A nonpolar solvent has the highest solubility with a(n) _____ solute. A. polar B. nonpolar C. ionic D. electrically charged solute. ​

1. The pH of the solution is 1.36

2. The pH of the solution is 10.47

3. the concentration of [H₃O⁺] in the solution is 6.3×10⁻¹⁴ M

This is simply a measure of the acidity / alkalinity of a solution.

The pH measures the hydrogen ion (hydronium ion) concentration while the poH measures the hydroxide ion concentration. It is expressed mathematically as

pH = –Log H₃O⁺

pH = –Log H₃O⁺

pH = –Log 4.32×10⁻²

pH = 1.36

pH = –Log H₃O⁺

pH = –Log 3.4×10⁻¹¹

pH = 10.47

pH = –Log H₃O⁺

13.2 = –Log H₃O⁺

Multiply through by -1

-13.2 = Log H₃O⁺

Take the anti-log of -13.2

H₃O⁺ = anti-log (-13.2)

H₃O⁺ = 6.3×10⁻¹⁴ M

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Amount of heat released:

Heat is released when 25.0 gm of liquid copper at its freezing point of 1,085°C changes to solid is -5125 J

Given:

Mass of liquid copper, m= 25 gm

Latent heat of fusion of copper, L = 205.0 J/g

The amount of heat needed to change one gram of a substance from solid to liquid is known as the latent heat of fusion.

However, in this instance, it is necessary to determine the quantity of heat emitted when liquid copper solidifies.

So, L = -205.0 J/g

Calculation:

We may use the unitary approach to calculate the heat released when 25.0 grams of liquid copper solidify at 1,085°C, which is its freezing point.

So,

Q = L x m

Q = -205 J/g x 25 g

Q = -5125 J

So, the correct answer is b) -5,130 J, which is much nearest to -5125 J as we have calculated above.

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Answer:

-5,130 J

Explanation:

I got 100% on the quiz edge 2022

Rather than being oxidized for energy, sugars are connected by glycosidic linkages to make a capsule. This is catabolism process.

It is a type of linkage in which covalent bond that joins a carbohydrate molecules to another group is called glycosidic bond or glycosidic linkage.

Example: Hemiacetal and hemiketal.

Catabolism is the part of metabolic process which uses energy to break down into smaller units.

Thus from the above conclusion we can say that Rather than being oxidized for energy, sugars are connected by glycosidic linkages to make a capsule. This is catabolism process.

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During photosynthesis, 5 moles of water are needed to produce 150 grams of glucose. The correct option is D.

It is the process by which green plants and some other organisms use sunlight to synthesize nutrients from carbon dioxide and water.

6 CO₂ + 6 H₂O ⇒ C₆H₁₂O₆ + 6 O₂

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

150 g × 1 mol/180.16 g = 0.833 mol

The molar ratio of H₂O to C₆H₁₂O₆ is 6:1.

0.833 mol C₆H₁₂O₆ × 6 mol H₂O/1 mol C₆H₁₂O₆ = 5.00 mol H₂O

During photosynthesis, 5 moles of water are needed to produce 150 grams of glucose. The correct option is D.

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Answer:

B) They are made up of particles that travel in a curved path when in motion.

Explanation:

We know that gas, being a state of matter, is made up of particles. John Dalton's solid-sphere model was proven to be incorrect by Ernest Rutherford's gold foil experiment.

These particles do NOT always move very slowly. There are multiple different factors that affect the speed of the particles such as temperature or concentration.

Answer:

0.296 mg

Explanation:

8.00/2.90 half lives have passed, so this means the remaining mass is

[tex]2.00 {(0.5)}^{8.00 \div 2.90} [/tex]

which is about 0.296 mg.

The amount of polonium remaining after 8.00 years is 0.296 mg. This is obtained by using rate law of first order kinetics.

The expression for the rate law for first order kinetics is

k = [tex]\frac{2.303}{t}log\frac{a}{a-x}[/tex] years⁻¹

Where,

k - rate constant

t - time passed by the sample

a - initial amount of the reactant

a-x  - final amount of the reactant

The expression for the half-life of the compound is

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex] years

Given that,

A Polonium-208 is an alpha emitter with a half-life of 2.90 years.

