Electrolysis of water produces its constituent elements, hydrogen and oxygen. This process is represented by the following balanced chemical equation:

2 H2O(l) → 2 H2(g) + O2(g).

If 8.05 grams of oxygen gas was produced in the reaction, how many molecules of water were decomposed?

Do enter the coefficient of the answer only. Do NOT include the unit in final answer.

_______________________________ × 1023 H2O molecules.

Answer:

A - Atom ---> molecule or compound.

Answer:

Water molecules are polar, so they form hydrogen bonds. This gives water unique properties, such as a relatively high boiling point, high specific heat, cohesion, adhesion and density.

EXPLANATION :

So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together.

Answer:

Number of neutrons

Explanation:

All have one single proton.  Hydrogen has no neutrons.  Hydrogen 2 or deuterium has 1 neutron.  Hydrogen 3 or tritium has 2 neutrons.

Answer: 0 I think

Explanation:

pretty sure its zero because  i learned it last year but im in middle school so you might want to look it up.

Answer:

C is for carbon and O is for oxygen

Answer:

2.28 atm

Explanation:

V₁ = 2.33L,  V₂ = 1.08L

T₁ = 289K,  T₂ = 304K

P₁ = 1.05 atm, P₂ = ?

Where V₁ and V₂ are initial and final volume respectively

T₁ and T₂ are initial and final temperature respectively

P₁ and P₂ are initial and final pressure respectively

The formula to be used here is the general gas equation:

P₁V₁/T₁=P₂V₂/T₂

1.05 × 2.23/289 = P₂ × 1.08/304

P₂ × 1.08 × 289 = 1.05 × 2.23 × 304

P₂ = (1.05 × 2.23 ×304) ÷ (1.08 × 289)

P₂ = 711.82 ÷ 312.12

P₂ = 2.28 atm

Answer:

Answer:

B. Potassium

Explanation:

The element with 19 electrons, 19 protons, and 20 neutrons is potassium

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial  molarity =  1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ =  0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

Answer:

3.55 L.

Explanation:

We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:

Molar mass of MnO2 = 55 + (16×2)

= 55 + 32

= 87 g/mol

Mass of MnO2 = 12 g

Mole of MnO2 =...?

Mole = mass /Molar mass

Mole of MnO2 = 12 / 87

Mole of MnO2 = 0.138 mole

Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)

From the balanced equation above,

1 mole of MnO2 reacted to produce 1 mole of Cl2.

Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.

Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:

Temperature (T) = 25 °C = 25 °C + 273 = 298 K

Pressure (P) = 0.950 atm

Number of mole (n) = 0.138 mole

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =.?

PV = nRT

0.950 × V = 0.138 × 0.0821 × 298

Divide both side by 0.950

V = (0.138 × 0.0821 × 298) / 0.950

V = 3.55 L

Therefore, 3.55 L of chlorine gas were obtained from reaction.

Answer:

2 double bonds 2 single bonds

Answer:

0.307

Explanation:

use the formula of n=mass /molar mass

Answer:

Boron and Aluminum

Explanation:

If you write the electron configuration for boron and aluminum, you get:

[tex]1s^22s^22p^1[/tex] for boron and [tex]1s^22s^22p^63s^23p^1[/tex] for aluminum. Both have 3 valance electrons and has 2 electrons in a s-orbital and 1 in a p-orbital. These valance electron similarities are based on the column/group the elements are. Therefore, Boron and Aluminum have similar chemical behaviours and similar arrangement of outer/valance electrons.

Answer:

Above 7

Explanation:

The equivalence point of any titration can be read off from the appropriate titration curve.

A titration curve is a plot of the pH of analyte against the volume of titrant added.

For a strong base and weak acid, the equivalence point lies above 7.

The pH of the resulting salt to be pH> 7 .

The equivalence point of any titration can be read off from the appropriate titration curve. A titration curve is a plot of the pH of analyte against the volume of titrant added. It is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration.

For a strong base and weak acid, the equivalence point lies above 7.

