Calculate the volume, in milliliters, of a 0.211 M solution of NaOH that will completely neutralize each of the following. 2.30 mL of a 0.820 M solution of H2SO4. Express the volume in milliliters to three significant figures.

Answer:

Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred

Explanation:

Answer:

  0.0480 M

Explanation:

The reaction is ...

  H₂SO₄ + 2NaOH ⇒ Na₂SO₄ +2H₂O

That is, 2 moles of NaOH react with each mole of H₂SO₄. Then the molarity of the H₂SO₄ is ...

  moles/liter = (0.186 M/2)(12.9 mL)/(25.0 mL) ≈ 0.0480 M

The sand that grained will be quickly chemically weathered away should be option b. ferromagnesian silicate minerals.

It should be considered as the Silicate minerals where cations of iron and the form of magnesium should be important for the chemical components. It is used for covering up the minerals. Also, calcite should be normal weather via the solution process so it required a lot of water that contains a high amount of carbonic acid.

Hence, the correct option is b.

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Answer:

[tex]n_{K^+}=0.0009molK^+[/tex]

Explanation:

Hello,

In this case, the first step is to compute the number of moles of potassium phosphate in 20.0 mL (0.020L) of the 0.015-M (mol/L) solution as shown below:

[tex]n=0.020L*0.015\frac{mol}{L}=0.0003mol[/tex]

Thus, these moles correspond to potassium phosphate moles, which molecular formula is K₃PO₄, therefore, one mole of this compound contains three moles of potassium ions as it has three as its subscript in the formula. Thereby, the moles of potassium ions result in:

[tex]n_{K^+}=0.0003molK_3PO_4*\frac{3molK^+}{1molK_3PO_4} \\\\n_{K^+}=0.0009molK^+[/tex]

Best regards.

Answer:

Kb = 1.77x10⁻⁵

Explanation:

When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

And the dissociation constant, Kb, for this equilibrium is:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

To find Kb you need to find the concentration of each species. The equilibrium concentrations are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate.

You can know [OH⁻] and, therefore, X, with pH of the solution, thus:

pH = -log [H⁺] = 11.612

[H⁺] = 2.4434x10⁻¹²

As 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]

4.0926x10⁻³ = [OH⁻] = X

Replacing, concentrations of the species are:

[NH₃] = 0.950M - X

[NH₄⁺] = X

[OH⁻] = X

[NH₃] = 0.9459M

[NH₄⁺] = 4.0926x10⁻³M

[OH⁻] = 4.0926x10⁻³M

Replacing in Kb expression:

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]

The branch of science which deals with the chemical bond is called Chemistry.

The correct answer to the question is  [tex]Kb = 1.77*10^{-5[/tex]

Explanation:

When NH₃, is acts as a  weak base it forms an equilibrium with water the reaction occurs is:

[tex]NH_3(aq) + H_2O(l) ---><NH_4^+(aq) + OH^-(aq)[/tex]

The formula we gonna use is as follows:-

[tex]Kb = \frac{[NH_4^+] [OH^-]}{[NH_3]}[/tex]

The data is given in the question is as follows:-

Where X stands for reaction coordinate.

After solving the ph of the compound the value is as follows:-

Putting the value in the formula.

[tex]Kb = \frac{[4.0926*10^{-3}M] [4.0926*10^{-3}M]}{[0.9459M]}[/tex]

After solving the equation the value of Kb is [tex]1.77*10^{-5[/tex]

Hence, the correct answer is [tex]1.77*10^{-5[/tex]

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Answer:

See explanation

Explanation:

In this case, we can start with the calculation of the total energy released by the 20 matches. One match release 838.2 J of energy ( 1 match = 838.2 J). So:

[tex]20~matches\frac{838.2~J}{1~match}=16764~J[/tex]

With this in mind, we have to find al the conversion ratios to "Jolues", so:

[tex]1~KJ=1000~J[/tex]

