In acidic solution, the nitrate ion can be used to react with a number of metal ions. One such reaction is NO3−(aq)+Sn2+(aq)→NO2(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: NO3−(aq)+Sn2+(aq)+ −−−→NO2(aq)+Sn4+(aq)+ −−− Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Answer:

1) The no. of grams of Na₂CO₃ = 1.96 g.

2) The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.

3) The no. of grams of Ag₂CO₃ = 3.998 g ≅ 4.0 g.

4) The no. of grams of NaNO₃ = 2.498 g ≅ 2.5 g.

Explanation:

Na₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃(s) + 2NaNO₃,

It is clear that 1 mol of Na₂CO₃ and 2 mol of AgNO₃ to produce 1 mol of Ag₂CO₃ and 2 mol of NaNO₃.

no. of moles of Na₂CO₃ = mass/molar mass = (3.5 g)/(105.9888 g/mol) = 0.033 mol.

no. of moles of AgNO₃ = mass/molar mass = (5.0 g)/(169.87 g/mol) = 0.0294 mol.

1) the no. of grams of sodium carbonate, and silver nitrate:

1 mol of Na₂CO₃ reacts completely with 2 mol of AgNO₃ with (1: 2 molar ratio).

∴ 0.0145 mol of Na₂CO₃ "the excess reactant, and the remaining is in excess (0.033 mol - 0.0145 mol = 0.0185 mol)" reacts completely with (0.0294 mol) of AgNO₃ "limiting reactant".

∴ The no. of grams of Na₂CO₃ = (no. of moles remaining)(molar mass) = (0.0185 mol)((105.9888 g/mol) = 1.96 g.

∴ The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.

2) the no. of grams of silver carbonate:

Using cross multiplication:

1 mol of Na₂CO₃ produces → 1 mol of Ag₂CO₃, from stichiometry.

∴ 0.0145 mol of Na₂CO₃ produces → 0.0145 mol of Ag₂CO₃.

∴ The no. of grams of Ag₂CO₃ = (no. of moles of Ag₂CO₃)(molar mass of Ag₂CO₃) = (0.0145 mol)(275.7453 g/mol) = 3.998 g ≅ 4.0 g.

3) the no. of grams of sodium nitrate:

Using cross multiplication:

2 mol of AgNO₃ produces → 2 mol of NaNO₃, from stichiometry.

∴ 0.0294 mol of AgNO₃ produces → 0.0294 mol of NaNO₃.

∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass of NaNO₃) = (0.0294 mol)(84.9947 g/mol) = 2.498 g ≅ 2.5 g.

After the reaction of 3.50 g sodium carbonate and 5.00 g silver nitrate, approximately 0.42 g of sodium carbonate, 0 g of silver nitrate, 7.8 g of silver carbonate, and 2.3 g of sodium nitrate will be present.

The reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) forms solid silver carbonate (Ag2CO3) and a solution of sodium nitrate (NaNO3). The balanced chemical equation for this reaction is 2AgNO3 + Na2CO3 yields 2Ag2CO3 (s) + 2NaNO3 (aq). To determine the amount of each compound present after the reaction, we first need to calculate the moles of sodium carbonate and silver nitrate. We find that 3.50 g Na2CO3 amounts to 0.033 moles and 5.00 g AgNO3 equals 0.029 moles. Since AgNO3 is the limiting reagent, it will be completely consumed in the reaction, producing 0.029 moles of Ag2CO3 and 0.029 moles of NaNO3. This results 0.004 moles of Na2CO3 left unreacted. Converting these moles back into grams gives approximately 0.42 g Na2CO3, 0 g AgNO3, 7.8 g Ag2CO3, and 2.3 g NaNO3 present after the reaction.

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Answer:

[tex]\boxed{\text{47.4 g}}[/tex]

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    17.03   32.00                 18.02  

           4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g:     70.1      70.1

Step 1. Calculate the moles of each reactant

[tex]\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}[/tex]

Step 2. Identify the limiting reactant  

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

[tex]\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}[/tex]

From O₂:  

The molar ratio of H₂O:O₂ is 6:5.  

[tex]\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}[/tex]

O₂ is the limiting reactant because it gives the smaller amount of H₂O.  

Step 3. Calculate the theoretical yield.

[tex]\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}[/tex]

The given equations (a)-(d) represent synthesis, decomposition, neutralization, and precipitation reactions, respectively.

(a) The given equation represents a synthesis reaction, where hydrogen gas (H2) reacts with carbon solid (C) to form carbon monoxide gas (CO) and hydrogen gas (H2).

(b) The given equation represents a decomposition reaction, where potassium chlorate solid (KClO3) decomposes to form potassium chloride solid (KCl) and oxygen gas (O2).

(c) The given equation represents a neutralization reaction, where aluminum hydroxide (Al(OH)3) reacts with hydrochloric acid (HCl) to form aluminum chloride (AlCl3) and water (H2O).

(d) The given equation represents a precipitation reaction, where lead nitrate (Pb(NO3)2) reacts with sulfuric acid (H2SO4) to form lead sulfate solid (PbSO4) and nitric acid (HNO3) in aqueous form.

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The given equations (a)-(d) represent, a) Synthesis reaction, b) Decomposition reaction, c) neutralization reaction and d) precipitation reaction.

