What does ozone do to make life on earth possible?

Answer:the answer is b

Explanation:got it right on ed

Using stoichiometry, we find that when 4.3 g of Al reacts, 8.14 g of Aluminium Oxide and 8.89 g of Iron are formed.

To calculate the grams of products formed in a reaction, we use stoichiometry, which is based on the law of conservation of mass. Here, for every 2 moles of Al (Aluminium, 26.98 g/mol) reacted, 1 mole of Al2O3 (Aluminium oxide, 101.96 g/mol) and 2 moles of Fe (Iron, 55.85 g/mol) are formed.

To calculate the amount in grams of each product for 4.3 g of Al reacted, we first need to convert this amount into moles of Al by dividing by its molecular weight. Then we use the stoichiometry of the reaction to find the amount of product formed.

(4.3 g Al) * (1 mol Al / 26.98 g Al) = 0.159 moles Al

For Aluminium oxide: (0.159 mol Al) * (1 mol Al2O3 / 2 mol Al) * (101.96 g Al2O3 / 1 mol Al2O3) = 8.14 g Al2O3

For Iron: (0.159 mol Al) * (2 mol Fe / 2 mol Al) * (55.85 g Fe / 1 mol Fe) = 8.89 g Fe

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The method for calculation of the mass of CO2 emitted per Kj of heat produced in a combustion reaction has been explained below..

To answer this, let us use the combustion reaction of methanol which is;

CH3OH (l) + (3/2)O2 (g) = CO2 (g) + 2H2O (g)

We can find the enthalpy of combustion of this reaction from tables or it will be given to us. Let us use; ΔH = -638.54 KJ

This means for every 1 mole of CO2, the heat released is -638.54 KJ

Now, Molar mass of CO2 is 44 g/mol

Thus;

Mass of CO2 per Kj of heat produced is;

44 g/mol × 1 mol/-638.54 KJ

>> 6.89 × 10^(-2) g/kJ

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only a few should pass

The result of Rutherford's experiment demonstrates that most of the positively charged particles should bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil. Thus, the correct option is A.

Ernest Rutherford was a British physicist who remarkably discovered the atomic nucleus first and proposed a nuclear model of the atom. He also performs an experiment on the Gold foil.

The nuclear model experiment of Rutherford showed that the atom is mostly empty space with a small, dense, positive charge. In the Gold foil experiment, most of the alpha particles did pass straight through the foil, while others bounce back at a range of angles.

Therefore, the correct option for this question is A.

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Answer:

The theoretical zero volume temperature of a gas.

Explanation:

The experimental data for effect of temperature on Volume of a gas is plotted in the graph. The graph is a straight line. It has been extrapolated where dash line meets the temperature axis. This point shows the temperature that the gas would have if it had zero volume. Experimentally, this data is not obtained but theoretically measured by extrapolating the volume versus temperature curve.

Thus, the point where blue line meets is the theoretical zero volume temperature of  a gas.

Answer : The amount of [tex]N_2H_4[/tex] will be, 64 g

Solution : Given,

Mass of [tex]N_2O_4[/tex] = 92 g

Molar mass of [tex]N_2O_4[/tex] = 92 g/mole

Molar mass of [tex]N_2H_4[/tex] = 32 g/mole

First we have to calculate the moles of [tex]N_2O_4[/tex].

[tex]\text{Moles of }N_2O_4=\frac{\text{Mass of }N_2O_4}{\text{Molar mass of }N_2O_4}=\frac{92g}{92g/mole}=1mole[/tex]

Now we have to calculate the moles of [tex]N_2H_4[/tex].

The balanced chemical reaction is,

[tex]2N_2H_4+N_2O_4\rightarrow 3N_2+4H_2O[/tex]

From the balanced reaction, we conclude that

1 mole of [tex]N_2O_4[/tex] react with 2 moles of [tex]N_2H_4[/tex]

and from the calculated moles,

1 mole of [tex]N_2O_4[/tex] react with 2 moles of [tex]N_2H_4[/tex]

Now we have to calculate the mass of [tex]N_2H_4[/tex]

[tex]\text{Mass of }N_2H_4=\text{Moles of }N_2H_4\times \text{Molar mass of }N_2H_4[/tex]

[tex]\text{Mass of }N_2H_4=(2mole)\times (32g/mole)=64g[/tex]

Therefore, the amount of [tex]N_2H_4[/tex] will be, 64 g

Answer: Option (b) is the correct answer.

