 78g / mol -  200 g × (1 mol / 78 g)

= 2.5641 moles de benceno

X = Moles de benceno/Moles de benceno + Moles de hexano

0.9211 = 2.5641 moles Benceno/2.5641 moles Benceno + Moles de Hexano

⇒ 0,9211 × moles de Hexano + 2,3618 = 2,5641    

⇒ 0.9211 × moles de Hexano = 0.2023

∵ Moles de Hexano = 0.2023/0.9211

[tex]= 0.2196 moles[/tex]

[tex]Hexane mass - Molar mass = 86g/mol - 0.2196 mole hexane[/tex] × [tex](86g/mole)[/tex]

[tex]= 18.9g[/tex]

Por tanto, 18,9 g es la respuesta correcta.

Learn more about "Benzene" here:

brainly.com/question/25798187

Answer:

The combination of ions most likely to produce a precipitate is a group of answer choices:

lead nitrate soluble in water

Mg2+ and C2H3O2-.

Fe3+ and OH-.

Li+ and PO43-.

Pb2+ and NO3-.

NH4+ and SO42-.

Explanation:

Among the given options,

magnesium acetate, lithium phosphate, lead nitrate, ammonium sulfate are soluble in water.

The only one which is insoluble in water is [tex]Fe^3+[/tex] and [tex]OH^-[/tex] combination.

[tex]Fe(OH)_3[/tex] is insoluble in water. It forms a precipitate.

Answer:

3 Protons, 2 Neutrons, & 3 Electrons. Is the charge of this atom positive, negative, or neutral?

Explanation:

Put these atoms in order from most negative overall charge to least negative

overall charge.

11 Atom X: 104 protons, 102 electrons

11 Atom B: 24 protons, 18 electrons

1 Atom Q: 15 protons, 16 electrons

11. Atom P: 7 protons, 10 electrons

Answer:

Explanation:

Because you only have one repeat unit, n=1. 2n-1 becomes 2(1)-1 which is equal to one, meaning one molecule of H2O is produced, as is shown by the top condensation polymerisation reaction.

If you had two repeat units, n=2 so 2n-1=3. Three H2O molecules are produced because you would need two molecules of each reactant so three condesation reactions would occur and three molecules of H2O would be released.

Answer:

[tex]\huge\color{cyan}{\colorbox{magenta}{Answer}}[/tex]

thank you

Answer:

λ = 2.38 × 10^(-7) m

Explanation:

We are given the work function for palladium as 503.7 kJ/mol.

Now let's convert this to KJ/electron.

We know from avogadro's number that;

1 mole of electron = 6.022 × 10^(23) electrons

Thus,

503.7 kJ/mol = 503.7 × 1/(6.022 × 10^(23)) = 8.364 × 10^(-22) KJ/electron = 8.364 × 10^(-19) J/electron

Formula for energy of a photon is;

E = hv

Where;

h is Planck's constant = 6.626 × 10^(-34) J.s

v is velocity

Now, v = c/λ

Where;

c is speed of light = 3 × 10^(8) m/s

λ is wavelength of light.

Thus;

E = hc/λ

Making λ the subject, we have;

λ = hc/E

λ = (6.626 × 10^(-34) × 3 × 10^(8))/(8.364 × 10^(-19))

λ = 2.38 × 10^(-7) m

Answer:

the number of electrons in the cell

[tex]\huge\fcolorbox{red}{pink}{Answer ♥}[/tex]

The carbon atom is unique among elements in its tendency to form extensive networks of covalent bonds not only with other elements but also with itself. ... Moreover, of all the elements in the second row, carbon has the maximum number of outer shell electrons (four) capable of forming covalent bonds.

Hope it helps uh ✌️✌️✌️

Gud mrng

ajdhrhahdjdjdndoxhdebdixhxb dbds ebenej

Answer: The mass density is 1166.36 [tex]kg/m^{3}[/tex].

Explanation:

Given: Weight of sample in air [tex](F_{air})[/tex] = 500 N

Weight of sample in alcohol [tex](F_{alc})[/tex] = 200 N

Specific gravity = 0.7 = [tex]0.7 \times 1000 = 700 kg/m^{3}[/tex]

Formula used to calculate Buoyant force is as follows.

[tex]F_{B} = F_{air} - F_{alc}\\= 500 - 200 \\= 300 N[/tex]

Hence, volume of the material is calculated as follows.

