HNO3 is a strong acid. Calculate the pH of an aqueous 0.60 M solution.

Answer:

[tex]\large \boxed{\text{250 lb}}[/tex]

Explanation:

1. Weight of oranges

An average box contains 102 oranges, so

Five boxes contain 510 oranges

[tex]\text{Weight of oranges} = \text{510 oranges} \times \dfrac{\text{7.2 oz}}{\text{1 orange}} = \text{3672 oz}\\\\\text{Weight} = \text{3672 oz} \times \dfrac{\text{1 lb}}{\text{16 oz}} = \text{230 lb}[/tex]

2. Weight of boxes

[tex]\text{Weight} = \text{5 boxes} \times \dfrac{\text{3.2 lb}}{\text{1 box}} = \text{16 lb}[/tex]

3. Total weight

[tex]\begin{array}{rcr}\text{Oranges} & = & \text{230 lb}\\\text{Boxes} & = & \text{16 lb}\\\text{TOTAL} & = & \textbf{250 lb}\\\end{array}\\\text{On average, the total weight is $\large \boxed{\textbf{250 lb}}$}[/tex]

Note: The answer can have only two significant figures because that is all you gave for the average weight of an orange.

Answer:

The sample tube holds 3.0 cc which is equal to 0.0030 Liters.

Explanation:

Volume of the blood sample in the sample tube = 3.0 cc

cc = cubic centimeter ; A unit to measure a volume

1 cubic centimeter is equal to 0.001 Liters of volume.

[tex]1 cm^3=0.001 L[/tex]

So in 3.0 cubic centimeter there will be:

[tex]3.0 cm^3=3.0\times 0.001 L=0.0030 L[/tex]

The sample tube holds 3.0 cc which is equal to 0.0030 Liters.

Answer:

(a) Utilization

Explanation:

Thermodynamic energy are:

Kinetic energy is that type of energy of a body which occurs due to the motion of body. Kinetic energy is always positive. In a bound system, the system remains bound as long as the kinetic energy is less than the potential energy due to the interaction of the body.

                            [tex]K.E = \frac{1}{2}mv^{2}[/tex]

Potential energy is defined as the energy in which the energy is possessed by a body or a system for doing some work, by virtue of its position above the ground level.

Therefore,

Potential energy = PE = mgh

This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 m s⁻².

Stored energy is a Potential energy.

Hence, the only energy which is not a thermodynamic energy is option (a) Utilization energy.

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

[tex]T_{solution}-T{solvent}=K_bm_solute[/tex]

Considering that the difference in the temperature is 0.284°C, and the given molality by:

[tex]m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m[/tex]

Now, solving for [tex]K_b[/tex], we get:

[tex]K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m[/tex]

Best regards.

Explanation:

Silly Putty -

It is an inorganic polymer , in which the molecules are linked via covalent bonding . And the bonds can easily be broken , by the application of even a small amount of pressure .

The composition of silly putty is , the Elmer's glue solution and the sodium borate in the ration , 4 : 1 .

silly putty is polar , hydrophilic and binding in nature .

Candle Wax -

The Wax of the candle is an ester of the ethylene glycol and fatty acids , and the chemical properties of was is , it is hydrophobic in nature and is insoluble in water as well .

The principles affecting the properties of silly putty and candle wax involve their hydrophobic, nonpolar nature, making them insoluble in water and more soluble in nonpolar solvents like heptane. Solubility depends on the balance of intermolecular forces and molecule polarity, with polar and ionic substances being hydrophilic. Heat transfer during solvation is dictated by the relative strength of solute-solvent interactions compared to original solute-solute and solvent-solvent interactions.

The chemistry principle for substances such as silly putty and candle wax as it relates to hydrophilic/hydrophobic, polar/nonpolar, and soluble/non-soluble properties is rooted in their intermolecular forces and molecular structure. Silly putty, which is a silicone polymer, is primarily hydrophobic and non-polar, making it insoluble in water and more soluble in nonpolar solvents. On the other hand, candle wax, typically composed of long-chain hydrocarbons, is also nonpolar and hydrophobic, displaying similar solubility characteristics. Things that are polar and can dissolve in water are referred to as being hydrophilic, which allows them to form hydrogen bonds with water. This property is not present in nonpolar substances like silly putty and candle wax, which resist mixing with water.

