How many grams of sodium iodide Nai must be used to produce 51.9 G of iodine I2

Answer:

They undergo photosynthesis which makes the carbon dioxide to be used rather than produced.

Explanation:

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Forests are widely known as the "Earth's lungs" due to the photosynthesis that vegetable life constantly perform as the carbon dioxide that is in the environment is used by them to produce energy, glucose and oxygen considering such metabolic pathway. Now, forest act as carbon sinks as the proportion between carbon dioxide consumers to producers is by far greater than 1 as long as there are more plants that use higher amounts of carbon dioxide than those that are released during the respiration of animals or any other natural process producing carbon dioxide.

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Explanation:

Relation between freezing temperature and molal concentration is as follows.

          [tex]\Delta T_{f} = k_{f} \times m[/tex]

The given data is as follows.

    [tex]\Delta T_{f}[/tex] = difference in temperature = [tex][0 - (-2.75)]^{o}C[/tex] = [tex]2.75^{o}C[/tex]

      [tex]k_{f} = 1.86^{o}C/mol[/tex]

            molality, (m) = ?  

Now, putting the given values into the above formula as follows.

                m = [tex]\frac{\Delta T_{f}}{k_{f}}[/tex]

                    = [tex]\frac{2.75^{o}C}{1.86^{o}C/mol}[/tex]

                    = 1.48 m

Therefore, we can conclude that molal concentration of glucose in the given solution is 1.48 m.

Kb(NH₂OH) = 1,8·10⁻⁵.
c₀(NH₂OH) = 0,0500 M = 0,05 mol/L.
c(NH₂⁺) = c(OH⁻) = x.

c(NH₂OH) = 0,05 mol/L - x.
Kb = c(NH₂⁺) · c(OH⁻) / c(NH₂OH).

0,0000000066 = x² /  (0,05 mol/L - x). 

solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.
pOH = -log(0,000018 mol/L) = 4,74.
pH = 14 - 4,74 = 9,23.

The pH of a 0.0500 M solution of hydroxylamine is 9.26.

To find the pH of a 0.0500 M solution of hydroxylamine, we need to find the hydroxide ion concentration [OH⁻] and then use the pOH-pH relationship.

Since hydroxylamine is a weak base, we can use the following equilibrium equation:

NH₂OH + H₂O ⇌ NH₃OH ⁺+ OH⁻

The base dissociation constant (Kb) is given as 6.6 x 10⁻⁹.

Let x be the concentration of hydroxide ions [OH⁻] formed. Then, the concentration of NH₃OH⁺ will also be x.

The initial concentration of hydroxylamine is 0.0500 M, and since it's a weak base, the amount of hydroxylamine that dissociates is very small compared to the initial concentration. Therefore, we can assume that the concentration of hydroxylamine remains approximately constant at 0.0500 M.

The equilibrium expression for Kb is:

Kb = [NH₃OH⁺][OH⁻] / [NH₂OH] = x² / 0.0500

Rearranging the equation to solve for x:

x² = Kb × 0.0500 = 6.6 x 10⁻⁹ × 0.0500 = 3.3 x 10⁻¹⁰

x = √(3.3 x 10⁻¹⁰) = 1.81 x 10⁻⁵ M

This is the concentration of hydroxide ions [OH⁻].

Now, we can find the pOH using the following equation:

pOH = -log[OH⁻] = -log(1.81 x 10⁻⁵) = 4.74

Finally, we can find the pH using the pOH-pH relationship:

pH + pOH = 14 pH = 14 - pOH = 14 - 4.74 = 9.26

Therefore, the pH of a 0.0500 M solution of hydroxylamine is 9.26.

Mercury is liquid at room temperature, expands consistently, and doesn't wet glass, making it ideal for accurate temperature measurements in thermometers despite its toxicity.

Mercury is the only metal to have been used extensively in thermometers for several reasons. Firstly, it is the only metal that is liquid at room temperature, allowing it to easily expand and contract with temperature changes, making it excellent for precise temperature measurements. Additionally, mercury has a high coefficient of expansion, meaning it expands and contracts uniformly, resulting in accurate and consistent readings.

Mercury also does not wet glass, maintaining a clear meniscus that makes it easy to read the temperature. Its high density compared to other liquids, such as water, allows for more compact and portable thermometers. Despite its hazardous nature, these unique physical properties have historically made mercury the preferred choice for use in thermometers until safety concerns led to the adoption of safer alternatives like alcohol-filled instruments.

The solubility of an amphoteric hydroxide in a buffered solution depends on the pH of the solution. It can act as both an acid and a base. The Henderson-Hasselbalch equation can be used to estimate the solubility in a buffered solution.