[tex]V = \frac{F_{B}}{\rho \times g}[/tex]

where,

[tex]F_{B}[/tex] = Buoyant force

[tex]\rho[/tex] = specific gravity

g = acceleration due to gravity = 9.81

Substitute the values into above formula.

[tex]V = \frac{F_{B}}{\rho \times g}\\= \frac{300}{700 \times 9.81}\\= \frac{300}{6867}\\= 0.0437 m^{3}[/tex]

Now, mass of the material is calculated as follows.

[tex]mass = \frac{F_{air}}{g}\\= \frac{500 N}{9.81}\\= 50.97 kg[/tex]

Therefore, density of the material or mass density is as follows.

[tex]Density = \frac{mass}{volume}\\= \frac{50.97 kg}{0.0437 m^{3}}\\= 1166.36 kg/m^{3}[/tex]

Thus, we can conclude that the mass density is 1166.36 [tex]kg/m^{3}[/tex].

Answer:

Todas son correctas

Explanation:

I) Los compuestos del carbono son solubles en disolventes no polares como el hexano, benceno, éter, etc. VERDADERO. La mayoría de los compuestos de carbono son apolares, y basados en la regla: Similar disuelve similar, podemos presumir que la mayoría de compuestos de carbono se disuelven en solventes no polares.

II) Los compuestos del carbono generalmente presentan puntos de fusión y de ebullición bajos. VERDADERO. Al ser sustancias apolares, sus fuerzas electrostáticas son bajas. De la misma manera, como su masa atómica es pequeña, las fuerzas de Van der Waals son despreciables haciendo que sus puntos de fusión y ebullición sean bajos respecto a sustancias de estrucutra similar.

III) La gran mayoría de los compuestos del carbono son combustibles, sean estos, gaseosos, líquidos o sólidos. VERDADERO. Los gases (Como el gas natural) son combustibles usados para cocina. Los líquidos (Como la gasolina) son combustibles y bastante inflamables. Los sólidos (Como la madera) se usan como combustibles para hacer asados o son el combustible en incendios forestales.

IV) El enlace covalente es característico de los compuestos del carbono. VERDADERO. La polaridad del carbono es neutral haciendo que la mayoría de los enlaces que forma sean covalentes.

Answer:

The frequency of a wave that has a wavelength of 28 m is 1.07*10⁷ Hz.

Explanation:

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave's disturbance propagates along its displacement. Relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f * λ.

Being:

Replacing:

2.998*10⁸ m/s= f* 28 m

Solving:

[tex]f=\frac{2.998*10^{8}m/s }{28m}[/tex]

f= 1.07*10⁷ Hz

The frequency of a wave that has a wavelength of 28 m is 1.07*10⁷ Hz.

Answer:

i believe energy is your answer here

Answer:

In nomenclature of simple molecular compounds, the more electropositive atom is written first and the more electronegative element is written last with an -ide suffix.

The Greek prefixes are used to dictate the number of a given element present in a molecular compound.

Prefixes can be shortened when the ending vowel of the prefix “conflicts” with a starting vowel in the compound.

Common exceptions exist for naming molecular compounds, where trivial or common names are used instead of systematic names, such as ammonia (NH3) instead of nitrogen trihydride or water (H2O) instead of dihydrogen monooxide.

Terms

nomenclatureA set of rules used for forming the names or terms in a particular field of arts or sciences.

electronegativeTending to attract electrons within a chemical bond.

electropositiveTending to not attract electrons (repel) within a chemical bond.

Chemical Nomenclature

The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning to what chemical compound the name refers. Each chemical name should refer to a single substance. Today, scientists often refer to chemicals by their common names: for example, water is not often called dihydrogen oxide. However, it is important to be able to recognize and name all chemicals in a standardized way. The most widely accepted format for nomenclature has been established by IUPAC.

Molecular compounds are made when two or more elements share electrons in a covalent bond to connect the elements. Typically, non-metals tend to share electrons, make covalent bonds, and thus, form molecular compounds.

Rules for Naming Molecular Compounds:

Remove the ending of the second element, and add “ide” just like in ionic compounds.

When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.

If there is only one of the first element, you can drop the prefix. For example, CO is carbon monoxide, not monocarbon monoxide.

If there are two vowels in a row that sound the same once the prefix is added (they “conflict”), the extra vowel on the end of the prefix is removed. For example, one oxygen would be monooxide, but instead it’s monoxide. The extra o is dropped.

Generally, the more electropositive atom is written first, followed by the more electronegative atom with an appropriate suffix. For example, H2O (water) can be called dihydrogen monoxide (though it’s not usually). Organic molecules (molecules made of C and H along with other elements) do not follow this rule.

