Calculate the boiling point of water if the atmospheric pressure is 630 mmhg

The spectator ions for the reaction between lithium sulfide and copper nitrate are [tex]\boxed{{\text{c}}{\text{. L}}{{\text{i}}^ + }{\text{ and NO}}_3^ - }[/tex].

Further Explanation:

Spectator ions:

These are the ions that exist in the same form on both sides of the reaction. These ions do not affect the equilibrium of any reaction.

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

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

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

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

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

In the reaction, [tex]{\text{L}}{{\text{i}}_2}{\text{S}}[/tex] reacts with [tex]{\text{Cu}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}[/tex] to form CuS and [tex]{\text{LiN}}{{\text{O}}_3}[/tex]. The balanced molecular equation of the reaction is as follows:

[tex]{\text{L}}{{\text{i}}_{\text{2}}}{\text{S}}\left( {aq} \right) + {\text{Cu}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\left( {aq} \right) \to {\text{CuS}}\left( s \right) + 2{\text{LiN}}{{\text{O}}_3}\left( {aq} \right)[/tex]

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

[tex]2{\text{L}}{{\text{i}}^ + }\left( {aq} \right) + {{\text{S}}^{2-}}\left({aq}\right)+{\text{C}}{{\text{u}}^{2+}}\left( {aq} \right) + 2{\text{NO}}_3^-\left({aq}\right)\to {\text{CuS}}\left(s\right)+2{\text{L}}{{\text{i}}^+}\left({aq}\right)+2{\text{NO}}_3^-\left( {aq}\right)[/tex]

Step 3: The common or spectator ions on both sides of the reaction get cancelled out to get the net ionic equation.

[tex]\boxed{2{\text{L}}{{\text{i}}^ + }\left( {aq} \right)} + {{\text{S}}^{2 - }}\left( {aq} \right) + {\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right) + \boxed{2{\text{NO}}_3^ - \left( {aq} \right)} \to {\text{CuS}}\left( s \right) + \boxed{2{\text{L}}{{\text{i}}^ + }\left( {aq} \right)} + \boxed{2{\text{NO}}_3^ - \left( {aq} \right)}[/tex]

[tex]{\text{L}}{{\text{i}}^ + }[/tex] and [tex]{\text{NO}}_3^ -[/tex] ions are present in the same form on both the reactant and the product side and therefore are known as spectator ions.

Therefore, the net ionic equation is as follows:

[tex]{{\text{S}}^{2 - }}\left( {aq} \right) + {\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right) \to {\text{CuS}}\left( s \right)[/tex]

Learn more:

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: net ionic equation, spectator ions, CuS, Cu2+, Li+, NO3-, S2-, 2 Li+, 2 NO3-.

The dissolved particles in a solution containing a molecular solute are the molecules of the solute themselves. When a molecular solute dissolves, its molecules distribute uniformly within the solvent, forming a homogeneous mixture at the molecular level. However, ionic solutes dissociate into individual cations and anions.

The dissolved particles in a solution containing a molecular solute like sugar (C12H22O11) would be molecules. In chemistry, a solution can be thought of as a mixture that is homogeneous at the molecular level. In these solutions, one component (the solvent) is usually present in a significantly larger concentration. Other components (solutes), are present in relatively lesser amounts. When a covalent solid like sugar dissolves in water, its molecules become uniformly distributed among the molecules of water creating an aquaeous solution.

However, not all solutions result in dissolved molecules. Solutions that include ionic compounds like sodium chloride (NaCl) will separate into individual cations and anions when dissolved in water. In contrast to molecular substances, ionic type compounds will break apart into separate ions. Hence, it's crucial to understand the type of solute in your solution.

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Answer:

207.2 g

Explanation:

The mass of 1 mole of lead is given by its molar mass, which  is 207.2 g/mol. This is an average molar mass, corresponding to the weighted average of the 4 lead isotopes, ²⁰⁴Pb, ²⁰⁶Pb, ²⁰⁷Pb, ²⁰⁸Pb, considering their abundance in nature. The mass of 1 mole of Pb is:

1 mol × (207.2 g/mol) = 207.2 g

Answer:c mobile home

Explanation:

A mobile home is preconstructed structure, that can be run and transported to other location according to the choice of living place. This can be prepared from wood or other kind of cheap material which can keep the weight of mobile home low.