Explanation:

When there are more number of hydroxide ions in a solution then there will be high concentration of [tex]OH^{-}[/tex] or hydroxide ions. As a result, more will be the strength of base in that particular solution.

A base is strong when it readily dissociate into its ions in the solution. When a base is strong, then it does not matter at what concentration it is dissolved in the solution because despite of its low concentration it will remain a strong base.

Thus, we can conclude that out of the given options, the statement even at low concentrations, a strong base is strong best relates the strength and concentration of a base.

Inhibitory neurotransmitter

Answer: Option (B) is the correct answer.

Explanation:

When we move from top to bottom in a group then size of elements keep of increasing due to the addition of more number of electrons into new shells.

But when we move across a period then there occurs a decrease in size of the elements due to the addition of electrons into the same shell.

So, if we move from bottom to top then it means there is decrease in size of atoms as number of electrons decreases then.

Thus, we can conclude that atomic size decreases as one moves from bottom to top in a group.

Answer:

It's slightly soluble in an aqueous solution of [tex]NaHCO_{3}[/tex], and almost insoluble in an aqueous solution of NaOH.

Explanation:

Sodium benzoate comes from benzoic acid, which is a weak acid. It means that in an aqueous solution benzoic acid does not ionize easily to form the ions [tex]H_{3}O^{+}[/tex] and [tex]C_{7}H_{5}O_{2} ^{-}[/tex]

It also implies, according to the Le Châtelier's principle, that the ion [tex]C_{7}H_{5}O_{2}^{-}[/tex] tends to form the acid [tex]C_{7}H_{6} O_{2}[/tex] more easily. It can be seen in the following equation:

[tex]C_{7}H_{6} O_{2}[/tex]  ⇔ [tex]C_{7}H_{5}O_{2}^{-}[/tex]  + [tex]H_{3}O^{+}[/tex]

In an aqueous solution, the equilibrium shifts to the left, thus letting water dissolve sodium benzoate.  But why? Because water in that case would produce enough [tex]H_{3}O^{+}[/tex] ions to facilitate the disolution of sodium benzoate. It's shown by its solubility in water at 15°C (62.78g/100mL, according to Wikipedia).

In contrast, the presence of NaOH or [tex]NaHCO_{3}[/tex], both chemical species producing the [tex]OH^{-}[/tex] ions in aqueous solution, would make the equilibrium shift to the right because it would be a higher need of [tex]H_{3}O^{+}[/tex] ions to offset the presence of [tex]OH^{-}[/tex].

However, the effect of NaOH is not the same due to [tex]NaHCO_{3}[/tex], because the first is a strong base and the other is a weak one. Thereby it is reasonable to think that solubility of sodium benzoate is greater in water than in [tex]NaHCO_{3}[/tex] and NaOH.

Solubility in water > solubility in  [tex]NaHCO_{3}[/tex]> solubility in NaOH.

Final answer:

The most important lab safety rule is to follow all safety instructions and guidelines given by your teacher and lab instructions.

Explanation:

The most important lab safety rule is to follow all safety instructions and guidelines given by your teacher and lab instructions.

By following safety rules, you ensure your own safety as well as the safety of others in the laboratory.

For example, one important safety rule is to never eat or drink in the laboratory and to avoid using laboratory glassware for eating or drinking. This is because table tops and counters could have dangerous substances on them and leftover substances in glassware could interact with future experiments.

The bond length of H-Br in the HBr molecule, estimated using the dipole moment and the percent ionic character, is approximately 159 picometers.

The dipole moment (μ) of a molecule is given by the product of the charge (q) and the distance between the charges (d). In this case, you have been provided with the dipole moment, the physical constants, and the percent ionic character. Let's use these details to estimate the bond length. According to the given data, the dipole moment (μ) is 0.797 D, which is equivalent to 0.797 * 3.34*10^-30 C.m (as 1 D = 3.34*10^-30 C.m).

Since HBr has 11.8% ionic character, the effective charge (q_eff) is 11.8% of the charge of one electron, which is 1.6*10^-19 C. Therefore, q_eff = 0.118 * 1.6*10^-19 C. Thus, using the definition μ = q_eff . d (d = bond length), we can solve for d = μ / q_eff. Substituting the given and calculated values, we get d is approximately equal to 159 picometers.

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The estimated bond length of the [tex]\( \text{H-Br} \)[/tex] bond in picometers is approximately [tex]21 Pm[/tex].

To estimate the bond length of [tex]\( \text{HBr} \),[/tex] we will use the given dipole moment and percent ionic character.

Given.

- Dipole moment [tex](\( \mu \))[/tex] of [tex]\( \text{HBr} \)[/tex] = 0.797 d [tex](debye)[/tex]

- Percent ionic character = 11.8%

First, convert the dipole moment from debye to SI units coulomb meters.

[tex]\[ \mu = 0.797 \text{ d} \][/tex]

Since [tex]\( 1 \text{ d} = 3.34 \times 10^{-30} \text{ C} \cdot \text{m} \),[/tex]

[tex]\[ \mu = 0.797 \times 3.34 \times 10^{-30} \text{ C} \cdot \text{m} \][/tex].

[tex]\[ \mu = 2.66198 \times 10^{-30} \text{ C} \cdot \text{m} \][/tex].

Calculate the bond moment in a purely ionic bond [tex](\( \mu_{\text{ionic}} \)):[/tex]

For a bond with 100% ionic character, the dipole moment is given by.

[tex]\[ \mu_{\text{ionic}} = \sqrt{q \cdot r^2} \][/tex]

where [tex]\( q \)[/tex] is the charge in coulombs and [tex]\( r \)[/tex] is the bond length in meters.

Given [tex]\( q = 1.6 \times 10^{-19} \) C (charge for 100% ionic character),\[ \mu_{\text{ionic}} = \sqrt{1.6 \times 10^{-19} \cdot r^2} \][/tex]

Solve for [tex]\( r \):[/tex]

[tex]\[ \mu_{\text{ionic}} = 2.66198 \times 10^{-30} \text{ C} \cdot \text{m} \][/tex]

[tex]\[ \sqrt{1.6 \times 10^{-19} \cdot r^2} = 2.66198 \times 10^{-30} \][/tex]

[tex]\[ 1.6 \times 10^{-19} \cdot r^2 = (2.66198 \times 10^{-30})^2 \][/tex]

[tex]\[ 1.6 \times 10^{-19} \cdot r^2 = 7.0865 \times 10^{-60} \][/tex]

Divide both sides by [tex]\( 1.6 \times 10^{-19} \)[/tex].

[tex]\[ r^2 = \frac{7.0865 \times 10^{-60}}{1.6 \times 10^{-19}} \][/tex]

[tex]\[ r^2 = 4.42906 \times 10^{-41} \][/tex]

Take the square root to find [tex]\( r \)[/tex]

[tex]\[ r = \sqrt{4.42906 \times 10^{-41}} \][/tex]

[tex]\[ r \approx 2.105 \times 10^{-21} \text{ m} \][/tex]

Convert meters to picometers  

[tex]\[ r \approx 2.105 \times 10^{-21} \text{ m} \times 10^{12} \text{ pm/m} \][/tex]    

[tex]\[ r \approx 21.05 \text{ pm} \][/tex]

1) The characteristics of nuclear fission reaction are [tex]\boxed{{\text{it involves splitting of an atom into two or more fragments}}}[/tex].

2) Both the nuclear fission and fusion reaction [tex]\boxed{{\text{release large amount of energy}}}[/tex].

3) The type of reaction shown in the diagram is [tex]\boxed{{\text{nuclear fission reaction}}}[/tex].