The solubility of an amphoteric hydroxide, M(OH)2, in a buffered solution depends on the pH of the solution. An amphoteric hydroxide can act as both an acid and a base. At low pH, the hydroxide ion concentration is low and the hydroxide ion reacts with the excess hydronium ions, reducing the solubility. At high pH, the hydronium ion concentration is low and the hydroxide ion concentration is high, increasing the solubility. In a buffered solution, the pH remains relatively constant due to the presence of a weak acid and its conjugate base. The solubility of the hydroxide in the buffered solution can be estimated using the Henderson-Hasselbalch equation.

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The complete ionic equation for the reaction is as follows:

[tex]\boxed{{{\mathbf{H}}^ + }\left( q \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right) + {\mathbf{O}}{{\mathbf{H}}^ - }\left( {aq} \right) \to {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\left( l \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right)}[/tex]

Further Explanation:

Double displacement reaction is defined as the reaction in which ions of two compound interchange with each other to form the product. For example, the general double displacement reaction between two compounds AX and BY  is as follows:

[tex]{\text{AX}} + {\text{BY}} \to {\text{AY}} + {\text{BX}}[/tex]

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Complete ionic equation

3. Net ionic equation

The reactants and products remain in undissociated form in molecular equation. In the case of complete ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of net ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the complete ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction, HI reacts with RbOH to form RbI and [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex]. The balanced molecular equation of the reaction is as follows:

 [tex]{\text{HI}}\left( {aq} \right) + {\text{RbOH}}\left( {aq} \right) \to {\text{RbI}}\left( {aq} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right)[/tex]

Step 2: Dissociate all the compounds with the aqueous phase to write the complete ionic equation. The compounds with solid and liquid phase remain same. The complete ionic equation is as follows:

[tex]{{\mathbf{H}}^ + }\left( q \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right) + {\mathbf{O}}{{\mathbf{H}}^ - }\left( {aq} \right) \to {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\left( l \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right)[/tex]

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1. Balanced chemical equation https://brainly.com/question/1405182

2. Oxidation and reduction reaction https://brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: Double displacement reaction, types of equation, molecular equation, complete ionic equation, net ionic equation, RbI, RbOH, H2O, HI, chemical reaction.

To write a balanced complete ionic equation, first write the balanced chemical equation and then break it down into its ionic components. Finally, combine the ions to form the complete ionic equation.

To write a balanced complete ionic equation for the reaction between HI(aq) and RBOH(aq), we need to first write the balanced chemical equation:

HI(aq) + RBOH(aq) -> HRB(aq) + H2O(l)

Now, we can break down the equation into its ionic components:

HI(aq) -> H+(aq) + I-(aq)

RBOH(aq) -> RB+(aq) + OH-(aq)

HRB(aq) -> H+(aq) + RB-(aq)

H2O(l)

Putting it all together, the balanced complete ionic equation is:

H+(aq) + I-(aq) + RB+(aq) + OH-(aq) -> H+(aq) + RB-(aq) + H2O(l)

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To calculate mass of  Fe₂O₃ we need to apply concept of stoichiometry. So according to this we need molar mass of  Fe₂O₃, mole ratio of  Fe₂O₃ to C. Therefore the mass of  Fe₂O₃ required to react with  19.0 g C is  67.4g.

Stoichiometry is a part of chemistry that help us in making relationship between reactant and product from quantitative aspects.

The balanced equation is

2Fe₂O₃+3C [tex]\rightarrow[/tex]  3CO₂+4Fe

The molar ratio of Fe₂O₃ to carbon is 2:3

2 moles of Fe₂O₃ needed to react with 3 moles of carbon

3 mole of carbon needed= 2 mole of Fe₂O₃

1 mole of carbon needed = 2÷3 mole of Fe₂O₃

(19÷12) = 1.58 mole of carbon needed=  (2÷3 )× 1.58 mole= 0.422 mole of  Fe₂O₃

mass of  Fe₂O₃  = moles of  Fe₂O₃ ×Molar mass of  Fe₂O₃

                           = 0.422 mole×159.70

                           = 67.4g

Therefore the mass of Fe₂O₃ required to react with  19.0 g C is 67.4g

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To convert the moles of Fe₂O₃  to grams using its molar mass.

To calculate the number of grams of Fe₂O₃ needed to react with 19.0 g C, we need to use the stoichiometric mole ratio from the balanced chemical equation. First, convert the given mass of C to moles using its molar mass. Then, use the mole ratio from the equation Fe₂O₃+ 3CO -> 2Fe + 3CO₂ to determine the moles of Fe₂O₃ needed. Finally, convert the moles of Fe₂O₃ to grams using its molar mass.

Given: 19.0 g C
Calculate: grams of Fe₂O₃

Therefore, 83.8 grams of Fe₂O₃ are needed to react with 19.0 grams of C.

The statement comparing mitosis and meiosis accurately indicates that mitosis results in two identical diploid cells, whereas meiosis produces four genetically distinct haploid cells.