Thus mobile home is the lowest maintainace cost home for a single family.

When 6.01 g of NH₃ reacts with 4.91 g of HCl, there are 3.71 grams of excess NH₃. HCl is the limiting reactant in this reaction based on the number of moles available.

The reaction between ammonia (NH₃) and hydrogen chloride (HCl) to form ammonium chloride (NH₄Cl) is described by the balanced equation:

NH₃ + HCl → NH₄Cl

First, we need to determine the number of moles of each reactant.

Molar mass of NH₃ = 14.01 (N) + 3 × 1.01 (H) = 17.04 g/mol

Moles of NH₃ = 6.01 g / 17.04 g/mol = 0.3526 moles

Molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g/mol

Moles of HCl = 4.91 g / 36.46 g/mol = 0.1347 moles

From the balanced equation, the mole ratio of NH₃ to HCl is 1:1, meaning they react on a 1:1 basis. Therefore, HCl is the limiting reactant because we have fewer moles of HCl than NH₃.

Since the reaction is 1:1, 0.1347 moles of HCl will fully react with 0.1347 moles of NH₃. We started with 0.3526 moles of NH₃ and used up 0.1347 moles:

Moles of NH₃ remaining = 0.3526 - 0.1347 = 0.2179 moles

To find the mass of the excess NH₃:

Mass of excess NH₃ = 0.2179 moles × 17.04 g/mol = 3.71 g

Therefore, when 6.01 g of NH₃ reacts with 4.91 g of HCl, there are 3.71 grams of excess NH₃.

The Heck reaction involves bond formation between an alkene and an aryl halide. With bromobenzene and propene, two isomers can be obtained, 1-propylbenzene and 2-propylbenzene. These differ in the attachment point of the propyl group to the benzene ring.

The Heck reaction involves a palladium-catalyzed carbon-carbon bond formation between an alkene and an aryl halide. When bromobenzene reacts with propene in a Heck reaction with pd(php3)4 and et3n, two constitutional isomers can be obtained. The difference between the two isomers would lie in the point of attachment of the propyl group to the benzene ring - it could either be at the end (1-propylbenzene) or at the middle (2-propylbenzene).

For 1-propylbenzene, the propyl group is attached to the benzene ring at one end, resulting in a straight chain configuration.

However, for 2-propylbenzene, the propyl group is attached to the benzene ring in the middle, resulting in a branched chain configuration.

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To find the molecular formula, divide the given molar mass by the empirical formula mass of the compound and multiply the empirical formula by the resulting value. In this case, the molecular formula is 2 times the empirical formula.

The molar mass of a compound can be used to determine its molecular formula. To find the molecular formula, we divide the given molar mass by the empirical formula mass of the compound. If the resulting value is a whole number, then the empirical formula is also the molecular formula. If it is not a whole number, we need to multiply the empirical formula by the integer value closest to the resulting value to obtain the molecular formula.

In this case, since the molar mass is given as 73.8 grams/mole, and the calculated empirical formula mass is 176.124 g/mol, we find that 73.8 g/mol divided by 176.124 g/mol is equal to 0.419. Since this is not a whole number, we need to multiply the empirical formula by 2 since 2 is the integer that is closest to 0.419. Therefore, the molecular formula is 2 times the empirical formula.

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Kc is approximately 0.00362, and Kp is approximately 2.167 based on molar concentrations and gas collected data.

To accurately calculate the equilibrium constants [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex] for the given reaction, we need to follow the steps correctly:

[tex]\[ \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \leftrightarrow \text{CO}(\text{g}) + 3\text{H}_2(\text{g}) \][/tex]

1. Molar Masses:

- [tex]\(\text{CH}_4\):[/tex] 16.04 g/mol

- [tex]\(\text{H}_2\text{O}\)[/tex] : 18.02 g/mol

- [tex]\(\text{CO}\):[/tex] 28.01 g/mol

- [tex]\(\text{H}_2\):[/tex] 2.02 g/mol

2. Calculate Moles of Each Substance:

[tex]\(\text{CH}_4\): \( \frac{42.3 \, \text{g}}{16.04 \, \text{g/mol}} = 2.638 \, \text{moles} \)\\ \(\text{H}_2\text{O}\): \( \frac{49.2 \, \text{g}}{18.02 \, \text{g/mol}} = 2.730 \, \text{moles} \)\\\(\text{CO}\): \( \frac{8.32 \, \text{g}}{28.01 \, \text{g/mol}} = 0.297 \, \text{moles} \)\\\(\text{H}_2\): \( \frac{2.63 \, \text{g}}{2.02 \, \text{g/mol}} = 1.301 \, \text{moles} \)[/tex]

3. Calculate Concentrations:

[tex]- \([ \text{CH}_4 ] = \frac{2.638 \, \text{moles}}{5.00 \, \text{L}} = 0.528 \text{M} \)\\\\- \([ \text{H}_2\text{O} ] = \frac{2.730 \, \text{moles}}{5.00 \, \text{L}} = 0.546 \, \text{M} \) \\\\- \([ \text{CO} ] = \frac{0.297 \, \text{moles}}{5.00 \, \text{L}} = 0.0594 \, \text{M} \)\\\\- \([ \text{H}_2 ] = \frac{1.301 \, \text{moles}}{5.00 \, \text{L}} = 0.260 \, \text{M} \)[/tex]

4. Calculate [tex]K_c[/tex]

[tex]\[ K_c = \frac{[ \text{CO} ][ \text{H}_2 ]^3}{[ \text{CH}_4 ][ \text{H}_2\text{O} ]} \][/tex]

[tex]\[ K_c = \frac{(0.0594) \times (0.260)^3}{(0.528) \times (0.546)} \][/tex]

[tex]\[ K_c = \frac{0.0594 \times 0.017576}{0.288288} \][/tex]

[tex]\[ K_c = \frac{0.0010431744}{0.288288} \][/tex]

[tex]\[ K_c \approx 0.00362 \][/tex]

5. Calculate [tex]K_p[/tex]

[tex]\[ K_p = K_c \left( RT \right)^{\Delta n} \][/tex]

- [tex]\(\Delta n = \text{moles of products} - \text{moles of reactants} = (1 + 3) - (1 + 1) = 2\)[/tex]

- Assuming [tex]\( T = 298 \, \text{K} \)[/tex] (standard temperature) and [tex]( R = 0.0821 \, \text{L atm/(mol K)} \):[/tex]

[tex]\[ K_p = 0.00362 \left( 0.0821 \times 298 \right)^2 \][/tex]

[tex]\[ K_p = 0.00362 \left( 24.4658 \right)^2 \][/tex]

[tex]\[ K_p = 0.00362 \times 598.5877 \][/tex]

[tex]\[ K_p \approx 2.167 \][/tex]

Conclusion:

[tex]\( K_c \approx 0.00362 \)\\\\ \( K_p \approx 2.167 \)[/tex]

In summary, Kc is approximately 0.00362, and Kp is approximately 2.167

Answer:

The rate will be doubled by a Factor of 18

Explanation:

Good luck to everyone doing this!! you got this

The rate of the reaction will increase by a factor of 18 if the concentration of C2O42- is tripled and the concentration of HgCl2 is doubled.

The question asks how the rate of the reaction 2 HgCl2(aq) + C2O42-(aq) → 2 Cl-(aq) + 2 CO2 (g) + Hg2Cl2 (s) will change if the concentration of C2O42- is tripled and the concentration of HgCl2 is doubled. Given the rate law rate = k[HgCl2][C2O42-]2, we can predict the effect on the rate. If the rate law is followed, tripling [C2O42-] will increase the rate by a factor of 32 or 9 because the concentration of C2O42- is squared in the rate law. Doubling [HgCl2] will double the rate of reaction. Therefore, tripling [C2O42-] and doubling [HgCl2] together will increase the reaction rate by a factor of 2 * 9 = 18.

The alkoxide ion (RO⁻) and the alkyl(aryl) bromide can be used in a Williamson ether synthesis to create ethers. The structural formulas of these compounds depend on the specific alkyl or aryl group involved in the synthesis.

In a Williamson ether synthesis, an alkoxide ion reacts with an alkyl (or aryl) halide to form an ether. To perform this synthesis, let's consider the given ether and determine the structural formulas for the alkoxide ion and the alkyl(aryl)bromide.

Given Ether:

The given ether is not explicitly mentioned in your question. To proceed, let's assume the ether you intend to synthesize has the following structure:

R-O-R

where "R" represents an alkyl or aryl group. This is a general representation for ethers.

Structural Formulas:

1. Alkoxide Ion (RO⁻):

  - The alkoxide ion (RO⁻) is formed by removing a hydrogen atom from an alcohol (R-OH). It has a lone pair of electrons on the oxygen atom. The structural formula is as follows:

    RO⁻

    |

    R

2. Alkyl (Aryl) Bromide:

  - The alkyl bromide or aryl bromide can be represented as "R-Br" or "Ar-Br," where "R" represents an alkyl group, and "Ar" represents an aryl group. The structural formula depends on the specific alkyl or aryl group you are using.

  Example 1 (Alkyl Bromide):

  - For an alkyl bromide, if we use "R" as a general alkyl group, the structural formula is:

    R-Br

  Example 2 (Aryl Bromide):

  - For an aryl bromide, if we use "Ar" as a general aryl group, the structural formula is:

    Ar-Br

In a Williamson ether synthesis, the alkoxide ion (RO⁻) is a strong nucleophile that attacks the alkyl (aryl) halide, leading to the formation of the desired ether by replacing the halogen atom. The specific alkyl or aryl group used will depend on the reactants and the ether you intend to synthesize.

For more such information on: alkoxide ion

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Answer:

              Volume =  100 cm³

Solution:

             Density measures and compares the amount of mass contained by a substance to its volume. It is expressed as,

                                           Density  =  Mass / Volume

Data Given;

                   Density  =  1.0 g/cm³

                   Mass  =  100 g  

                   Volume  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Solving for Volume,

                   Volume  =  Mass ÷ Density  

Putting values,

                   Volume  =  100 g ÷ 1.0 g.cm⁻³

                   Volume =  100 cm³

Norfenefrine (C₈H₁₁NO₂).

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

[tex]\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }[/tex]

Given:

A mysterious white powder could be,

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

Question: What is the identity of the white powder?

The Process:

Let us identify the solute, the solvent, initial, and final temperatures.

Prepare masses of solutes and solvents.

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

[tex]\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }[/tex]

Now we combine it with the formula of freezing point depression.

[tex]\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }[/tex]

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

[tex]\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }[/tex]

[tex]\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }[/tex]

[tex]\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }[/tex]

We get [tex]\boxed{ \ Mr = 153.65 \ }[/tex]

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

The identity of the white powder is [tex]\boxed{\text{norfenefrine}(\text{C}_{8}\text{H}_{11}\text{NO}_{2})}[/tex] .

Further Explanation:

Colligative properties

These are the properties that depend on the number of solute particles and not on their mass or identities. Following are the four colligative properties:

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

The temperature where a substance in its liquid form is converted into the solid state is known as freezing point. Freezing point depression is a colligative property because it depends on the number of moles of solute particles.

The expression for the freezing point depression is as follows:

[tex]\Delta\text{T}_\text{f}=\text{K}_\text{f}\,\text{m}[/tex]       …… (1)

Here,

[tex]\Delta\text{T}_\text{f}[/tex] is the depression in freezing point.