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Answer:

The percentage yield of sodium carbonate is 91.47%

Explanation:

we start by writing the reaction equation:

2NaCl + CaCO3 ——-> Na2CO3 + CaCl2

From the reaction we can see that 2 moles of sodium chloride produced 1 mole of sodium carbonate

Let us calculate the actual number of moles of sodium chloride produced from 112 g of it

Mathematically,

number of moles = mass/molar mass

Molar mass of sodium chloride is 23 + 35.5 = 58.5 g/mole

So the number of moles of sodium chloride produced will be 112/58.5 = 1.91 moles

The number of moles of sodium carbonate produced is half of this = 1.91/2 = 0.955

The mass of sodium carbonate produced from 0.955 moles of it will be;

number of moles * molar mass

The molar mass of sodium carbonate is 106 g/mol

So the number of moles is = 0.955 * 106 = 101.23 g

Mathematically;

percentage yield = actual yield/theoretical yield * 100%

Percentage yield = 92.6/101.23 * 100% = 91.47%

Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

[tex]p(\%) = \frac{1.6}{19.3} \times 100 \\ = 8.290155440...[/tex]

We have the final answer as

Hope this helps you

Answer:

Linear molecule with two domains

Explanation:

Complete Question

The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).

(a) before addition of any HCl (b) after addition of 25.0 mL of HCl

Answer:

a The value  is  [tex]pH =12.81[/tex]

b [tex]pH  = 11.9[/tex]

Explanation:

From the question we are told that

  The first pKb value  for B is [tex]pK_b_1  =  2.10[/tex]

   The second pKb value  for B is [tex]pK_b_2  =  7.54[/tex]

     The volume is  [tex]V =   50.0 mL   =[/tex]

     The  concentration  of  B is  [tex][B]  =  0.60 M[/tex]

     The concentration of [tex]C_A =  0.60 M[/tex]

Generally the reaction equation showing the first dissociation of B is  

[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)}  +  OH^- _{(aq)} }[/tex]

Here the ionic  constant for B is mathematically represented as

      [tex]K_i  =  \frac{[BH^+] [OH^-]}{[B]}[/tex]

Let denot the concentration of  [BH^+]  as  z  and  since [tex][BH^+] =  [OH^-][/tex] then [tex][OH^-][/tex] is also  z

So  [B] =  0.60  -  z  

Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is

   [tex]K_i  =  7.94 *10^{-3}[/tex]

So

      [tex] 7.94 *10^{-3}=  \frac{z^2}{ 0.60 - z}[/tex]

=>   [tex]z^ 2 + 0.00794 z - 0.00476[/tex]

using quadratic formula to solve this equation

     [tex]z = 0.0651[/tex]

Hence the concentration of  [tex]OH^{-}[/tex] is   [tex][OH^-] =0.0651[/tex]

Generally  [tex]pOH =  -log [OH^-][/tex]

=>    [tex]pOH =  -log (0.065)[/tex]

=>    [tex]pOH = 1.187 [/tex]

Generally the pH is mathematically represented as

    [tex]pH = 14 - 1.187[/tex]

      [tex]pH =12.81[/tex]

Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is   [tex] 50 mL [/tex]

The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL

Now the pOH at half way to the first dissociation of the base is  

     [tex]pOH  =  -log(K_i)[/tex]

=>   [tex]pOH  =  -log(0.00794)[/tex]

=>   [tex]pOH  =  2.100[/tex]

Generally the pH after addition of 25.0 mL of HCl is  

    [tex]pH  =  14 -  2.100[/tex]\

=>   [tex]pH  = 11.9[/tex]

The first dissociation's equation is as follows:

[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]

The second dissociation of the base equation is

[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]

[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]

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Answer:

The mass of sodium chloride in the mixture is 18.75%

Explanation:

Here, we want to calculate the percentage of sodium chloride in the mixture.

The total mass of the mixture is 52 g + 12 g = 64 g

So the percentage mass of sodium chloride will be;

mass of sodium chloride/ Total mass * 100%

That will be: 12/64 * 100 = 18.75%

Answer : The mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.

Explanation :

Balanced chemical reaction : It is a chemical reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

The balanced chemical reaction is:

[tex]PCl_3+Cl_2\rightarrow PCl_5[/tex]

By the stoichiometry of the reaction we can say that 1 mole of [tex]PCl_3[/tex] reacts with 1 mole of [tex]Cl_2[/tex] to give 1 mole of [tex]PCl_5[/tex].

From this we conclude that the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.

Hence, the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.

Answer:

A

Explanation:

Answer:

P₄(l) + 5 O₂(g) + 6 H₂O(l) ⇒ 4 H₃PO₄(aq)

Explanation:

Phosphoric acid is synthesized using a two-step thermal process.