Reaction (a): H₂O(g) + C(s) ⟶ CO(g) + H₂(g) is a synthesis reaction, also known as a combination reaction, where two or more reactants combine to form a single product. In this case, hydrogen gas (H₂) and carbon monoxide (CO) are produced from the combination of water vapor (H₂O) and carbon (C).

Reaction (b): 2KClO₃(s) ⟶ 2KCl(s) + 3O₂(g) is a decomposition reaction, where a single compound breaks down into two or more simpler substances.

Reaction (c): Al(OH)₃(aq) + 3HCl(aq) ⟶ AlCl₃(aq) + 3H₂O(l) is an acid-base neutralization reaction, in which an acid (HCl) reacts with a base (Al(OH)₃) to produce a salt (AlCl₃) and water.

Reaction (d): Pb(NO₃)₂(aq) + H₂SO₃(aq) ⟶ PbSO₄(s) + 2HNO₃(aq) is a precipitation reaction, where two solutions react to form an insoluble solid (precipitate) and a soluble compound.

To calculate the mass of the solid CaCO3 in the vessel once equilibrium is reached, we need to use the equilibrium constant (K) and the given information. The mass of the solid CaCO3 can be calculated by subtracting the mass of the solid CaO from the initial mass of CaCO3. The molar concentration of CaCO3 can be calculated using the given mass and volume, and the equilibrium constant can be used to determine the concentrations of CaO and CO2.

To calculate the mass of the solid CaCO3 in the vessel once equilibrium is reached, we need to use the equilibrium constant (K) and the given information. The equilibrium constant (K) is 1.16 at 800 °C. We are given a 5.00 L vessel containing 10.0 g of CaCO3(s) and the volume occupied by the solid is ignored. To calculate the mass of the solid in the vessel, we can use the equation n = m/M to calculate the number of moles of CaCO3 and then multiply it by the molar mass of CaCO3 to get the mass.

Given that the volume of the vessel is 5.00 L, we can calculate the initial concentration of CaCO3 (Molar concentration = moles/volume) as [CaCO3] = (10.0g / 100.09 g/mol) / 5.00 L = 0.02 mol/L.

Now we can use the equation Q = [CaO][CO2] / [CaCO3] = 1.16 (as it's at equilibrium) and substitute the concentrations as 0.02 mol/L and calculate the concentration of [CaO] and [CO2]. Since the volume occupied by the solid is ignored, the concentrations of CaO and CO2 will be the same. Now we can multiply the concentration of CaO by the volume to get the moles of CaO. Finally, multiply the moles of CaO by its molar mass to calculate the mass of the solid CaO.

Therefore, the mass of the solid CaCO3 in the vessel once equilibrium is reached is the initial mass of CaCO3 minus the mass of the solid CaO, which is calculated above.

Answer:

see explanation...

Explanation:

                               Mg⁺²-24                     Co⁺³-60                     Clˉ-35    

Protons (p⁺)                12                                 27                             17              

Neutrons (n⁰)             12                                 33                             18              

Electrons (eˉ)             10                                 24                             18              

                                  (c)                                 (b)                            (a)

                         12/2 : 12/2 : 10/2      27/3 : 33/3 : 24/3        #n⁰ = 18

                             6    :    6   :    5          9  :   11  :   8             #eˉ = 18

The specs that match the conditions are 24Mg2+ for equal neutrons and electrons, 47Cr for the 9:11:8 proton:neutron:electron ratio, and 35Cl- for having neutrons equal to protons plus one-half the electrons.

The question is asking for the species that match specific sets of conditions related to the number of its protons, electrons, and neutrons. In atomic structure, the number of protons defines the atomic number of an element, while the number of electrons defines its charge. The number of neutrons can be found by subtracting the atomic number from the atomic mass.

(a) For a species with equal numbers of neutrons and electrons, we can choose 24Mg2+. Its atomic number is 12, so it has 12 protons. Since it’s a +2 ion, it has lost 2 electrons thus leaving it with 10 electrons. Also, it has 24-12=12 neutrons.

(b) For a species that has protons, neutrons, and electrons in the ratio 9:11:8, we can choose 47Cr. Chromium (Cr) has an atomic number of 24 so it has 24 protons, and 47-24=23 neutrons, and as neutral atom it has 24 electrons. Although not exact, it’s the closest among the choices given.

(c) Finally, a species with a number of neutrons equal to the number of protons plus one-half the number of electrons would be 35Cl-. Chlorine ordinarily has 17 protons and 17 electrons, but as a -1 ion, it has gained 1 extra electron, giving a total of 18 electrons. Half of this is 9. The 17 protons plus the 9 gives 26, which is the number of neutrons in chlorine-35 (35-17=18).

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In the given equation, 2 moles of N2O5 will react to form 4 moles of NO2. Therefore, 1 mole of N2O5 will form 2 moles of NO2, and 5.4 moles of N2O5 will form 11 moles of NO2.

When determining the amount of a product made from a reactant in a chemical equation, we use stoichiometry, guided by the coefficients in the balanced chemical equation. In this equation, 2N2O5(g)→4NO2(g)+O2(g), the coefficient in front of N2O5 is 2 and in front of NO2 is 4. This tells us that for every 2 moles of N2O5, 4 moles of NO2 will be produced.

Part A: If 1 mole of N2O5 fully reacts, it will form 2 moles of NO2. (Ratio is 2:4, therefore halved to 1:2)

Part B: If 5.4 moles of N2O5 fully reacts, it will form 10.8 (which rounds to 11 with two significant figures) moles of NO2. (Ratio 5.4 * 2 moles of NO2/mole of N2O5)

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Answer:

52.0004 grams of mass of potassium superoxide  is required

Explanation:

Let moles carbon dioxide gas be n at 22.0 °C and 767 mm Hg occupying 8.90 L of volume.

Pressure of the gas,P = 767 mm Hg = 0.9971 atm

Temperature of the gas,T = 22.0 °C = 295.15 K

Using an ideal gas equation to calculate the number of moles.

[tex]PV=nRT[/tex]

[tex]n=\frac{0.9971 atm\times 8.90 L}{0.0821 atm L/mol K\times 295.15 K}[/tex]

n = 0.3662 mol

[tex]4KO_2(s)+2CO_2(g)\rightarrow 2K_2CO_3(s)+3O_2(g)[/tex]

According to reaction, 2 moles of carbon-dioxide reacts with 4 moles of potassium superoxide.

Then 0.3662 mol of  of carbon-dioxide will react with:

[tex]\frac{4}{2}\times 0.3662 mol=0.7324 mol[/tex] of potassium superoxide.

Mass of 0.7324 mol potassium superoxide:

0.7324 mol × 71 g/mol = 52.0004 g

52.0004 grams of mass of potassium superoxide is required.

To find the mass of KO2 required to react with 8.90 L of CO2, we can use the stoichiometry of the balanced equation.

To determine the mass of KO2 required to react with 8.90 L of CO2, we need to use the stoichiometry of the balanced equation. From the equation, we can see that 4 moles of KO2 reacts with 2 moles of CO2. We can convert the volume of CO2 to moles using the ideal gas law and then use the mole ratio to find the mass of KO2:

1. Convert 8.90 L of CO2 to moles (using the ideal gas law)

2. Use the mole ratio to find moles of KO2

3. Convert moles of KO2 to grams using the molar mass of KO2

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Answer: The mass of aluminium chloride that can be formed are 46.3 g

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  ....(1)  

Given mass of aluminium = 32 g  

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol[/tex]

Given mass of chlorine = 37 g  

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol[/tex]

For the given chemical equation:

[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 0.521=0.347moles[/tex] of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So,  0.521 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 0.521=0.347moles[/tex] of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

[tex]0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g[/tex]

Hence, the mass of aluminium chloride that can be formed are 46.3 g

The quantity of the substance is given by the mass. The mass of aluminium chloride formed when reacting aluminium with chlorine will be 46.3 gm.

Mass is a quantitative factor that determines the amount of substance or matter present in the sample.

The chemical reaction can be shown as,

[tex]\rm 2 Al + 3Cl_{2} \rightarrow 2AlCl_{3}[/tex]

Calculate the number of moles of Aluminium:

[tex]\begin{aligned}\rm Moles & = \dfrac {\rm Mass }{\rm Molar\; mass}\\\\& = \dfrac{32}{26.98}\\\\& = 1.186\;\rm moles\end{aligned}[/tex]

Calculate the number of moles of Chlorine:

[tex]\begin{aligned}\rm Moles & = \dfrac {\rm Mass }{\rm Molar\; mass}\\\\& = \dfrac{37}{71}\\\\& = 0.521 \;\rm moles\end{aligned}[/tex]

From the stoichiometry of the reaction above:

2 moles of aluminium reacts with 3 moles of chlorine

So, moles of aluminium will react with 0.521 moles chorine is,

[tex]\dfrac{2}{3} \times 0.521 = 0.347 \;\rm moles[/tex]

From this, it can infer that chlorine gas is a limiting reagent and aluminium is an excess reagent.

From the stoichiometry of the reaction,

3 moles of chlorine =  2 moles of aluminium chloride

So, 0.521 moles chorine will give,

[tex]\dfrac{2}{3} \times 0.521 = 0.347 \;\text{ moles of aluminium chloride.}[/tex]

Calculate the mass of the aluminium chloride as:

[tex]\begin{aligned} \rm mass &= \rm moles \times molar\; mass\\\\&= 0.347 \times 133.34\\\\&= 46.3 \;\rm g\end{aligned}[/tex]

Therefore, the mass of the aluminium chloride is 46.3 gm.

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Final answer:

The letters in parentheses after each substance represent their states of matter in the reaction equation. In this reaction, sodium in solid state reacts with oxygen in gas state to form sodium oxide in solid state.

Explanation:

In the balanced equation 4 Na(s) + O2(g) → 2 Na2O(s), the letters in parentheses after each substance represent their states of matter. (s) stands for solid, (g) stands for gas. In this reaction, sodium (Na) in solid state reacts with oxygen (O2) in gas state to form sodium oxide (Na2O) in solid state.

The letters in parentheses denote states of matter: (s) for solid, (g) for gas, at room temperature and standard pressure.

The letters in parentheses after each substance in the chemical equation represent the state of matter for each reactant and product at room temperature and standard pressure. Here is what they stand for:

Na(s) stands for sodium in the solid state.

O₂(g) stands for oxygen in the gaseous state.

Na₂O(s) stands for sodium oxide in the solid state.

The balanced chemical equation provided is:

[tex]\[ 4 \text{Na}(s) + \text{O}_2(g) \rightarrow 2 \text{Na}_2\text{O}(s) \][/tex]

This equation indicates that four moles of solid sodium react with one mole of gaseous oxygen to produce two moles of solid sodium oxide.

Answer:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

α =2

β = 19

γ = 12

δ = 14

53.2moles of O₂

Explanation:

Proper equation of the reaction:

                    αC₆H₁₄ + βO₂ → γCO₂ + δH₂O

This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:

         CₓHₙ + (x + [tex]\frac{n}{4}[/tex])O₂ → xCO₂ + [tex]\frac{n}{2}[/tex]H₂O

From the given combustion equation, x = 6 and n = 14

Therefore:

β = x + [tex]\frac{n}{4}[/tex] = 6 + [tex]\frac{14}{4}[/tex] = 6 + 3.5 = 9[tex]\frac{1}{2}[/tex]

γ = 6

δ = [tex]\frac{n}{2}[/tex] = [tex]\frac{14}{2}[/tex] = 7

The complete reaction equation is therefore given as:

                   C₆H₁₄ + 9[tex]\frac{1}{2}[/tex]O₂ → 6CO₂ + 7H₂O

To express as whole number integers, we multiply the coefficients through by 2:

                  2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

Problem 2

           From the reaction:

2 moles of hexane are required to completely react with 19 moles of O₂

∴ 5.6 moles of hexane would react with k moles of O₂

This gives:     5.6 x 19 = 2k

                        k = [tex]\frac{5.6 x 19}{2}[/tex]

                        k = 53.2moles of O₂

Answer:

K(net) = K₁ x K₂ = 1.51 x 10⁻²⁶

Explanation:

D. radioactive isotopes are one of the environmental waste products of nuclear energy.

One of the environmental side effects of nuclear energy is radioactive isotopes.

A chemical element in an unstable state that emits radiation as it decomposes and becomes more stable.

Radioisotopes can be created in a lab or in the natural world. They are utilized in imaging studies and therapy in medicine.

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The enthalpy change of the reaction per mole of limiting reactant is -568 kJ/mol.

The enthalpy change of a reaction can be calculated using the amount of reactant used. In this case, 0.0500 mol of HCl was used, and the reaction absorbed 104 J of heat. Since AH is an extensive property, it is proportional to the amount of acid neutralized. Therefore, the enthalpy change for 1 mol of HCl is -2.9 kJ. To find the enthalpy of the reaction per mole of limiting reactant, we can use the molar mass of the metal and the stoichiometric ratio between HCl and the metal in the balanced equation.

To calculate the enthalpy change per mole of limiting reactant, we need to determine the number of moles of the metal. Since the molar mass of the metal is 57.78 g/mol, we can calculate the number of moles by dividing the mass of the metal (0.295 g) by its molar mass: 0.295 g ÷ 57.78 g/mol = 0.00510 mol.

Next, we use the stoichiometric ratio in the balanced equation to relate the moles of HCl reacted to the moles of metal reacted. From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of metal. Therefore, the moles of HCl reacted is also 0.00510 mol. Now we can calculate the enthalpy change per mole of limiting reactant: -2.9 kJ ÷ 0.00510 mol = -568 kJ/mol (rounded to three significant figures).

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Answer:

[tex]\boxed{\text{2.17 g/cm}^{3}}[/tex]

Explanation:

1. Ions per unit cell

(a) Chloride

8 corners + 6 faces

[tex]\text{No. of Cl$^{-}$ ions}\\\\= \text{8 corners} \times \dfrac{\frac{1 }{ 8} \text{ ion}} {\text{1 corner}} + \text{6 faces}\times \dfrac{\frac{1}{2} \text{ ion}}{\text{1 face}} = \text{1 ion + 3 ions = 4 ions}}[/tex]

(b) Chloride

12 edges + 1 centre

[tex]\text{No. of Na$^{+}$ ions}\\\\= \text{12 edges} \times \dfrac{\frac{1 }{ 4} \text{ ion}} {\text{1 edge}} + \text{1 centre}\times\dfrac{\text{1 ion}}{\text{1 centre}} = \text{3 ions + 1 ion = 4 ions}[/tex]

There are four formula units of NaCl in a unit cell.

2. Mass of unit cell

m = 4 × NaCl = 4 × 58.44 u = 233.76 u

[tex]m = \text{233.76 g} \times \dfrac{\text{1 g} }{6.022 \times 10^{23} \text{ u} } = 3.882 \times 10^{-22}\text{ g}[/tex]

3. Volume of unit cell

(a) Edge length

[tex]a = 2d_{\text{Na-Cl}} = 2 \times \text{2.819 \AA} = 5.638 \times10^{-10} \text{ m} = 5.638 \times10^{-8} \text{ cm}[/tex]

(b) Volume

[tex]V = a^{3} = \left( 5.638 \times (10^{-8} \text{ cm}\right)^{3} = 1.792 \times10^{-22} \text{ cm}^{3}[/tex]

4. Density

[tex]\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{3.882 \times 10^{-22}\text{ g}}{1.792 \times10^{-22} \text{ cm}^{3}}} = \text{2.17 g/cm}^{3}\\\\\text{The density of NaCl is }\boxed{\textbf{2.17 g/cm}^{3}}[/tex]

Answer:

[tex]\boxed{\text{28.0 s}}[/tex]

Explanation:

Whenever a question asks you, "How long does it take to reach a certain concentration?" or something like that, you must use the appropriate integrated rate law expression.

The integrated rate law for a first-order reaction is  

[tex]\ln \left (\dfrac{[A]_{0}}{[A]} \right ) = kt[/tex]

Data:

[A]₀ = 1.28 mol·L⁻¹

[A] = 0.17 [A]₀

  k = 0.0632 s⁻¹

Calculation:

[tex]\begin{array}{rcl}\ln \left (\dfrac{[A]_{0}}{0.170[A]_{0}} \right ) & = & 0.0632t\\\\\ln \left (5.882) & = & 0.0632t\\1.772 & = & 0.0632t\\\\t & = & \dfrac{1.772}{0.0632}\\\\t & = & \textbf{{28.0 s}}\\\end{array}\\\text{It will take } \boxed{\textbf{28.0 s}} \text{ for [HI] to decrease to 17.0 \% of its original value.}[/tex]

Answer:

ΔH°rxn = 39013.33 J/mol = 39.013 kJ/mol.

Explanation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃ = (M)(Volume of the solution (L)) = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn = Qrxn/(no. of moles of AgNO₃) = (1170.4 J)/(0.03 mol) = 39013.33 J/mol = 39.013 kJ/mol.

The heat of reaction (ΔH°rxn) for this chemical reaction in a coffee-cup calorimeter is -19.508 kJ/mol, indicating an exothermic reaction.

In this question, we need to calculate the heat change (ΔH°rxn) of a chemical reaction in a coffee cup calorimeter. The reaction has a temperature rise from 21.80 °C to 23.20 °C. This increase in temperature represents an exothermic reaction, which means heat is released in the process.

We can calculate the heat absorbed by using the equation q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature. Given that the density of the solution is 1.00 g/mL and the volume is 200 mL (100 mL of AgNO3 + 100 mL of HCl), our m = 200 g. The specific heat c is given as 4.18 J/g ∙ °C. With ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.40 °C, we find that q = (200 g)(4.18 J/g ∙ °C)(1.40 °C) = 1170.48 J.

ΔH°rxn is given per mole of reaction, so let's convert the heat released to kJ (1.17048 kJ) and the reaction to moles using the 0.300 M concentration. In this reaction, 1 mol of AgNO3 reacts with 1 mol of HCl. So the molar quantity in 200 mL solution is 0.3 mol/L * 0.2 L = 0.06 mol. Then the ΔH°rxn = 1.17048 kJ / 0.06 mol = 19.508 kJ/mol (Note that this value is negative as it is an exothermic reaction).

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Answer:

63.52°C is the final temperature

Explanation:

1) 131.0 mL of water at 26.0°C

Mass of water = m

Volume of the water =131.0 mL

Density of the water = 1.00 g/mL

[tex]Density=1.00 g/mL=\frac{m}{131.0 mL}[/tex]

m = 131.0 g

Initial temperature of the water = [tex]T_i[/tex] = 26.0°C

Final temperature of the water = [tex]T_f[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i[/tex]

Heat absorbed 131.0 g of water = Q

[tex]Q=m\times c\times \Delta T[/tex]

2) 81.0 mL of water at 85.0°C

Mass of water = m'

Volume of the water =81.0 mL

Density of the water = 1.00 g/mL

[tex]Density=1.00 g/mL=\frac{m'}{81.0 mL}[/tex]

m' = 81.0 g

Initial temperature of the water = [tex]T_i'[/tex] = 85.0°C

Final temperature of the water = [tex]T_f'[/tex]

Change in temperature ,[tex]\Delta T'=T_f'-T_i'[/tex]

Heat lost by 81.0 g of water = Q'

[tex]Q'=m'\times c\times \Delta T'[/tex]

After mixing both liquids the final temperature will become equal fro both liquids.

[tex]T_f=T_f'[/tex]

Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.

Q=-Q' (Law of conservation of energy.)

Let the specific heat of water be c

[tex]m\times c\times \Delta T=m'\times c\times \Delta T'[/tex]

[tex]131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))[/tex]

[tex]T_f=63.52^oC[/tex]

63.52°C is the final temperature

According to conservation law, the amount of energy in a closed and isolated system remains constant and does not get reduced or added. The final temperature of the system is 63.52°C.

First, calculate the mass of the 131 mL of water:

Given,

[tex]\begin{aligned}\rm Mass &= \rm Volume \times \rm density\\\\\rm m &= 131.0 \;\rm g\end{aligned}[/tex]

Given,

Initial temperature [tex]\rm (T_{i})[/tex] =  26.0°C

The final temperature of the water = [tex]\rm (T_{f})[/tex]

Change in the temperature is calculated as,

[tex]\rm \Delta T = T_{f} - T_{i}[/tex]

Heat absorbed by 131.0 g of water = Q

The formula to calculate heat absorbed is,

[tex]\rm Q = m \times c \times \Delta T[/tex]

Second, calculate the mass of the 81 mL of water:

Given,

[tex]\begin{aligned}\rm Mass' &= \rm Volume \times \rm density\\\\\rm m' &= 81.0 \;\rm g\end{aligned}[/tex]

Given,

Initial temperature [tex]\rm (T'_{i})[/tex] = 26.0°C

The final temperature of the water = [tex]\rm (T'_{f})[/tex]

Change in the temperature is calculated as,

[tex]\rm \Delta T' = T'_{f} - T'_{i}[/tex]

Heat lost by 81.0 g of water = Q'

The formula to calculate heat lost is,

[tex]\rm Q' = m' \times c \times \Delta T'[/tex]

Both the liquids are mixed and the final temperature will be equivalent and given as, [tex]\Delta \rm T_{f} = \Delta T'_{f}[/tex]

According to the law of conservation,

Heat lost by the water (Q') = Heat absorbed by the water (-Q)

[tex]\begin{aligned}\rm m \times c \times \Delta T &= \rm m' \times c \times \Delta T'\\\\131 \times\rm c(T_{f} -26) &= -(81 \times\rm c(T_{f}- 85))\\\\\rm T_{f} &= 63.52 \circC\end{aligned}[/tex]

Therefore, 63.52°C is the final temperature.

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Answer:

The oxygen atom has a greater attraction for electrons than the hydrogen atom does.

The electrons of the covalent bond are not shared equally between the hydrogen and oxygen atoms.

Its molecular geometry is bent.

Water is polar because the oxygen atom has a greater electronegativity than the hydrogen atoms, leading to unequal sharing of electrons and partial charges on the atoms.

Water is known as a polar molecule for various reasons. First, the oxygen atom in water has a greater attraction for electrons than the hydrogen atoms, which is due to its higher electronegativity. This results in the electrons of the covalent bond between hydrogen and oxygen not being shared equally, leading to a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms. The polarity of water is a critical factor in its ability to form hydrogen bonds, which are stronger than conventional dipole-dipole forces and contribute to water's unique properties such as high surface tension and the ability to dissolve many substances.

Answer:

-112 Kj/mole (diprotic acid)

Explanation:

The enthalpy of reaction is -55 KJ/mol.

We must first write down the equation of the reaction;

2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + 2H2O(l)

Then we compute the number of moles of H2SO4 = 23.6/1000 × 0.500 M  = 0.012 moles

And the number of moles of KOH = 23.6/1000 × 1.00 M = 0.024 moles

From what we know in the reaction equation;

2 moles of KOH produces 2 moles of water

Therefore, 0.0026 moles of KOH produces 0.024 moles of water.

The total volume of solution is obtained by adding  = 23.6 mL + 23.6 mL = 47.2 mL

Mass of water = density × volume = 1.00 g/mL ×  47.2 mL =47.2 g

Using the formula;

ΔH = mcθ

Mass of solution (m) = 47.2 g

Specific heat capacity of solution (c) =  4.184 J/g·°C

Temperature difference(θ) =  30.17°C - 23.50°C = 6.67°C

Substituting values;

ΔH = -( 47.2 g × 4.184 J/g·°C  × 6.67°C)/ 0.024 moles  

ΔH =  -(1.32 KJ/0.024 moles)

ΔH = -55 KJ/mol

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Answer : The correct option is, (A) [tex]1.8\times 10^{-10}M[/tex], basic.

Explanation : Given,

Concentration of [tex]OH^-[/tex] ion = [tex]5.5\times 10^{-5}M[/tex]

First we have to calculate the pOH.

[tex]pOH=-\log [OH^-][/tex]

[tex]pOH=-\log (5.5\times 10^{-5})[/tex]

[tex]pOH=4.26[/tex]

Now we have to calculate the pH.

[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.26=9.74[/tex]

Now we have to calculate the [tex]H_3O^+[/tex] concentration.

[tex]pH=-\log [H_3O^+][/tex]

[tex]9.74=-\log [H_3O^+][/tex]

[tex][H_3O^+]=1.8\times 10^{-10}M[/tex]

As we know that, when the pH value is less than 7 then the solution acidic in nature and when the pH value is more than 7 then the solution basic in nature.

From the pH value, 9.74 we conclude that the solution is basic in nature because the value of pH is greater than 7.

Therefore, the [tex]H_3O^+[/tex] concentration is, [tex]1.8\times 10^{-10}M[/tex], basic.

Answer:I can't art but I envision a comic of a swimming pool with chlorine water in it. The bottom of the pool is black. The chlorine is happy and excited to protect the people going to swim in it. Then the sun comes out, warms the black tile, the water is heated and the chlorine is boiled into gas form. Unable to control its movement through the atmosphere, the large amount of chlorine from the in-ground pool infiltrates the home of its lovely owners, and they die from chlorine gas inhalation, as well as half the neighborhood.

The end.

Here is a comic strip created to explain the transformation of a substance as it changes phases.

Title: The Adventures of Mr. Ice Cube

Introduction: Mr. Ice Cube is a solid block of water. He lives in a freezer, where it is very cold.