[tex]\text{K}_\text{f}[/tex] is the cryoscopic constant.

m is the molality of the solution.

The formula to calculate the density of substance is as follows:

[tex]\text{Density of substance}=\dfrac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]                                                      …… (2)

Rearrange equation (2) for the mass of substance.

[tex]\text{Mass of substance}=(\text{Density of substance })(\text{Volume of substance})[/tex]                                 …… (3)

The volume of the substance is to be converted into  . The conversion factor for this is,

[tex]1\,\text{ml}=1\,\text{cm}^3[/tex]  

Therefore the volume of the substance can be calculated as follows:

[tex]\begin{aligned}\text{Volume}&=(1.50\,\text{mL})\left(\frac{1\,\text{cm}^3}{1\,\text{mL}}\right)\\&=1.50\,\text{cm}^3\end{aligned}[/tex]  

Substitute [tex]1.50\,\text{cm}^3[/tex] for the volume of substance and [tex]0.789\,\text{g/gm}^3[/tex] for the density of the substance in equation (3) to calculate the mass of solvent.

[tex]\begin{aligned}\text{Mass of solvent}&=\left(\dfrac{0.789\,\text{g}}{1\,\text{cm}^3}\right)(1.50\,\text{cm}^3)\\&=1.1835\,\text{g}\end{aligned}[/tex]  

This mass is to be converted into kg. The conversion factor for this is,

[tex]1\,\text{g}=10^{-3}\,\text{kg}[/tex]  

Therefore the mass of solvent can be calculated as follows:

[tex]\begin{aligned}\text{Mass of solvent}&=(1.1835\,\text{g})\left(\dfrac{10^{-3}\,\text{kg}}{1\,\text{g}}\right)\\&=0.0011835\,\text{kg}\end{aligned}[/tex]  

The mass of solute is to be converted into g. The conversion factor for this is,

 [tex]1\,\text{mg}=10^{-3}\,\text{g}[/tex]

Therefore the mass of solute can be calculated as follows:

[tex]\begin{aligned}\text{Mass of solute}&=(82\,\text{mg})\left(\dfrac{10^{-3}\,\text{g}}{1\,\text{g}}\right)\\&=0.082\,\text{g}\end{aligned}[/tex]  

The freezing point depression can be calculated as follows:

[tex]\begin{aligned}\Delta\text{T}_\text{f}&=-114.6\,^\circ\text{C}-(-115.5\,^\circ\text{C})\\&=0.9\,\circ\text{C}\end{aligned}[/tex]

The formula to calculate the molality of solution is as follows:

[tex]\text{Molality of solution}=\dfrac{\text{Amount (mol) of solute}}{\text{Mass (kg) of solvent}}[/tex]                                                        …… (4)

The formula to calculate the amount of solute is as follows:

[tex]\text{Amount of solute}=\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}[/tex]                                                              …… (5)

Incorporating equation (5) into equation (4),

[tex]\text{Molality of solution}=\dfrac{\text{Mass of solute}}{(\text{Mass of solvent})(\text{Molar mass of solute})}[/tex]                               …… (6)

Incorporating equation (6) into equation (1),

[tex]\Delta\text{T}_\text{f}=\text{k}_\text{f}\left(\dfrac{\text{Mass of solute}}{(\text{Mass of solvent})(\text{Molar mass of solute})}\right)[/tex]                                           …… (7)

Rearrange equation (7) to calculate the molar mass of solute.

[tex]\text{Molar mass of solute}=\dfrac{\text{k}_\text{f}}{\Delta\text{T}_\text{f}}\left(\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\right)[/tex]                                                        …… (8)

Substitute [tex]1.99\,^\circ\text{C/m}[/tex] for [tex]\text{k}_\text{f}[/tex], [tex]0.9\,^\circ\text{C}[/tex] for [tex]\Delta\text{T}_\text{f}[/tex], 0.082 g for the mass of solute and 0.0011835 kg for the mass of solvent in equation (8).