Here are the prefixes in naming molecular compounds:

Mono- 1

Di- 2

Tri- 3

Tetra- 4

Penta- 5

Hexa- 6

Hepta- 7

Octa- 8

Nona- 9

Deca- 10

Molecular compounds are named using a systematic approach of prefixes to indicate the number of each element present in the compound.

I hope it helps ●~●

Answer:

Explanation:

If a substance contains ionic bonds, then it’s properties would include high melting and boiling points, a crystalline structure, the ability to dissolve quite easily in water, the ability to conduct electricity, have good insulation and lastly be quite hard and brittle.

If my answer was helpful, would you consider giving me a “Brainliest”?. Thank you :)

Answer:

CH2O

Explanation:

According to this question:

C = 2.04g

H = 0.34g

O = 2.73g

First, we divide the mass value of each element by its atomic mass to convert to moles.

C = 2.04g ÷ 12 = 0.17mol

H = 0.34g ÷ 1 = 0.34mol

O = 2.73g ÷ 16 = 0.17mol

Next, we divide by the smallest mole value (0.17):

C = 0.17mol = 0.17 = 1

H = 0.34mol ÷ 0.17 = 2

O = 0.17mol ÷ 0.17 = 1

The whole number ratio of C,H,O is 1:2:1, hence, the empirical formula is CH2O.

Answer:

C. x⁴ + 6·x³ - 12·x  - 72

Explanation:

The given functions are;

[tex]f(x) =\sqrt{x^2 + 12 \cdot x + 36}[/tex]

g(x) = x³ -12

We have that [tex]f(x) =\sqrt{x^2 + 12 \cdot x + 36}[/tex] = [tex]f(x) =\sqrt{(x + 6)^2}[/tex] =  (x + 6)

Therefore;

f(x)·g(x) = [tex]\sqrt{x^2 + 12 \cdot x + 36}[/tex] × (x³ - 12) = (x + 6) × (x³ - 12)

(x + 6) × (x³ - 12) = x⁴ - 12·x + 6·x³ - 72 = x⁴ + 6·x³ - 12·x  - 72

∴ f(x)·g(x) = [tex]\sqrt{x^2 + 12 \cdot x + 36}[/tex] × (x³ - 12) = x⁴ + 6·x³ - 12·x  - 72

Answer:

Equilibrium concentrations of the gases are

[tex]H_2S=0.596M[/tex]

[tex]H_2=0.004 M[/tex]

[tex]S_2=0.002 M[/tex]

Explanation:

We are given that  for the equilibrium

[tex]2H_2S\rightleftharpoons 2H_2(g)+S_2(g)[/tex]

[tex]k_c=9.0\times 10^{-8}[/tex]

Temperature, [tex]T=700^{\circ}C[/tex]

Initial concentration of

[tex]H_2S=0.30M[/tex]

[tex]H_2=0.30 M[/tex]

[tex]S_2=0.150 M[/tex]

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

[tex]H_2S=0.30+2x[/tex]

[tex]H_2=0.30-2x[/tex]

[tex]S_2=0.150-x[/tex]

At equilibrium

Equilibrium constant

[tex]K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]

Substitute the values

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

By solving we get

[tex]x\approx 0.148[/tex]

Now, equilibrium concentration  of gases

[tex]H_2S=0.30+2(0.148)=0.596M[/tex]

[tex]H_2=0.30-2(0.148)=0.004 M[/tex]

[tex]S_2=0.150-0.148=0.002 M[/tex]

Answer:

18.5g Na2CO3

Explanation:

Chromium (III) chloride, CrCl3, reacts with Na2CO3 as follows:

2CrCl3 + 3Na2CO3 → Cr2(CO3)3(s) + 6NaCl

Where 2 moles of CrCl3 react with 3 moles of Na2CO3 to produce 1 mole of Cr2(CO3)3 -The precipitate-

To solve this question we need to find the moles of CrCl3 added. With the chemical equation we can find the moles of Na2CO3 and its mass as follows:

Moles CrCl3:

520mL = 0.520L * (0.224mol/L) = 0.116 moles CrCl3

Moles Na2CO3:

0.116 moles CrCl3 * (3 mol Na2CO3 / 2mol CrCl3) = 0.175 moles Na2CO3

Mass Na2CO3 -Molar mass: 105.99g/mol-

0.175 moles Na2CO3 * (105.99g/mol) = 18.5g Na2CO3