I.e., [tex]t_{\frac{1}{2}}[/tex] = 2.90 years

Then,

2.90 = (0.693)/k

⇒ k = (0.693)/(2.90) = 0.2389 years⁻¹

Then,

for t = 8.00 years,

Since we have rate law of first order kinetics as

k =  [tex]\frac{2.303}{t}log\frac{a}{a-x}[/tex]

⇒ k = (2.303/8.00) log(a/a-x)

Where the initial amount of the reactants is a = 2.00 mg

So,

k = 0.287 log (2.00/a-x) (a-x is the final amount of the reactants)

⇒ 0.238 = 0.287 log(2.00/a-x)

⇒ (0.238/0.287) = log(2.00/a-x)

⇒ 0.829 = log(2.00/a-x)

⇒ (2.00/a-x) = [tex]10^{0.829}[/tex] = 6.745

⇒ (a-x) = 2.00/6.745 = 0.296 mg

Therefore, the remaining polonium is 0.296 mg after 8.00 years.

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Answer:

m =1,182.8g

Explanation:

there you go

neutralization reaction = acid+base = salt + water

molecular equation = balanced equation

2HI(aq) + Ba(OH)₂(aq) ⇒ BaI₂(aq) + 2H₂O(l)

[tex]6h2o + 6co2 = 6o2 + c6h12o6[/tex]

[tex]n = n \\ \frac{ mass}{molar \: mass} = \frac{mass}{molar \: mass} [/tex]

[tex] \frac{55}{44} = \frac{m}{32} \\ m = \frac{55 \times 32}{44} = 40 \: grams[/tex]

Photosynthesis reactions are the process that occurs in the plants to yield oxygen from food. In the process, 55 gm of carbon dioxide produces 40 gm of oxygen. Thus, option B is correct.

Photosynthesis is a metabolic reaction that takes place in the plant cell where the sun energy converts the water and carbon dioxide into sugar (glucose) and oxygen.

The photosynthetic reaction is given as,

6H₂O + 6 CO₂ → C₃H₁₂O₆ + 6O₂

In the above reaction, there are 6 moles of carbon dioxide needed to produce 6 moles of oxygen. Hence, moles of carbon dioxide consumed = moles of oxygen produced.

Mass from moles is calculated as,

Moles = mass ÷ molar mass

Equating moles of both CO₂ and O₂,

mass ÷ molar mass = mass ÷ molar mass

55 ÷ 44 = m ÷ 32

m = 55 × 32 ÷ 44

= 40 gm

Therefore, 40 gm of oxygen is produced from 6 moles of carbon dioxide.

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Answer:

0.5 mol/dm3

Explanation:

e.g. g/dm3, g/cm3 and mol/dm3

= g dm-3, g cm-3 and mol dm-3

The chemical formula of aluminum chloride is AlCl₃ and the  chemical formula of aluminum oxide is Al₂O₃.

The chemical formula of aluminum chloride is written as;

Al³⁺  + Cl⁻   → AlCl₃

The chemical formula of aluminum oxide is written as;

Al³⁺  + O²⁻   → Al₂O₃

Thus, the chemical formula of aluminum chloride is AlCl₃ and the  chemical formula of aluminum oxide is Al₂O₃.

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Answer:

gg mate

Explanation:

Answer:

A solution of NaOH has concentration 1.2M. Calculate the mass of NaOH in g/dm3 in this solution. = 1.2x 40x 1 = 48 g 3.

The mass of NaOH in the solution having concentration of 1.2 M is 48.00 grams per decimeter cubed (g/dm³).

Concentration refers to the amount of a solute dissolved in a given amount of solvent or solution. It is a measure of how much solute is present in a specific volume or mass of the solution.

Concentration can be expressed in various units, such as molarity (moles of solute per liter of solution), mass percent (mass of solute per 100 grams of solution), parts per million (ppm), and others.