[tex]1~Cal=4.184~J[/tex]

[tex]1~KCal=4184~J[/tex]

[tex]0.000239~food~calories=1~J[/tex]

Now, we can do the conversions:

[tex]16764~J\frac{1~KJ}{1000~J}=16.76~KJ[/tex]

[tex]16764~J\frac{1~Cal}{4.184~J}=4.005x10^3~Cal[/tex]

[tex]16764~J\frac{1~KCal}{4184~J}=4.007~KCal[/tex]

[tex]16764~J\frac{0.000239~food calories}{1~J}=4.006~food calories[/tex]

I hope it helps!

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

[tex]n_i=8[/tex]

[tex]\lambda=3745\ nm[/tex]

The amount of energy change during the transition is given by :

[tex]\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]

And

[tex]\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]

Plugging all the values we get :

[tex]\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5[/tex]

So, the final level of the electron is 5.

Explanation:

Before proceeding we have to understand what electrolytes and non electrolytes are;

An electrolyte is a substance that produces an electrically conducting solution when dissolved. An electrolyte is a compound that can dissociate into ions.

Non electrolytes: A substance whose molecules in solution do not dissociate to ions and thus do not conduct an electric current

Going through the options;

1. LiOH

This is a compound of hat would dissociate into Li+ and OH-.  This is an electrolyte.

2. C4H2O4

This is an organic compound. Gnerally organic acids are non electrolytes, with the exception og the acids. This is a nonelectrolyte.

3. LiBr

This is an electrolyte because it would dissociate into Li+ and Br-  ions.

4. HNO3

HNO3 is a strong acid. Because it is a strong acid it will dissociate completely into its ions (H+ and NO3-). Therefore we consider HNO3 to be a strong electrolyte.

Answer:

[tex]Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]

Explanation:

[tex]Fe_2O_3+CO\Rightarrow \:2Fe+CO_2\\\\Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]

Best Regards!

Answer:

1. 0.421 g (C₄H₁₀), 1.51 g (O₂), 1.28 g (CO₂), 0.653 g (H₂O)

2. 4.92 g (C₄H₁₀), 17.6 g (O₂), 14.9 g (CO₂), 7.63 g (H₂O)

3. 6.63 g (C₄H₁₀), 23.7 g (O₂), 20.12 g (CO₂), 10.3 g (H₂O)

4. 7.12 g (C₄H₁₀), 12.1 g (O₂), 10.2 g (CO₂), 8.84 g (H₂O)

5. 252 mg (C₄H₁₀), 903 mg (O₂), 763 mg (CO₂), 390 mg (H₂O)

6. 65 mg (C₄H₁₀), 234 mg (O₂), 198 mg (CO₂), 101 mg (H₂O)

Explanation:

1. First of all, we determine the moles of each reactant.

For the first case:

1.51 g . 1 mol/32 g = 0.0472 moles

Ratio is 13:2, 13 moles of oxygen needs 2 moles of C₄H₁₀ for the combustion,

Therefore 0.0472 mol will react with (0.0472 . 2)/13 = 7.26×10⁻³ mol.

Now we convert the moles to mass:

7.26×10⁻³ mol . 58 g/ 1mol = 0.421 g

Now we use stoichiometry to find the mass of the products.

Ratio is 13:8:10.

13 moles of oxygen can produce 8 moles of CO₂ and 10 moles of water

Then, 0.0472 mol would produce:

(0.0472 . 8)/13 = 0.0290 mol

We convert the moles to mass → 0.0290 mol . 44g /mol = 1.28 g

(0.0472 . 10)/13 = 0.0363 mol

We convert the moles to mass → 0.0363 mol . 18 g /1mol = 0.653 g

2. 4.92 g / 58 g/mol = 0.0848 moles of C₄H₁₀

2 moles of C₄H₁₀ react with 13 moles of O₂

So, 0.0848 moles will react with (0.0848 . 13) / 2 = 0.551 moles

We convert to mass: 0.551 mol . 32 g /mol = 17.6 g

Now we use stoichiometry to find the mass of the products.