Source of Thermal Energy: One day, Mr. Ice Cube is taken out of the freezer and placed in a hot cup of coffee. The hot coffee transfers thermal energy to Mr. Ice Cube, causing him to melt.

Particle Transformation from Solid to Liquid Phase: As Mr. Ice Cube melts, the particles in his solid structure start to move faster. They move so fast that they break free from the solid structure and become liquid particles.

Particle Transformation from Liquid to Gas Phase: As Mr. Ice Cube continues to heat up, the liquid particles start to move even faster. They move so fast that they escape from the liquid state and become gas particles.

Conclusion: Mr. Ice Cube has now transformed from a solid to a liquid to a gas. He is now a cloud of water vapor, floating in the air above the hot cup of coffee.

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Answer:

I believe it is the overseer of the operation

Answer:

The reaction releases energy

Explanation:

The products of an exergonic reaction have a lower energy state (Delta-G) compared to the reactants. Therefore there is a negative delta –G between products and reactants after the reactions. This means some energy is lost into the environment usually through light or heat.

Exergonic reactions are characterized by a net release of energy but they still require a small initial energy input to start, referred to as the 'activation energy'. The speed or direction of the reaction is not determined by whether it's exergonic.

In the context of chemical reactions, the true statement for all exergonic reactions is that such reactions result in a net release of energy. However, even exergonic reactions, which are characterized by energy release, require a small initial input of energy to get started. This initial energy demand is referred to as the 'activation energy'. Also, it's important to note that the speed of the reaction or its directionality (whether it proceeds only in a forward direction) are not inherently determined by whether a reaction is exergonic. These aspects depend on other reaction conditions and catalysis.

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Answer:

see explanation ...

Explanation:

5.00ml(0.100M NaOH) + 0.100M HOAc/NaOAc Bfr

5.00ml(0.100M NaOH) = 0.005(0.100) mole NaOH = 0.0005 mole NaOH = 0.0005 mole OH⁻ in 55 ml Bfr solution (50ml + 5ml)=> [OH⁻] = (0.0005/0.055)M OH⁻ = 0.0091M OH⁻ ≈ 0.010M OH⁻ added into Bfr solution. The amount of OH⁻ added must be removed by H⁺ in the HOAc equilibrium; that is,  H⁺ + OH⁻ → H₂O leaving a void at the H⁺ position in the HOAc equilibrium. HOAc then decomposes to replace the H⁺ removed by the excess OH⁻ giving new H⁺ and OAc⁻ concentrations. Reaction shifts right => subtract 0.01M from HOAc side of equilibrium and add 0.01M to OAc⁻side of equilibrium and recompute the H⁺ concentration and new pH.

                 HOAc  ⇄        H⁺     +    OAc⁻

C(i)             0.10M         ~0M*         0.10M =>pH=-log(Ka)=-log(1.85x10⁻⁵)=4.73

ΔC              -0.01M          +x          +0.01M

C(eq)          0.09M            x             0.11M  => New HOAc equil. conc.

Ka = [H⁺][OAc⁻]/[HOAc]

=> x(0.11)/(0.09) = 1.85x10⁻⁵

=> x = [H⁺] = 0.09(1.85x10⁻⁵)/0.11 = 1.52x10⁻⁵M (after adding NaOH)

=> pH = -log[H⁺] -log(1.52x10⁻⁵) = 4.82 ( pH shifts to more basic value b/c of OH⁻ addition )

If you add 5.00 mL of 0.100 M sodium hydroxide to 50.0 mL of acetate buffer. The pH of the resulting solution is 4.82.

pH is a measurement scale, used to measure the acids and the bases

The pH of the resulting solution

[tex]\rm Ka = \dfrac{ [H^+][OAc^-]}{[HOAc]}[/tex]

[tex]x \dfrac{ (0.11)}{(0.09)} = 1.85 \times 10^-^5[/tex]

[tex]\rm x = [H^+] = 0.09 \dfrac{(1.85x10^-^5)}{0.11} = 1.52 \times 10^-^5 M[/tex]

pH = -log[H⁺] -log(1.52x10⁻⁵) = 4.82

( pH shifts to more basic value b/c of OH⁻ addition )

Thus, the pH of the resulting solution is 4.82.

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Answer:

A) the same molecular formula but different chemical properties.

Explanation:

Isomerism is the existence of a compound with the same molecular formula but different molecular structures due to the difference in the arrangement of atoms or spatial orientation of the atoms. In such a compound, they differ in physical and/or chemical properties. Such possible structures are known as isomers.

Propanol and isopropanol differs only in the arrangement of atoms but they both have the same molecular formula. Their physical and chemical properties differ.

Propanol and isopropanol are structural isomers, which means they have the same molecular formula but different chemical properties due to different arrangement of atoms within the molecules.

Propanol and isopropanol are indeed isomers, specifically they are structural isomers. Structural isomers are compounds that share the same molecular formula but have a different spatial arrangement of their atoms, leading to different chemical properties. This means, option A) the same molecular formula but different chemical properties, is correct.

A classic example of this are the compounds n-butane and 2-methylpropane. Both compounds have the same molecular formula, C4H10, but n-butane contains an unbranched chain of carbon atoms, while 2-methylpropane has a branched chain. This difference in structure results in different chemical properties for each compound.