Further explanation:

Nuclear fission is defined as the splitting of heavy nucleus into two lighter ones. For example, a sample of uranium is bombarded with neutrons in an attempt to produce new elements with Z greater than 92. The lighter elements such as barium (Z = 56) were formed during the reaction and such products originate from the neutron-induced fission of uranium-235.

The reaction that shows nuclear fission is as follows:

 [tex]_{{\text{92}}}^{{\text{235}}}{\text{U}} + _{\text{0}}^{\text{1}}{\text{n}} \to _{{\text{56}}}^{{\text{141}}}{\text{Ba}} + _{{\text{36}}}^{{\text{92}}}{\text{Kr}} +{\text{3}}_{\text{0}}^{\text{1}}{\text{n}}[/tex]

The krypton-92 is the fission product. In this, the nucleus usually divides asymmetrically rather than symmetrically dividing into two equal parts, and the fission of a given nuclide does not give the same products.

In a nuclear fission reaction, more than one neutron is released by each nucleus on division. When these neutrons collide and induce fission in other nuclei, a series of nuclear fission reactions known as a nuclear chain reaction occurs.

In nuclear fusion, two light nuclei join to produce a heavier and more stable nucleus. Nuclear fusion is the opposite phenomenon of nuclear fission. The reaction that shows nuclear fusion is as follows:

[tex]{\text{2}}_{\text{1}}^{\text{2}}{\text{H}} \to _{\text{2}}^{\text{3}}{\text{He}} + _{\text{0}}^{\text{1}}{\text{n}}[/tex]

Nuclear fission and fusion reactions both produce large amount of energy.

1)

The characteristics of nuclear fission reaction are that it involves splitting of an atom into two or more fragments because it is a process in which a heavy nucleus splits into two lighter ones.

2)

Both the nuclear fission and fusion reaction involves splitting and collision of atoms that produce large amount of energy. Hence, both the reaction releases a large amount of energy.

3)

The type of reaction shown in the diagram is nuclear fission. In the diagram, a large atom split into two small atoms and again the atom forms split into two smaller atoms include a series of reactions called nuclear chain reaction. The nuclear chain reaction is a series of nuclear fission reactions in which neutrons collide and induce fission in other nuclei.

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1. Identify the precipitate of compounds https://brainly.com/question/2094744

2. How many covalent bonds does nitrogen forms https://brainly.com/question/2094744

Answer details

Grade: Senior School

Subject: Chemistry

Chapter: Nuclear binding energy

Keywords: Nuclear fission, nuclear fusion, splits, bombardment, uranium, krypton, heavy nucleus, nuclear chain reaction.

The box exerts a pressure of 10 pounds per square inch (lbs/in²) on the floor, calculated by dividing the weight of the box (120 lbs) by the area of its bottom (12 in²).

The pressure exerted by the box on the floor can be calculated using the formula for pressure, which is the force applied (in this case, the weight of the box) divided by the area over which the force is distributed.

The weight of the box is given as 120 lbs, and the area of the bottom of the box in contact with the floor is 12 in². To find the pressure, we divide the weight by the area:

Pressure = Weight / Area

Pressure = 120 lbs / 12 in² = 10 lbs/in²

Therefore, the box exerts a pressure of 10 pounds per square inch (lbs/in²) on the floor.

This system describes, the involvement of both energy transformation and energy transfer. Hence option C is correct.

Energy transfer is nothing but the phenomenon of gaining or losing the amount of energy by a point from which we are taking or giving the energy. Energy transfer is the movement of energy from one point to other. for example, when we want to increase the temperature of the body from room temperature to certain temperature, then we have to transfer certain amount of energy to gain that temperature.

According to law of thermodynamic, energy neither be created nor be destroyed, it can be transferred from one form into another form. Energy transformation is when energy changes from one type to another. for example when mechanical energy gets converted into electric energy in generator, there is transformation of energy from mechanical to electric energy. Vice verse is true for electric fan.

In this system, everything is gaining temperature hence it is gaining energy, means there is transfer of energy. and first chemical energy is changed to thermal energy, thermal energy is transferred  to kinetic energy and last into electric energy, hence there is of course transformation of energy from one form into another form.

Hence option C is Correct.

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