Ratio is 13:8:10.

0.551 moles of O₂ will produce:

(0.551 . 8)/13 = 0.339 mol of CO₂

We convert to mass: 0.339 mol . 44g / mol =  14.9 g

(0.551 . 10)/13 = 0.424 mol of H₂O

0.424 mol . 18 g /mol = 7.63 g

3. In this case, we have the mass of one of the product

20.12 g . 1mol / 44 g = 0.457 moles of CO₂

According to stoichiometry:

8 moles of CO₂ are produced by the reaction of 13 moles of O₂ and 2 moles of C₄H₁₀

Then, 0.457 moles of CO₂ would be produced by:

(0.457 . 13)/ 8 = 0.743 moles of O₂

We convert to mass: 0.743 mol . 32 g/1mol = 23.7 g

(0.457 . 2)/8 = 0.114 moles of C₄H₁₀

We convert to mass: 0.114 mol . 58g/mol = 6.63g

Now we can determine, the mass of produced water:

(0.743 . 10)/13 = 0.571 mol of H₂O . 18g /mol = 10.3 g

4. We convert the moles of water:

8.84 g / 18g/mol = 0.491 moles

According to stoichiometry: 10 moles of water are produced by 13 moles of O₂ and 2 moles of C₄H₁₀

Then 0.491 moles will be produced by:

(0.491 . 10)/ 13 = 0.378 moles of O₂

We convert to mass: 0.378 mol . 32 g/1mol = 12.1 g

(0.491 . 2)/8 = 0.123 moles of C₄H₁₀

We convert to mass: 0.123 mol . 58g/mol = 7.12g

Now we can determine, the mass of produced carbon dioxide:

(0.378 . 8)/13 = 0.232 mol of CO₂ . 44g /mol = 10.2 g

5. Mass of mg, must be converted to grams

252 mg . 1 g/1000 mg = 0.252 g

It is the same as 2.

0.252 g of C₄H₁₀ . 1mol/58 g = 4.34×10⁻³ mol

2 mol of C₄H₁₀ react to 13 moles of O₂ then,

4.34×10⁻³ mol will react with (4.34×10⁻³ mol . 13) / 2 = 0.0282 mol

We convert the grams → 0.0282 mol . 32 g/mol = 0.903 g (903 mg)

0.0282 mol of oxygen will produced:

(0.0282 . 8)/13 = 0.0173 mol of CO₂

We convert to mass: 0.0173 mol . 44g / mol =  0.763 g (763 mg)

(0.0282 . 10)/13 = 0.0217 mol of H₂O

0.0217 mol . 18 g /mol = 0.390 g  (390 mg)

6. We define the mass of CO₂ → 198 mg . 1g/1000 mg = 0.198 g

0.198 g / 44g/mol = 4.5×10⁻³ moles of CO₂

According to stoichiometry:

8 moles of CO₂ are produced by the reaction of 13 moles of O₂ and 2 moles of C₄H₁₀

Then, 4.5×10⁻³ moles of CO₂ would be produced by:

(4.5×10⁻³ . 13)/ 8 = 7.31×10⁻³ moles of O₂

We convert to mass: 7.31×10⁻³ . 32 g/1mol = 0.234 g (234 mg)

(4.5×10⁻³ . 2)/8 = 1.125×10⁻³ moles of C₄H₁₀

We convert to mass: 1.125×10⁻³ mol . 58g/mol = 0.065 g (65 mg)

Now we can determine, the mass of produced water:

(7.31×10⁻³ . 10)/13 = 5.62×10⁻³ mol of H₂O . 18g /mol = 0.101 g (101 mg)

Answer:

pH= 12

Explanation:

Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:

KOH⇒ K⁺ + OH⁻

Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:

[OH⁻]= [KOH]

In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):