Given:

Concentration of NaOH solution = 1.2 M

The molar mass of NaOH (sodium hydroxide) is:

Na: 22.99 g/mol

O: 16.00 g/mol

H: 1.01 g/mol

Total molar mass of NaOH = 22.99 + 16.00 + 1.01

= 40.00 g/mol

1 mole of NaOH = 40.00 g

1.2 moles of NaOH = 1.2 moles × 40.00 g/mole

= 48.00 g/L

1 L = 1 dm³

So, 48.00 g/L = 48.00 g/dm³

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Answer: 50%

Explanation:

To find the percent composition of oxygen in SO₂, we need to find out the molar masses of both elements and add them.

First, figure out how many atoms are in each element of this compound. Then figure out the molar mass by multiplying it by the molar mass you'd see on the periodic table.

[tex]1atomS*32.06g=32.06g/molS\\2atomO*15.999g=31.998g/molO_{2}[/tex]

Now that we have that information, we must add up the numbers to get the molar mass of SO₂.

[tex]32.06g/molS+31.998g/molO_{2} =64.058g/molSO_{2}[/tex]

Now we have the molar mass of SO₂. To find the percent composition of oxygen in SO₂, we must divide the molar mass of O₂ by that of SO₂, then multiply that by 100.

[tex]\frac{31.998g/molO_{2} }{64.058g/molSO_{2} } *100=49.95; 50[/tex]

Given this, the percent composition of oxygen in SO₂ is 50%.

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Colloidal solution has large size particles hence it is able to scatter light.

A colloidal solution is a solution that is composed of particles that are larger in size than the particles that compose a true solution.

For question 6;

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

For question 7;

Assertion (A) is true but reason (R) is false.

For question 8;

Assertion (A) is false but reason (R) is true.

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Answer:

9.9363 x 10^23  atoms

Explanation:

1.65   x   Avagadro's Number

1.65 * 6.022 x 10^23 =

In ground-state electron configuration, the principal quantum number (n), the orbital (s, p, d, or f), and the total number of electrons are used to represent electron configurations.

Case 1 Ground-state electron configuration of elements.

A) Chlorine: 1s²2s²2[tex]p^{6}[/tex]3s²3[tex]p^{5}[/tex]

B) Cesium: 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s¹

C) Vanadium: 1s²2s²2p⁶3s²3p⁶3d³4s²

D) Rhenium: 1s² 2s² 2p⁶ 3s² 3p⁶3d¹⁰ 4s²4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d⁵ 6s²

Case 2 : Number of valence electron in atoms

A) Sn -Tin - Four valence electrons

B) La - Lanthanum - Three valence electrons

C) Mn - Manganese - Seven valence electrons

D) Zn- Zinc - Two valence electrons

Case 3 Ground-state electronic configuration of ions

A) Co⁺³ : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹3d⁵

B) Mo⁺² : 1s² 2s² 2p⁶3s²3p⁶ 4s² 3d¹⁰ 4p⁶5s² 4d²

C) Ra⁺² : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6p⁶

D) I⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶

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The density = 1.77 x [tex]10^{-2}[/tex] where Mass =4.7033 x [tex]10^4[/tex] g and Volume=2.6540 x [tex]10^6[/tex]mL.

What is density?

Density is the measurement of how tightly a material is packed together. It is defined as the mass per unit volume. Density Symbol: D or ρ Density Formula: ρ = m/V, where ρ is the density, m is the mass of the object and V is the volume of the object.

Given:

Density = (4.7033 x [tex]10^4[/tex] g/ 2.6540 x [tex]10^6[/tex] mL)

Mass =4.7033 x [tex]10^4[/tex] g

Volume=2.6540 x[tex]10^6[/tex] mL

Density = (4.7033 x [tex]10^4[/tex] g/ 2.6540 x [tex]10^6[/tex] mL)

Density = 1.772155237 x [tex]10^{-2}[/tex] = 1.77 x [tex]10^{-2}[/tex]

Hence, the density = 1.77 x [tex]10^{-2}[/tex].

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Answer:

Explanation:

Sounds like oxygen acts as a 'catalyst'

A solution containing 1 mol of a strong electrolyte will have a greater effect on the colligative properties of the solution than 1 mol of a nonelectrolyte. A strong electrolyte will dissociate or ionize in an aqueous solution, and therefore the total number of dissolved solute particles is greater for 1 mol of a strong electrolyte than for 1 mol of a nonelectrolyte solute.