[tex]\begin{aligned}\text{Molar mass of solute}&=\left(\dfrac{1.99}{0.9}\right)\left(\dfrac{0.082}{0.0011835}\right)\\&=153.199\,\text{g/mol}\\&=\approx153.2\,\text{g/mol}\end{aligned}[/tex]  

The molar mass of powdered sugar is 342.3 g/mol.

The molar mass of cocaine is 303.4 g/mol.

The molar mass of codeine is 299.4 g/mol.

The molar mass of norfenefrine is 153.2 g/mol.

The molar mass of fructose is 180.2 g/mol.

The calculated molar mass of solute is similar to that of norfenefrine. So the identity of the white powder is norfenefrine.

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Answer details:

Grade: Senior School

Chapter: Solutions

Subject: Chemistry

Keywords: colligative properties, depression in freezing point, cryoscopic constant, freezing point, 153.199 g/mol, norfenefrine.

In the Ideal Gas law, the letter R represents the universal gas constant. Its value depends on the units being used for pressure, volume, moles, and temperature. The ideal gas law equation is PV=nRT.

In the Ideal Gas law, the letter R stands for the universal gas constant. This constant is derived from the Ideal Gas Equation: R = 0.08206 L atm mol-¹ K-¹ or 8.314 L kPa mol-¹ K-¹. The ideal gas law itself is represented as PV = nRT, where P represents pressure, V is volume, n is the number of moles of the gas, T is temperature in Kelvin, and R is the universal gas constant.

The value of R, 0.08206 L atm mol-¹ K-¹ or 8.314 J/mol · K, depends on the units being used. For instance, when using SI units, R would be 8.31 J/mol · K.

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The concentration of drug after 4.5 hoursis [tex]\boxed{{\text{0}}{\text{.25 mg/100 mL}}}[/tex].

Further Explanation:

Radioactive decayis responsible to stabilizeunstable atomic nucleusaccompanied by the release of energy.

Half-life is the duration after which half of the original sample has been decayed and half is left behind. It is represented by [tex]{t_{{\text{1/2}}}}[/tex].

The relation between rate constant and half-life period for first-order reaction is as follows:

[tex]k = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex]                                                                                                    …… (1)

Here,

[tex]{t_{{\text{1/2}}}}[/tex] is half-life period.

k is rate constant.

Substitute 90 min for [tex]{t_{{\text{1/2}}}}[/tex] in equation (1).

[tex]\begin{aligned}k &= \frac{{0.693}}{{90{\text{ min}}}} \\&= 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} \\\end{aligned}[/tex]  

Radioactive decay formula is as follows:

[tex]{\text{A}} = {{\text{A}}_0}{e^{ - kt}}[/tex]                                        …… (2)

Here

A is the concentration of sample after time t.

t is the time taken.

[tex]{{\text{A}}_0}[/tex] is the initial concentration of sample.

k is the rate constant.

The time for the process is to be converted into min. The conversion factor for this is,

[tex]1{\text{ hr}} = 60\min[/tex]  

Therefore time taken can be calculated as follows:

[tex]\begin{aligned}t&= \left( {4.5{\text{ hr}}} \right)\left( {\frac{{60{\text{ min}}}}{{1{\text{ hr}}}}} \right) \\&= 270{\text{ min}} \\\end{aligned}[/tex]  

Substitute 2 mg/100 mL for [tex]{{\text{A}}_0}[/tex] , [tex]7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}[/tex] for k and 270 min for t in equation (2).

[tex]\begin{aligned}{\text{A}} &= \left( {\frac{{2{\text{ mg}}}}{{100{\text{ mL}}}}} \right){e^{ - \left( {7.7 \times {{10}^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}} \right)\left( {270{\text{ min}}} \right)}} \\&= \frac{{0.25{\text{ mg}}}}{{100{\text{ mL}}}} \\\end{aligned}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, 0.25 mg/100 mL, 2 mg/100 mL, A, k, t, rate constant, 4.5 h, 270 min, radioactive decay formula.

Answer is: The heat is unusable and lost to the surroundings.

