"Calculate the pH of an aqueous solution at 25°C that is (a) 0.15 M in HCl. 0.82 (b) 3.7 M in HNO3. -0.57 (c) 6.9 × 10−4 M in HClO4."

Answer:

From fastest to slowest its: (4)A to B, (1)E to F, (3)C to D, (2)D to E

Explanation:

The steeper the line is the faster she went. D to E she didn't make any progress because the line is straight. Sry I'm terrible at explaining things.

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

Please mark me brainliest! hope this helped!

God bless!

Answer:

Explanation:

In order to find the original pressure , we use the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

[tex]V_1 = \frac{P_2V_2}{P_1} \\[/tex]

From the question

P1 = 250020 Pa

P2 = 120870 Pa

V2 = 6 L

We have

[tex]V_1 = \frac{120870 \times 6}{250020} = \frac{725220}{250020} \\ = 2.90064794...[/tex]

We have the final answer as

Hope this helps you

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:

Answer:

B

Explanation:

I did the question before and got it right.

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:

41.7 kJ/mol

Explanation:

ln(k) = ln(A) − Eₐ/(RT)

Pick any two points.  I'll choose 100°C and 400°C.

When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:

ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)

When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:

ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)

Subtract the two equations and solve:

ln(4.40×10⁻⁷) −  ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)

5.991 = 0.00120 Eₐ/R

Eₐ/R = 5013.4

Eₐ = 41700 J/mol

Eₐ = 41.7 kJ/mol

Answer:

175

Explanation:

Answer:

Answer:

B. Potassium

Explanation:

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

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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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:

I am sure they are why won't they

Answer:

In this example, a 3 kilogram mass, at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s 2 * 5m = 147.15 J. 9.81 meters per second squared (or more accurately 9.80665 m/s 2 ) is widely accepted among scientists as a working average value for Earth's gravitational pull.

Explanation:

Answer:

6.7

⋅

10

23

atoms of H

Explanation:

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:

29.8

Explanation:

The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.

Answer:

Salt domes storage has advantages in cost, security, environmental risk, and maintenance. Salt formations offer the lowest cost, most environmentally secure way to store crude oil for long periods of time. Stockpiling oil in artificially-created caverns deep within the rock-hard salt costs historically about $3.50 per barrel in capital costs. Storing oil in above ground tanks, by comparison, can cost $15 to $18 per barrel - or at least five times the expense. Also, because the salt caverns are 2,000-4,000 feet below the surface, geologic pressures will sea; any crack that develops in the salt formation, assuring that no crude oil leaks from the cavern. An added benefit is the natural temperature differential between the top of the caverns and the bottom - a distance of around 2,000 feet; the temperature differential keeps the crude oil continuously circulating in the caverns, giving the oil a consistent quality.

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:

about 0.13 mol

Explanation:

To find number of mols when given grams you first have to find the molar mass of the compound. This is done by adding up the atomic masses of the element in the compound. So Sr= 88 g/mol S=32 g/mol and O=16 g/mol. Then 88+32+(16x4)=184. Then using this you can convert from grams to mols by dividing the grams by the molar mass. So, 24.23/184 equals about 0.13 mol.

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:

b

Explanation:

Answer:

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

Explanation:

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: 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:

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  .

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:

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:

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.