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Answer:

184.46

Explanation:

(84.7°c x 9/5)+32=184.46 °F

In the balanced molecular equation between CuCl₂(aq) and K₃PO₄(aq), the cations and anions exchanged, resulting in 6KCl(aq) and Cu₃(PO₄)₂(s). The net ionic equation, focusing only on participating substances is: 3Cu²⁺(aq) + 2PO₄³⁻(aq) → Cu₃(PO₄)₂(s).

The reaction between aqueous copper(II) chloride (CuCl2) and aqueous potassium phosphate (K3PO4) is a double displacement reaction, where the cation of one compound combines with the anion of the other compound. The balanced molecular equation for this reaction is: 2CuCl2(aq) + 3K3PO4(aq) → 6KCl(aq) + Cu3(PO4)2(s).

Now, for the net ionic equation, you would only include the species that participate in the reaction and exclude the spectator ions (ions that do not participate in the reaction). After breaking down the molecular equation to ions (excluding spectator ions) the net ionic equation of the reaction would be: 3Cu^2+(aq) + 2PO4^3-(aq) → Cu3(PO4)2(s).

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The balanced molecular equation is 3CuCl₂(aq) + 2K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq). The net ionic equation is 3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s). This represents the formation of copper(II) phosphate precipitate.

Complete and Balance the Molecular Equation

Let's complete and balance the molecular equation for the reaction of aqueous copper(II) chloride (CuCl₂) and aqueous potassium phosphate (K₃PO₄).

Molecular Equation:

CuCl₂(aq) + K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)

First, we need to balance this equation:

Balance the copper (Cu) atoms: 3 CuCl₂(aq) + K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)

Balance the potassium (K) atoms: 3 CuCl₂(aq) + 2 K₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6KCl(aq)

Write the Net Ionic Equation

To write the balanced net ionic equation, first write the complete ionic equation:

3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 K⁺(aq) + 2 PO₄³⁻(aq) ⟶ 6 K⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)

Next, identify and remove the spectator ions (K⁺ and Cl⁻), which are ions that do not participate in the reaction:

Net Ionic Equation:

3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

This net ionic equation represents the formation of copper(II) phosphate precipitate from copper(II) ions and phosphate ions in solution.

Final answer:

Using the half-life of Potassium-40, which is 1.25 billion years, and given the 8:1 ratio of K-40 to Ar-40, the rock sample is determined to be approximately 3.75 billion years old.

Explanation:

To date the rock sample to the time it contained only Potassium-40 (K-40), we must analyze the ratio of K-40 to Argon-40 (Ar-40). Given the information that there are eight times as many K-40 atoms as Ar-40 atoms in the sample, we can use the concept of half-lives to date the sample. One half-life (1.25 billion years) would result in an equal number of K-40 and Ar-40 atoms.

Since we have eight times as many K-40 atoms, the sample has gone through three half-lives (since 2³ = 8), which equates to 3.75 billion years (three half-lives of 1.25 billion years each). The rock sample would therefore be approximately 3.75 billion years old, which is the time it would have contained only K-40 atoms.

The solution with a pH of 11 has a hydrogen ion concentration of 1 x 10^-11. Hence, the correct option is B.

The question involves understanding the relationship between pH and hydrogen ion concentration ([H+]) in solutions. The pH of a solution is a measure of its acidity or basicity, with values less than 7 being acidic, around 7 being neutral, and greater than 7 being basic. The [H+] is related to pH through the formula [H+] = 10-pH.

For a solution with a pH of 11, we can calculate its [H+] concentration using the formula mentioned. Converting the pH into the hydrogen ion concentration gives us [H+] = 10-11. Thus, the correct answer is B. 1 x 10-11.

This formula demonstrates the logarithmic relationship between pH and hydrogen ion concentration, highlighting that small changes in pH equate to exponential changes in [H+]. For instance, a pH difference of 1 corresponds to a tenfold difference in hydrogen ion concentration. The question essentially tests the understanding of this critical concept in Chemistry.

By equating the heat gained by the melting ice to the heat lost by the water, and using the enthalpy of fusion for ice and the specific heat of water, the calculation reveals that there were approximately 595 grams of water in the sample.

To find out how many grams of water were in the sample before adding the ice, we need to equate the heat gained by the melting ice to the heat lost by the water when its temperature decreased from 7.4 °C to 0.0 °C. The enthalpy of fusion of ice, which is the amount of heat required to melt the ice without changing its temperature, is a key concept in this problem. The specific heat of water is also an important factor as it influences the amount of heat water can lose or absorb.

The enthalpy of fusion of ice is typically given as 334 J/g. Using this, the heat gained by the ice as it melts can be calculated by multiplying the mass of the ice (48 g) by the enthalpy of fusion. The heat lost by the water can be calculated using the specific heat of water (4.184 J/g°C), the change in temperature (7.4 °C), and the unknown mass of water.

Heat gained by ice = Heat lost by water (Qice = Qwater)

334 J/g × 48 g = 4.184 J/g°C × (mass of water) × 7.4 °C

By rearranging the equation, we find the mass of water:

mass of water = (334 J/g × 48 g) / (4.184 J/g°C × 7.4 °C)

After solving, the mass of the water is approximately 595 g.