Gasoline is a mixture  of many different hydrocarbons: alkanes (paraffins), cycloalkanes  and alkenes (olefins).

Balanced chemical reaction of gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O + energy.

One part of energy produced in combustion of gasoline is used for car ungine (kinetic energy) and other part of that energy (in form of heat) is lost to the surronding and engine does not use it.

Waves produced a diffraction pattern.

Results supported the wave theory of light.

Answer:

Waves produced a diffraction pattern.

Results supported the wave theory of light.

Explanation:

i got it right

The statements that describe phase changes are particles in a liquid need to move more slowly in order to freeze and attractive forces between the particles in a liquid are broken when a liquid boils.

For better understanding let's explain what it means

From the above we can therefore say that the answer The statements that describe phase changes are particles in a liquid need to move more slowly in order to freeze and attractive forces between the particles in a liquid are broken when a liquid boils, is correct

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The correct statements about phase changes are Option 1 & 3 : particles in a liquid need to move more slowly to freeze, and attractive forces between particles in a liquid are broken when the liquid boils.

Phase changes refer to the transformation of a substance from one state of matter to another: solid, liquid, or gas. Let's examine the given statements:

Based on the explanation, the correct statements are 1 and 3.

The balanced chemical equation for the reaction between phosphoric acid and cesium hydroxide to produce cesium dihydrogen phosphate and water is 3 CsOH(aq) + H3PO4(aq) → Cs3PO4(aq) + 3 H2O(l).

The balanced chemical equation for the reaction between aqueous solutions of phosphoric acid (H3PO4) and cesium hydroxide (CsOH) to produce aqueous cesium dihydrogen phosphate (CsH2PO4) and liquid water (H2O) is as follows:

3 CsOH(aq) + H3PO4(aq) → Cs3PO4(aq) + 3 H2O(l)

This reaction is an example of an acid-base neutralization, resulting in the formation of a salt (cesium dihydrogen phosphate) and water.

The answer would be false because the true answer would be astronomy

Final answer:

To produce a single molecule of water (H₂O), exactly two hydrogen atoms are required. This 2:1 ratio of hydrogen to oxygen atoms is consistent no matter how many molecules of water are being made.

Explanation:

To produce a single molecule of water, which has the chemical formula H₂O, you need two hydrogen atoms. The subscript '2' in H₂ indicates that there are two hydrogen atoms for every oxygen atom in a water molecule. Therefore, for each molecule of water you create, you must have this 2:1 ratio of hydrogen to oxygen atoms.

For example, if you want to create two water molecules, you would need a total of 4 hydrogen atoms (2 molecules × 2 atoms/molecule). Similarly, to produce five water molecules, you would require 10 hydrogen atoms in all (5 molecules × 2 atoms/molecule). No matter the number of water molecules you are looking to produce, you will always need twice as many hydrogen atoms as you have water molecules because of this consistent ratio.

Answer:The green house gases are: carbon dioxide, nitrous oxide

Explanation:

Green house gases are the atmospheric gases which absorbs and emit radiant energy. These gases are responsible green house effect and rise in temperature of planet earth.For example: carbon-dioxide,nitrous oxide, methane, , water vapors etc. There also man made green house gases like: chloro-fluorocarbons , hydro-fluorocarbons etc.

Answer: 2.4 grams

Explanation:

To calculate the moles, we use the equation:

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

For [tex]O_2[/tex]

Given mass = 19 g

Molar mass of  [tex]O_2[/tex] = 32 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of}O_2=\frac{19}{32}=0.60moles[/tex]

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

1 mole of [tex]O_2[/tex] reacts with 2 moles of [tex]H_2[/tex]

0.60 moles of [tex]O_2[/tex] wil react with =[tex]\frac{2}{1}\times 0.60=1.20[/tex] moles of [tex]H_2[/tex]

mass of [tex]H_2=moles\times {\text {molar mass}}=1.20moles\times 2g/mol=2.4g[/tex]

Thus 2.4 grams of hydrogen will react completely with 19 g of oxygen.