In terms of solubility predictions for various substances:
Vegetable oil is nonpolar and more soluble in a nonpolar solvent like heptane.

Isopropyl alcohol is polar due to the presence of an -OH group and more soluble in water.

Potassium bromide is ionic and highly soluble in water which is a polar solvent.

When solutions form, heat may be released or absorbed depending on the intermolecular forces involved. In solvation where the solute-solvent interactions are stronger than the solute-solute and solvent-solvent interactions, heat is released (exothermic). Conversely, if breaking the original interactions requires more energy than is released during new interaction formation, the process is endothermic and absorbs heat.

Answer:

Ion-ion force between Na+ and Cl− ions

London dispersion force between two hexane molecules

Explanation:

"Ion-dipole force between Na+ ions and a hexane molecule " does not exist since hexane has only non-polar bonds and therefore no dipole.

"Ion-ion force between Na+ and Cl− ions " exists since both are ions.

"Dipole-dipole force between two hexane molecules " does not exist since hexane molecules do not have a dipole.

"Hydrogen bonding between Na+ ions and a hexane molecule " does not exist since the hydrogen in the hydrogen bond must be bonded directly to an electronegative atom, which hexane does not have since it is a hydrocarbon.

"London dispersion force between two hexane molecules" exist since hexane is a molecular compound.

Answer:

Ion-ion force between Na⁺ and Cl⁻ ions.

London dispersion force between two hexane molecules.

Explanation:

Suppose that NaCl is added to hexane (C₆H₁₄) instead of water. Which of the following intermolecular forces will exist in the system?

Check all that apply.

Answer:

66622.653 mi/hr

Explanation:

The average speed at which the Earth rotates around the Sun = 29.783 km/s

Also, The conversion of km to miles is shown below:

1 km = 0.621371 miles

The conversion of s to hr is shown below:

1 s = 1 / 3600 hr

So,

[tex]29.783\ km/s=\frac {29.783\times 0.621371\ miles}{\frac {1}{3600}\ hr}[/tex]

Thus,

The average speed at which the Earth rotates around the Sun in miles per hour =  66622.653 mi/hr

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

[tex]4NH_3+7O_2\rightarrow 4NO_2+6H_2O[/tex]

The expression for rate of reaction for the reactant :

[tex]\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}[/tex]

[tex]\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}[/tex]

The expression for rate of reaction for the product :

[tex]\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}[/tex]

[tex]\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}[/tex]

From this we conclude that, all the options are correct.

Based on the equation of the reaction, all of the given expressions are valid expressions of the reaction rate.

The rate of a chemical reaction is the rate at which reactants are used up or the rate at which products are formed.

The rate of a reaction is obtained from the balanced equation of the reaction.

The equation of the reaction is given below:

Rate of disappearance of NH3 = - 1/4 × Δ[NH3]/Δt

Rate of disappearance of O2 = - 1/7 × Δ[O2]/Δt

Rate of formation of NO2 = 1/4 × Δ[NO2]/Δt

Rate of formation of H2O = 1/6 × Δ[H2O]/Δt

Therefore, all of the above are valid expressions of the reaction rate.

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Answer:

The answer is 916.67 g

Explanation:

48.0 wt% NaOH means that there are 48 g of NaOH in 100 g of solution. With this information and the molecular weight of NaOH (40 g/mol), we can calculate the number of mol there are in 100 g of this solution:

[tex]\frac{48 g NaOH}{100 g solution}[/tex] x [tex]\frac{1 mol NaOH}{40 g}[/tex] = 0.012 mol NaOH/100 g solution

Finally, we need 0.11 mol in 1 liter of solution to obtain a 0.11 M NaOH solution.