Mw (KOH)= K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol

moles KOH: mass/Mw= 0.140 g/(56 g/mol) = 2.5 x 10⁻³ moles

The molality of the solution is the number of moles of KOH per liter of solution:

V= 250.0 ml x 1 L/1000 ml= 0.250 L

M = (2.5 x 10⁻³moles)/(0.250 L)= 0.01 M

Now, we calculate pOH:

pOH = -log [OH⁻]= - log [KOH]= -log (0.01) = 2

Finally, we calculate pH from pOH:

pH + pOH = 14

⇒pH = 14 - pOH= 14 -2 = 12

Answer:

I think we will know a lot more about the universe and the things around us. We may also know a lot more about other planets, such as Mars and Saturn, and we might also know a lot more about other stars in the universe.

Explanation:

Answer:

remain the same

Explanation:

The pH of the buffer system remain the same when 0.030 moles of strong acid are added because buffer system has the property to resist any change in the pH  when acid or base is added to the solution. In buffer system, one molecule is responsible for neutralizing the pH of the solution by giving H+ or OH-.This molecule is known as buffer agent. If more base is added, the molecule provide H+ and when more acid is added to the solution, then the molecule add OH- to the solution.

Answer:

calcium chloride dihydrate

Answer:

14.3 mL  

Explanation:

Assume the student used 0.113 g ascorbic acid  and 0.0900 mol·L⁻¹ NaOH.

1. Balanced chemical equation.

The formula of ascorbic acid is H₂C₆H₆O₆ (MM = 176.12 g/mol).

However, for the balanced equation, let's write it as H₂A.

[tex]\rm H_{2}A + 2NaOH \longrightarrow Na_{2}A + 2H_{2}O[/tex]

2. Moles of ascorbic acid

[tex]\text{Moles of H$_{2}$A} =\text{0.113 g H$_{2}$A} \times \dfrac{\text{1 mmol H$_{2}$A}}{\text{0.176 12 mg H$_{2}$}A} = \text{0.6416 mmol H$_{2}$A}[/tex]

3. Moles of NaOH

The molar ratio is 2 mmol NaOH:1 mmol H₂A.

[tex]\text{Moles of NaOH}= \text{0.6416 mmol H$_{2}$A} \times \dfrac{\text{2 mmol NaOH}}{\text{1 mmol H$_{2}$A}} =\text{1.283 mmol NaOH}[/tex]

4. Volume of NaOH

[tex]V = \text{1.283 mmol NaOH}\times \dfrac{\text{1 mL NaOH}}{\text{0.0900 mmol NaOH}} = \textbf{14.3 mL NaOH}\\\\\text{The student will need $\large \boxed{\textbf{14.3 mL NaOH}}$}[/tex]

Answer:

The heat of the reaction or standard enthalpy of the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) is ΔH(rxn) = -41.2 kJ

Explanation:

The reaction whose standard enthalpy is required, as obtained from the internet is

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

The formation reaction for some of the reactants and products are given as

C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ

2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ

To find the standard enthalpies of the given reaction, we need the heat of formation of each of the species involved in the reaction

ΔH (CO₂(g)) = -393.5 kJ

ΔH (H₂O(g)) = -483.6 kJ ÷ 2 = -241.8 kJ (Because 2 moles of H₂O(g) are formed in the given formation reaction)

ΔH (O₂(g)) = ΔH (H₂(g)) = 0 kJ (No heat of formation for elements)

ΔH (CO(g)) = ? (This isn't given)

But it can be calculated from the second given reaction

2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ

Heat of reaction = ΔH(products) - ΔH(reactants)

Heat of reaction = -566.0 kJ

ΔH (products) = 2 × ΔH (CO₂(g)) = 2 × -393.5 = -787 kJ

ΔH (reactants) = [2 × ΔH (CO(g))] + [1 × ΔH (O₂(g))] = 2 × ΔH (CO₂(g))

Hence, we have

-566 = -787 - [2 × ΔH (CO₂(g))]

2 × ΔH (CO₂(g)) = -787 + 566 = -221 kJ

ΔH (CO₂(g)) = -221 ÷ 2 = -110.5 kJ

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Heat of reaction = ΔH(products) - ΔH(reactants)

ΔH (products) = [ΔH (CO₂(g))] + [ΔH (H₂(g))]

= -393.5 + 0 = -393.5 kJ

ΔH (reactants) = [1 × ΔH (CO(g))] + [1 × ΔH (H₂O(g))] = -110.5 - 241.8 = -352.3 kJ

Heat of reaction = -393.5 - (-352.3) = -41.2 kJ

Hope this Helps!!!

Answer:

Here's what I find  

Explanation:

Sodium vapour produces mostly yellow light.

Mercury vapour produces mostly blue light.

A white car reflects all colours back to our eyes.

A coloured car, say green, absorbs all the other colours and reflects the green.

Under a sodium light, a white or a yellow car will appear yellow. A red car will be nearly black because there is no red light to reflect.

Under a mercury light, a white or a blue car will appear blue. A yellow car will be quite dark, but there will be a yellowish tinge because there is some yellow in the Hg spectrum for it to reflect.  

The picture below shows a red car and a black car under sodium light. Can you tell which is which?

Answer:

Explanation:

(R)-5-octyne-4-ol

C8H140

Correct name for 5-octyne is 3-octane.

Structure is attached below.

Octane is a hydrocarbon and an alkane with the chemical formula C₈H₁₈, and the condensed structural formula CH₃(CH₂)₆CH₃. Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.

The structure for 5-octyne is given below.

Correct name for it is 3-octane.

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Answer:

distance = 7 miles

displacement = 5 miles

Explanation:

Distance is a scalar quantity as it takes account of magnitude traveled but not the direction traveled from starting point.

The distance traveled is the sum total of distances moved

distance = 4 + 3 = 7 miles

Displacement however, is a vector and measure the shortest possible distance traveled in a given direction from the starting point.

The path of Elvis' walking forms a right-angle triangle with the hypotenuse being the displacement and the other two sides being the distance traveled west and south.

Using Pythagoras' theorem; c² = a² + b²

where c = hypotenuse and a and b are the other two sides

c² = 4² + 3²

c² = 16 + 9 = 25

√c² = √25

c = 5

Therefore, displacement = 5 miles

Answer:

e) 4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

Explanation:

The missing image of the the  compound we are to name is attached below.

Before we can name an organic compound; It is crucial we know the guiding rules in naming them.

1. Select the longest continuous carbon chain as the root hydrocarbon and name according to the number of carbon atoms it contains, adding appropriate suffix to indicate the principal substituent group.

2. Number the carbon atoms in the root hydrocarbon from the end which gives the lowest number to the substituents.

3. If the same substituent is present two  or more times  in a molecule; the number of this substituent is indicated by the prefix di -(2), tri - (3) , tetra - (4) etc attached to the substituent name.

4. If there is more than one type of substituent in the molecule ; the substituents are named according to the alphabetical order but where  there are mixed substituents ; the inorganic are named first.

5. In selecting and numbering the longest continuous chain, the functional groups are given preference over substituents., i.e the functional group is given the smallest possible number.

In the light of the above guiding rules; we were able to name the given compound because the compound contains nine carbons in the ring form which result to root name nonane. The two methyl are on the third and eight carbon; bromine is on the fourth carbon ; there are two cyclic ring present in the compound where we have 5 carbons in one structure, another 2 carbons in the second structure and zero carbon in the bridge structure which eventually result to the correct  name:

4-bromo-3,8-dimethylbicyclo[5.2.0]nonane

Answer:

TiO2 = Nothing ledt

C = 7g

Ti = 5.98g

CO =3.5g

Explanation:

TiO2(s) +2C(s) → Ti(s) +CO(g)

Initial reactant masses of 10g

This is a limiting reactant problem. We have to find the limiting reactant; this is the reactant that determines the amount of product formed.

From the stoichiometry of the reaction;

1 mol of TiO2 reacts completely with 2 moles of C

Converting the starting masses to molres, we have;

Moles = mass /  molar mass

For TiO2

Moles = 10g / 79.866 g/mol = 0.125 moles

For C

Moles = 10g / 12g/mol = 0.8333 moles

Time to find the limiting reactant;

If all the 0.125 moles of TiO2 were used, it would require 2* 0.125 = 0.250 moles of C. But we have 0.833 moles of C. This means C is in excess, hence TiO2 is our limiting reactant.

All of TiO2 would be used up.

Moles left of C = 0.8333 - 0.250 moles = 0.58333

Mass left = Moles * Molar mass = 0.5833 * 12 = 6.99 = 7g

Mass of Ti and Co formed;

Moles formed is 0.125 moles .

Mass of Ti = Moles * Molar mass = 0.125 * 47.867 = 5.98 g

Mass of CO = Moles * Molar mass = 0.125 * 28 = 3.5g

-lg[H+]=1.5

which means that [H+]=0.03M

pOH=14-pH=14-1.5=12.5

-lg[H+]=1.5

which means that [H+]=0.03M

pOH=14-pH=14-1.5=12.5

the hydroxide ion concentration. is 12.5.

The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH. [H3O+] = 10-pH or [H3O+] = antilog (- pH) Example: What is the hydronium ion concentration in a solution that has a pH of 8.34? On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34).

For a 0.025 M HCl solution the concentration of the hydrogen ions, [H+] , is 0.025 moles/liter (i.e. 0.025 M). pH is defined as −log[H+] . Substituting 0.025 into the equation for pH we find that the 0.025 M HCl solution has a pH of -log(0.025) = 1.60.

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Answer:

E) n=4 l=2 ml=1 ms= 1/2

Explanation:

Arsenic is a member of group 15 in the periodic table. Its electronic configuration is;

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3. Its condensed electronic configuration can be written as [Ar]4s2 3d10 4p3. This electronic configuration shown here can now enable us to consider each option given in the question in order to meaningfully arrive at a logical answer.

If we look at option E, the data given for that electron is; n=4 l=2 ml=1 ms= 1/2. This refers to an electron in a 4d orbital. In the ground state configuration of arsenic shown above, there is no 4d orbital, hence option E must be the correct answer.

Answer:

...

Explanation:

in saturated fats there is no double bond between the acids and are tightly packed and unsaturated fats arent tight and loosely packed/put together

saturated- solid at room temperature

unsaturated= liquid at room temperature

two types of unsaturated fats, Polyunsaturated fats and Monounsaturated fats

Answer:

A.Electrolysis

Explanation:

A.Electrolysis

For example,  electrolysis of solution of NaCl in water gives H2 and O2.

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

Answer:

Volume is 0.00233mL

Explanation:

Hello,

The density of a substance is the mass per unit volume of that substance.

Density = mass / volume

To solve this question, we need to get our data first.

Data;

Density = 19.3g/mL

Mass = 45mg = 0.045g

Volume = ?

Density = mass / volume

Density (ρ) = mass (m) / volume (v)

ρ = m / v

v = m / ρ

v = 0.045 / 19.3

v = 0.00233mL

The volume of the substance is 0.00233mL

Answer:

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Explanation:

According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The complete ionic equation includes all the ions and molecular species.

C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Answer:

Coals

Explanation:

The land feature that supports the theory of continental drift is ; ( C ) coal fields

Continental drift is the gradual shift in position of the earth tectonic plates ( i.e. gradual shift in the continents in relation to ocean basins) and this due to the heat from the earths' mantle.

Coal fields supports this theory because the it is an area where coal is found in large quantities and mined for commercial purposes. coal fields areas are found as a result of continental drift.

Hence we can conclude that the land feature that supports the theory of continental drift is coal fields

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