An electrolyte is a material that separates into charged ions when it is in contact with water. Cations are positively charged ions. Anions are ions that are negatively charged. A substance that may conduct an electric current when melted or dissolved in water is known as an electrolyte.

A substance is referred to as a strong electrolyte if the physical or chemical process that produces the ions is virtually 100% effective (all of the dissolved molecules give ions). Weak electrolytes are those in which only a very small portion of the dissolved substance goes through the ion-producing process.

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Answer:

Explanation:

I'm assume that 3 Br2 + 6 OH¹-5 Br¹ + BrO3¹ + 3 H₂O is meant to read:

3 Br2 + 6 OH^-1    =   5 Br^-¹ + BrO3^-¹ + 3 H₂O

The balanced equation tells us that 3 moles of Br2 will react with 6 moles of OH^-1 to produce 5 moles of Br^-1.

The first step is to determine whether the Br2 or OH^-1 are limiting reagents.  That is, is there enough of each to complete the reaction, with none left over.

We need 3 moles of Br2 for every 6 moles of OH^-1, a molar ratio of 1/2 (Br2/OH).

We are given 4.68 moles of Br2 and 8.12 moles of OH^-1.  That is a ratio of 4.68/8.12 or 0.5764.  This is higher than the ratio of 1/2 or 0.5 that is required.  That means we have more than enough Br2.  The limiting reagent is the OH^-1.  Once it is consumed, the reaction stops and we are left with some unreacted Br2.

So we need the molar ratio of the OH^-1 to the Br from the balanced equation:  We see that 6 moles of OH^-1 are required to produce 5 moles of Br, a 6/5 molar ratio.

Therefore, we may assume all 8.12 moles of the limiting reagent, OH^-1, will be consumed to produce *6/5) that amount of Br.

(8.12 moles OH^-1)*((6 moles Br)/(5 moles OH^-1)) = 9.75 moles of Br.

For curiosity's sake, we can determine the amount of unreacted Br2.  8.12 moles of OH^-1 would require (8.12 moles OH)*(1/2) = 4.06 moles of Br2.

4.68 moles starting Br2

4.06 moles consumed

0.62 moles remaining Br2

The Answer to the question is 14 grams approximately.

Atomic mass of Na = 23 u

Given mass of Na = 3.4 gm

Moles of Na = given mass of Na / molar mass of Na

Moles of Na = 3.4 / 23 moles = 0.15 moles approximately

So, 0.15 moles of Na approximately

Molar mass of Cl₂ = 35.45 × 2 = 70.9  u

Given mass of Cl₂ = 8.9 gm

Moles of Cl₂ = given mass of Cl₂ / molar mass of Cl₂

Moles of Cl₂ = 8.9 / 70.9 moles = 0.12 moles approximately

So, 0.12 moles of Cl₂ approximately

Balanced chemical equation

2Na   +     Cl₂ → 2NaCl

Ratio of Moles to Stoichiometric coefficients

0.15/2     0.12/1  

0.075  <  0.12

So, as ratio of Na is less then it will be our Limiting reagent.

Which means that Cl₂ will consume complete in the reaction.

It is clear from the reaction that 1 moles of Cl₂ are making 2 mole of NaCl

1 moles of Cl₂ → 2 mole of NaCl

0.12 moles of Cl₂ → 2 × 0.12 mole of NaCl

0.12 moles of Cl₂ → 0.24 mole of NaCl

So, 0.12 moles of Cl₂ will make 0.24 mole of NaCl

0.24 mole NaCl = mass formed NaCl / molar mass of NaCl

mass formed NaCl = 0.24 × molar mass of NaCl

mass formed NaCl = 0.24 × 58.44

mass formed NaCl = 0.24 × 58.44

mass formed NaCl = 14 grams    (approximately)

14 grams of NaCl will be formed.

So, approximately 14 grams NaCl is formed from 3.4 grams of Na and 8.9 grams of Cl.

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The main combustible part of wood is cellulose, a compound of carbon, hydrogen, and oxygen, which, when it burns, breaks down to carbon dioxide, water, and heat.

The substances which catches fires easily in air and produce heat and light upon burning is known as Combustible Substances.

Example: Wood, Oil, Kerosene, Hydrogen gas, Cloth etc.

When these compounds carbon, hydrogen and oxygen burns completely the carbon oxidizes to carbon dioxide , water and heat.

Thus from the above conclusion we can say that The main combustible part of wood is cellulose, a compound of carbon, hydrogen, and oxygen, which, when it burns, breaks down to carbon dioxide, water, and heat.

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Heat produced is given by -0.813 kJ

Calculation of heat produced:

The reaction:

[tex]3NaOH + H_3PO_4 \rightarrow Na_3PO_4 + H_2O[/tex]     ΔH[tex]_{rxn}[/tex]= -173.7 kJ (Assumed)

Step 1: Calculation of moles using Molarity.

Volume of NaOH= 31.2 mL

Concentration= 0.45M

[tex]M=\frac{n}{V}\times1000\\\\n=\frac{M\times V}{1000} \\\\n= \frac{31.2\times0.45}{1000}\\n= 0.01404[/tex]of NaOH

Volume of [tex]H_3PO_4[/tex] = 65.4 mL

Concentration= 0.088M

[tex]n=\frac{0.088\times 65.4}{1000}\\\\n= 0.005755[/tex] mole of [tex]H_3PO_4[/tex]

Step 2: Limiting reagent

According to the above reaction

1 mole of [tex]H_3PO_4[/tex] reacts with 3 mole of NaOH

0.005755 moles will react with [tex]3\times 0.005775 \, NaOH[/tex]= 0.017 moles of NaOH is required. But we have 0.01404 moles of NaOH.

So NaOH is the limiting reagent.

Step3: Calculate ΔH:

We will limiting reagent to calculate the heat produced-

[tex]0.01404 \,mole \times\frac{-173.7}{3}= -0.813\, kJ[/tex]

Heat produced= -0.813 kJ

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Answer:

0.3011 X10²³

Explanation:

hope this helps u

Answer: C. 20

Explanation: One equation for the equilibrium constant is [tex]\frac{[Products]}{[Reactants]}[/tex]. The products are typically on the left side and the reactants on the right.

Given this and the values of each, we can easily plug this in. However, we must multiply the products instead of adding them.

[tex]\frac{0.69M*0.98M}{0.034M} =19.888, or 20M[/tex]

19.888 can be easily rounded to 20, meaning that C is the best answer.

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When electrons in an atom transition from a lower energy state to a higher energy state it has to absorb some energy in order to transition to the higher energy state.

When properly stimulated, the electrons in these materials move from lower energy levels to higher energy levels and occupy different orbits. Then, at some point, these high-energy electrons emit "extra" energy in the form of photons and return to their original energy levels.

Electrons jump to higher energy levels when excited by external energy gains such as B. Large heating, the presence of an electric field, or collision with another electron.

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Butane is a gaseous molecule which is part of the organic compounds called  hydrocarbon,

Butane is a gaseous molecule which is generally called a hydrocarbon.

Hydrocarbons are composed of only carbon and hydrogen. They are of organic origin.

Butane is a low molecular weight hydrocarbon which is used as a fuel.

In conclusion, butane is a gaseous hydrocarbon,

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The reactant in this reaction is 2-fluoro-5-propyl - heptane while the product is  2-hydroxy-5-propyl - heptane.

We define a substitution reaction as one in which a given moiety is replaced by another. This is an example of a substitution reaction.

The name of the reactant in this reaction is 2-fluoro-5-propyl - heptane while the product is  2-hydroxy-5-propyl - heptane.

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Answer:

The answer is A

Explanation:

Answer:

Most fire extinguishers work by separating the fuel from the oxygen. The oxygen comes from the air. It is the same oxygen we breathe. Since the oxygen has to be in contact with the fuel, if you can coat the fuel with something that keeps the oxygen away, the fire will go out. Some ways of using fire extinguisher are:

Explanation:

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