0.012 mol NaOH ------------ 100 g solution

0.11 mol NaOH------------------------- X

X= 0.11 mol NaOH x 100 g/ 0.012 mol NaOH= 916.67 g

We have to weigh 916.67 g of 48.0%wt NaOH and dilute it in a final volume of 1 L of water to obtain a 0.11 M NaOH solution.

Answer:

fluid pressure = 132590 pascal

Explanation:

Fluid pressure can be  calculated by using following relation:

[tex]P_o - P_g = g\time p\times (Z_b - Z_a )[/tex]

Where

Pb = fluid pressure at a distance of 3.2 m below from surface

Pa = fluid pressure at surface = atmospheric pressure = 101325 Pascal

g = gravitational constant = 9.8 m/sec2

[tex]\rho = density of fluid = 997 kg/m³[/tex]

Z = location depth from surface of lake

Therefore , [tex]Z_a = 0 meter[/tex]

[tex]Z_b = 3.2 meter[/tex]

[tex]P_b - P_a =g\times\rhop\times (Z_a - Z_b)[/tex]

               [tex]= 9.8 \times 997\times (3.2-0)[/tex]

               [tex]= 31265 kg/m-sec^2 [/tex]

1 Pascal = 1 Kg/(m.Sec)

[tex]P_b = P_a +31265[/tex]

       = 101325 + 132590 Pascal

       =  132590 pascal

Answer:

The atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

Explanation:

Reaction involved: MSO₄ + BaCl₂ → BaSO₄ + MCl₂

Atomic mass (g/mol): oxygen (O)=16, sulphur (S)=32,

Molar mass of SO₄²⁻ = 96 g/mol and molar mass of BaSO₄ = 233.38 g/mol

Let the atomic mass of M be m g/mol.

Therefore, molar mass of MSO₄= (m + 96) g/mol

If 0.1131 g of MSO₄ gives 0.2193 g of BaSO₄ on reaction

Then, (m + 96) g/mol of MSO₄ gives 233.38 g/mol of BaSO₄

Therefore, (m + 96) g/mol of MSO₄ = [(233.38 g/mol) × (0.1131 g)] ÷ (0.2193 g)

⇒(m + 96) g/mol = [26.395] ÷ (0.2193)

⇒(m + 96) g/mol = 120.36 g/mol

⇒m = 120.36 g/mol - 96 g/mol = 24.36 g/mol ≈ 24.3 g/mol

Since, the atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

Answer: Option (c) is the correct answer.

Explanation:

Entropy is defined as the degree of randomness that is present within the particles of a substance.

As [tex]NH_{4}NO_{3}[/tex] is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ([tex]NH^}{+}_{4}[/tex]) and nitrate ions ([tex]NO^{-}_{3}[/tex]).

Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.

Thus, we can conclude that ammonium nitrate ([tex]NH_{4}NO_{3}[/tex]) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.

Answer:

4687.5 mL

Explanation:

Given that the density of the substance = 8.96 g/mL

Mass = 42.0 kg

The conversion of kg into g is shown below as:

1 kg = 1000 g

So, mass = 42.0 ×  1000 g = 42000 g

Volume = ?

So, volume:  

[tex]Volume=\frac {{Mass}}{Density}[/tex]  

[tex]Volume=\frac {42000\ g}{8.96\ g/mL}[/tex]  

The volume of the 42.0 kg of the substance = 4687.5 mL

Answer: The concentration of water at equilibrium is 0.3677 mol/L

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by [tex]K_{c}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The [tex]K_{c}[/tex] is written as:

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

The chemical equation for the conversion of methane to carbon monoxide and hydrogen gas follows:

[tex]CH_4+H_2O\rightleftharpoons 3H_2+CO[/tex]

The [tex]K_{c}[/tex] is represented as:

[tex]K_{c}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]      ....(1)

To calculate the concentration, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

[tex][CO]=\frac{0.145mol}{0.379L}=0.383mol/L[/tex]

[tex][H_2]=\frac{0.218mol}{0.379L}=0.575mol/L[/tex]

[tex][CH_4]=\frac{0.25mol}{0.379L}=0.660mol/L[/tex]

[tex]K_c=0.30[/tex]

Putting values in equation 1, we get:

[tex]0.30=\frac{(0.575)^3\times 0.383}{0.660\times [H_2O]}[/tex]

[tex][H_2O]=\frac{(0.575)^3\times 0.383}{0.660\times 0.30}=0.3677[/tex]

Hence, the concentration of water at equilibrium is 0.3677 mol/L

Answer:

a) p = 1.10 * 10⁻²⁷ kg·m/s

b) p = 9.46 * 10⁻²⁴ kg·m/s

c) p = 3.31 * 10⁻³⁶ kg·m/s

Explanation:

To solve this problem we use the de Broglie's equation, which describes the wavelenght of a photon with its momentum:

λ=h/p

Where λ is the wavelength, h is Planck's constant (6.626 * 10⁻³⁴ J·s), and p is the linear momentum of the photon.

Rearrange the equation in order to solve for p:

p=h/λ

And now we proceed to calculate, keeping in mind the SI units:

a) 600 nm= 600 * 10⁻⁹ m

p=(6.626 * 10⁻³⁴ J·s) / (600*10⁻⁹m) = 1.10 * 10⁻²⁷ kg·m/s

b) 70 pm= 70 * 10⁻¹² m

p=(6.626 * 10⁻³⁴ J·s) / (70*10⁻¹²m) = 9.46 * 10⁻²⁴ kg·m/s

c) 200 m

p=(6.626 * 10⁻³⁴ J·s) / (200m) = 3.31 * 10⁻³⁶ kg·m/s

Answer:

option D= temperature and volume

Explanation:

Definition:

Charle's law stated that ,

" At constant pressure, the volume of given amount of gas is directly proportional to its absolute temperature"

V∝ T

V= kT

OR

V/T = k     (k is proportionality constant)

if the temperature is changed from T1 to T2 then the volume of gas is also changed from V1 to V2. Then expression will be:

V1/T1= k  and   V2/T2=k

V1/T1=V2/T2

suppose a cylinder is filled with a gas having volume V1 at temperature T1. When the gas is heated its temperature raises from T1 to T2 and its volume also increased with increase of temperature from V1 TO V2.

In Charles' law, the two changing properties are temperature and volume, with pressure maintained constant. The law indicates a direct proportionality between temperature and volume for a given amount of gas.

The two changing properties in Charles' law are temperature and volume. Charles' law states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. This means if the temperature of the gas increases, its volume increases as well, provided the pressure and the amount of gas remain constant. In other words, doubling the absolute temperature at constant pressure will double the volume. It is important to note that in gas laws, temperatures must always be expressed in kelvins.

Answer:

Dissolve the impure solid in a minimal amount of boiling solvent, cool the solution to form crystals, vacuum filter the solution to collect the pure crystals.

Explanation:

Recrystallization is a process when a solid with impurities is purified. To do this a solvent of the compound we want must be used. We need to use only the quantity necessary to dissolve the compound of interest, otherwise, the solvent will dissolve the impurities or it will interfere in the crystallization.

For most of the solids, the solubility increases with the increase of the temperature, so to speed up the process, heat must be added at the system, or the solvent must be boiling. Then, the solution will be cooled to form the crystals of the compound purified, and then it must be filtered in a vacuum because the crystals can slow down the filtration.

Answer:

The hydrogen cannot participate in hydrogen bonding with other molecules.

Explanation:

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus, hydrogen bonding is present in alcohols as the oxygen atom is linked to hydrogen

Partially positive end of the hydrogen atom is attracted to partially negative end of the oxygen atom. It is strong force of attraction between the molecules. Thus, hydrogen acquires slight positive charge and become electron deficient or acidic.

Hence, option C is incorrect.

When 5.0 moles of C₄H₁₀ react with an excess of O2, 20.0 moles of CO₂ are formed.