In the first step, phosphorus and oxygen react to form diphosphorus pentoxide. The corresponding chemical equation is:

P₄(l) + 5 O₂(g) ⇒ 2 P₂O₅(g)

In the second step, diphosphorus pentoxide and water react to form phosphoric acid. The corresponding chemical equation is:

P₂O₅(g) + 3 H₂O(l) ⇒ 2 H₃PO₄(aq)

We can get the net chemical equation by adding the first step, the second step multiplied by 2, and canceling what is repeated on both sides.

P₄(l) + 5 O₂(g) + 2 P₂O₅(g) + 6 H₂O(l) ⇒ 2 P₂O₅(g) + 4 H₃PO₄(aq)

P₄(l) + 5 O₂(g) + 6 H₂O(l) ⇒ 4 H₃PO₄(aq)

Answer:

C

Explanation:

Because sodium is basically always soluble with any compound, it is between a and c. a is part of the reactant so it cant be A. So C.

The statement, that describes the compounds would be the precipitate that forms in the reaction is "Ba3(PO4)2"

Precipitate is a solid generated by a change in a solution, usually due to a chemical reaction or a change in temperature that reduces a solid's solubility.

The combination of more than one element will be identified ad compound.

When cations and anions in aqueous solution combine to create an insoluble ionic solid called a precipitate, double displacement reactions occur, resulting in the formation of a solid form residue. Except for salts of Group 1 metals and ammonium, salts of phosphates and carbonates ions are insoluble, according to the solubility flow chart. The production of a solid white precipitate is used to demonstrate the insoluble nature of barium phosphate.

Hence the correct option is c.

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Answer:

Explanation:

moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles

1 mole of potassium dichromate reacts with 6 moles of Fe⁺²

65 x 10⁻⁵ moles of potassium dichromate will react with

6 x 65 x 10⁻⁵ moles of Fe⁺²

= 390 x 10⁻⁵ moles

390 x 10⁻⁵ moles are contained in 25 mL of solution

molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³

= 15.6 x 10⁻² M  .

Given :

A volume of 80.0 mL of a 0.690 M [tex]HNO_3[/tex] solution is titrated with 0.790 M KOH.

To Find :

The volume of KOH required to reach the equivalence point.

Solution :

We know, at equivalent point :

moles of [tex]HNO_3[/tex] = moles of KOH

[tex]M_{HNO_3}V_{HNO_3}=M_{KOH}V_{KOH}\\\\0.690\times 80 = 0.790\times V_{KOH}\\\\V_{KOH}=\dfrac{0.690\times 80 }{ 0.790}\ ml\\\\V_{KOH}=69.87\ ml[/tex]

Therefore, volume of KOH required is 69.87 ml.

Hence, this is the required solution.

Answer:

b

Explanation:

Answer:

Gravity attracts the moon Earth, and Inertia keeps the moon in motion.

Explanation:

Answer:

Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.)

(a)

the electron affinity of Ar2+

kJ/mol

(b)

the electron affinity of S+

kJ/mol

(c)

the ionization energy of Cl−

kJ/mol

(d)

the ionization energy of Cl

kJ/mol

(e)

the electron affinity of Cl+

kJ/mol

Successive Ionization Energies in Kilojoules per Mole for the Elements in Period 3 General increase 13 14 Element 11 12 I. 16

Explanation:

Answer:

B

Explanation:

I did the question before and got it right.

Answer:

Kps = 3.07 x 10⁻⁸

pKsp= 7.51

Explanation:

First, we calculate the molar solubility of  silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):

2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s

Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:

     Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻

I                               0              0

C                            +3s           +s

E                            3s               s

Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴

Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:

Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸

The pKsp value is:

pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

[tex]volume = \frac{mass}{density} \\ [/tex]

From the question

mass = 15.5 g

density = 6.89 g/mL

We have

[tex]volume = \frac{15.5}{6.89} \\ = 2.249637155...[/tex]

We have the final answer as

Hope this helps you

Answer:

acceleration = force/mass

                 = (68.6+mg)/7

                 = 19.6 m/s²

Explanation:

9.8 m/s² is the acceleration acting on 7 kg mass if force of 68.6 N  is used to move it towards earth.

Force is defined as a cause which is capable of changing the motion of an object. It can cause an object which has mass to change it's velocity. It is also simply a push or a pull . It has both magnitude as well as direction.Hence, it is a vector quantity.

It has SI units of Newton and is represented by'F'.Newton's second law states that force which acts on an object is equal to momentum which changes with time. If mass of object is constant, acceleration is directly proportional to net force acting on an object.

The concepts which related to force are thrust and torque .Thrust increases the velocity of an object and torque produces change in rotational speed of an object.

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