The correct statement that compares mitosis and meiosis is:

A) The result of mitosis is two identical diploid cells. The result of meiosis is four genetically different haploid cells.

Both processes are preceded by one round of DNA replication. However, while mitosis includes one cellular division resulting in two identical diploid cells, meiosis includes two nuclear divisions which result in four genetically distinct haploid cells. Mitosis is essential for growth and repair, and meiosis provides genetic diversity through sexual reproduction.

2) So, after every period of one half-life the concentration of the reactant will decrease by half.

Based on the number of half-lives undergone by the substance,  the mass of the substance remaining after three half-lives is 2.25 g.

The half-life of a substance is the time it will take for half the amount of the substance to decay or decompose.

The initial mass of the substance is 18.0 g

The substance undergoes three half-lives.

After the first half-life, mass remaining = 18/2 = 9.0 g

After the first half-life, mass remaining = 18/2 = 9.0 g

After the third half-life, mass remaining = 4.5/2 = 2.25 g

Therefore, the mass of the substance remaining after three half-lives is 2.25 g.

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Answer:

True

Explanation:

When you are responsible for paying your own expenses, you start to take more responsibility for how your money will be spent. At this point you begin to understand the importance and begin to value money management. Money management is the activity, where you decide how much of the money you have will be spent and what will be spent. When you pay your own expenses, you must manage your money well and ensure that all your needs, such as home, food, taxes, gasoline and other things are paid. If you do not manage your money you will end up spending on needless things and you may have some unpaid necessary expenses, which will cause problems for your life.

Answer: The molecular equation is written below.

Explanation:

Every balanced chemical equation follows law of conservation of mass.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

A molecular equation is the balanced chemical equation where the ionic compounds are expressed as molecules rather than constituent ions.

The chemical equation for the reaction of copper (II) chloride and potassium phosphate follows:

[tex]3CuCl_2(aq.)+2K_3PO_4(aq.)\rightarrow Cu_3(PO_4)_2(s)+6KCl[/tex]

By Stoichiometry of the reaction:

3 moles of copper (II) chloride reacts with 2 moles of potassium phosphate to produce 1 mole of copper (II) phosphate and 6 moles of potassium chloride.

Hence, the molecular equation is written above.

The student is not provided with the direct answers to their K12 3.10 Unit Assessment (Chemistry). Instead, they are advised to review key concepts and principles from the unit, apply them to different contexts during the assessment, and utilise computational and analytical skills.

While it's not appropriate to provide the direct answers to your K12 3.10 Unit Assessment: Solutions, Part 1, I can help you understand how to arrive at correct solutions. The assessment likely includes both multiple-choice and short-response questions from various topics you've studied during the unit.

Critical Thinking Questions usually require you to apply concepts and principles you've learned to different contexts or situations. You might have to use analytic and computational skills to solve some problems. For instance, an example of a chemistry AP question could be asking you to calculate molarity of a solution, given the mass of the solute and volume of the solution.

Finally, review the materials from your textbooks and lessons, particularly the areas you feel less confident about. Practice using the concept to solve problems and try to understand the underlying principles. Good luck with your assessment!

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The mass of CO₂ produced in the reaction is 4.10 grams.

To find the mass of CO₂ produced, we can use the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

The total mass of the reactants must equal the total mass of the products.

We start with the given masses of the reactants:

The total mass of the reactants is the sum of the masses of CaCO₃ and the HCl solution:

After the reaction, we have the following:

 The negative sign indicates that 4.10 grams of gas (CO₂) have been released from the solution, as the mass of the products is less than the mass of the reactants.

This is consistent with the observed bubbles, which are CO₂ gas being produced and escaping from the solution.

Therefore, the mass of CO₂ produced is 4.10 grams.

Ionization energy is the property of an atom that describes the ease with which an atom gives up an electron to form a positive ion.

The ionization energy of a chemical element is expressed in joules or electron volts. It is commonly measured inside an electric discharge tube where fast-moving electrons are generated due to an electric current collision with a gaseous atom of the element.

This causes the ejection of one of its electrons. In the case of a hydrogen atom, which has only one orbiting electron which is in turn bound to a nucleus with only one proton, the ionization energy of 2.18 × 10^−18 joule or 13.6 electron volts is needed to move the electron from its lowest energy level out of the atom.

The ionization energy magnitude is dependent on the element and the combined effects of the electric charge of its nucleus, atomic size, and also its electronic configuration. Electron removal is also the hardest for noble gases and easiest for alkali metals.

The ionization energy required for the removal of electron removal is the hardest as the electron number decreases progressively. Because as the atom loses electrons, the positive charge on the nucleus of the atom does not change; thus, as each electron is removed, the remaining ones are held more firmly.

Therefore, Ionisation energy is the property of an atom that describes the ease with which an atom gives up an electron to form a positive ion.

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Answer: 8 x 10^7

Explanation: