An atom has the following isotopes in the the following abundances. 20 10Z 30% , 19 10Z10% , 18 10Z 60% What is the average atomic mass

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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I2 (s) has very low solubility in H2O (l) because the weak dipole-induced dipole forces between I2 and H2O are not strong enough to replace the hydrogen bonds between the H2O molecules.

Dipole interactions are susceptible interactions that get up from the close association of permanent or precipitated dipoles. Collectively these forces are called Van der Waals interactions. Proteins incorporate a massive number of these interactions, which vary substantially in energy.

Solubility depends on the solvent's capacity to overcome the intermolecular forces in a solid. Red blood cells have an excessive concentration of dissolved ions. When positioned into pure water they rupture.

Polar species are soluble in water, whilst nonpolar species are soluble in oils and fats. Covalent solubility uses the like dissolves like a rule. This means that materials with an equal form of polarity might be soluble in one another. moreover, compounds with differing polarities might be insoluble in one another.

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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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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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This refers to the material that is made when different substances mix up physically and causes a reaction.

You can make on the project of Mixture -

This refers to the chemical bond that holds different atoms tightly

You can make on the project of Compound -

Based on the fact that the atom is the smallest indivisible part of an element, elements like phosphorous cannot be further broken down.

You can make on the project of Elements -

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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:

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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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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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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Answer:

4170 j

Explanation:

water has low melting point

This chapter deals with chemical equilibrium, or how far chemical reactions proceed. Some reactions convert reactants to products with near 100% efficiency but others do not.

Chemical equilibrium is a dynamic process and occurs when the rates of two opposing processes (reactions) become the same. Unless conditions are somehow changed, the concentrations of both reactants and products remain constant after equilibrium is achieved.

Consider the decomposition reaction in which N2O4 (g) ' 2 NO2(g) What happens to the concentration of each, based upon the rate equations, as the reaction proceeds

Assuming reversible elementary processes, Rate (forward) = k (forward) x [N2O4]  and rate (reverse) = k (reverse) x [NO2]2As [N2O4] decreases, rate forward decreases, and, as [NO2] increases, rate reverse increases, but more quickly.

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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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To be effective, collisions between atoms must occur with the proper orientation and also the proper amount of energy.

An inelastic collision is a collision in which there is a loss of kinetic energy. While the momentum of the system is conserved in an inelastic collision, the kinetic energy is not. This is because some kinetic energy has been transferred to something else.

In this case, is necessary have energy to be effective, so collisions between atoms must occur with the proper orientation.

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Answer:

energy

Explanation:

[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:

0.296 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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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:

The answer is A

Explanation:

[tex]\\ \tt\leadsto -log[H^+]=5.31[/tex]

[tex]\\ \tt\leadsto log[H^+]=-5.31[/tex]

[tex]\\ \tt\leadsto [H^+]=antilog(-5.31)[/tex]

[tex]\\ \tt\leadsto [H^+]=4.89\times 10^{-6}M[/tex]

Answer:

option c

Explanation:

They are fast-moving particles that can penetrate the skin.

Answer:

discuss the successes and the short comings of the conservative reaction to the french rvolution as seen in the actions of the congress of vienna and the holy aliance?

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)

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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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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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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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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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

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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