According to the balanced chemical equation:

2C₄H₁₀(g) + 13O₂(g) ==> 8CO₂(g) + 10H₂O(L)

The stoichiometric coefficient of CO₂ is 8. This means that for every 2 moles of C₄H₁₀ reacted, 8 moles of CO₂ are formed. Therefore, when 5.0 moles of C₄H₁₀  react, the number of moles of CO₂ formed can be calculated using the stoichiometric ratio:

Number of moles of CO₂ = (5.0 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀ )

                       = 20.0 moles CO₂

So, when 5.0 moles of C₄H₁₀  react with an excess of O₂, 20.0 moles of CO₂ are formed.

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In the combustion of butane, 2 moles of C4H10 react to produce 8 moles of CO2, making a 1:4 ratio. Therefore, when 5.0 moles of butane reacts it produces 20 moles of CO2 due to this stoichiometric ratio.

In the balanced chemical equation for the combustion of butane, we see that '2 moles of C4H10' react to produce '8 moles of CO2'. This is a ratio of 1:4. So, for each mole of butane that reacts, 4 moles of carbon dioxide are produced.

If in a particular reaction, we have '5.0 moles of C4H10', then according to the stoichiometric ratio (1:4) we can calculate the number of moles of CO2 formed. By multiplying the moles of C4H10 by 4, i.e., 5.0 x 4, we get '20 moles of CO2'. Therefore, when 5.0 moles of butane reacts with excess oxygen, 20 moles of CO2 are produced.

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Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

Answer : The value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

[tex]A+B+C\rightarrow D+E[/tex]

Rate law expression for the reaction:

[tex]\text{Rate}=k[A]^a[B]^b[C]^c[/tex]

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

[tex]1.24\times 10^{-4}=k(0.40)^a(0.40)^b(0.40)^c[/tex] ....(1)

Expression for rate law for second observation:

[tex]3.6\times 10^{-4}=k(0.40)^a(0.40)^b(1.20)^c[/tex] ....(2)

Expression for rate law for third observation:

[tex]4.8\times 10^{-4}=k(0.80)^a(0.40)^b(0.40)^c[/tex] ....(3)

Expression for rate law for fourth observation:

[tex]4.8\times 10^{-4}=k(0.80)^a(0.80)^b(0.40)^c[/tex] ....(4)

Dividing 1 from 2, we get:

[tex]\frac{3.6\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(1.20)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\3=3^c\\c=1[/tex]

Dividing 1 from 3, we get:

[tex]\frac{4.8\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\4=2^a\\a=2[/tex]

Dividing 3 from 4, we get:

[tex]\frac{4.8\times 10^{-4}}{4.8\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.80)^a(0.80)^b(0.40)^c}\\\\1=2^b\\b=0[/tex]

Thus, the rate law becomes:

[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

[tex]1.24\times 10^{-4}=k(0.40)^2(0.40)^0(0.40)^1[/tex]

[tex]k=1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Hence, the value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Explanation:

The given data is as follows.

       P = 28.3 mm = [tex]\frac{28.3}{760}[/tex],            V = 600 mL = [tex]600 ml \frac{0.001 L}{1 ml}[/tex] = 0.6 L

      R = 0.082 L atm/mol K,                 T = (28 + 273) K = 301 K

Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.

                                    PV = nRT

                 [tex]\frac{28.3}{760} \times 0.6 L[/tex] = [tex]n \times 0.082 L atm/mol K \times 301 K[/tex]            

                       n = [tex]9.03 \times 10^{-4}[/tex] mol  

As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)

               No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

        [tex]9.03 \times 10^{-4}[/tex] mol = [tex]\frac{mass}{18 g/mol}[/tex]

                            mass = [tex]162.5 \times 10^{-4}g[/tex]

                                      = [tex]163 \times 10^{-4}g[/tex] (approx)

Since, 1 g = 1000 mg. Therefore, [tex]163 \times 10^{-4}g[/tex] will be equal to  [tex]163 \times 10^{-4}g \times \frac{10^{3}mg}{1 g}[/tex]

                          = 16.3 mg

Thus, we can conclude that mass of water vapor present is 16.3 mg.

To find the mass of water vapor in a given vapor volume at a certain temperature, you can use the ideal gas law. Convert the temperature from Celsius to Kelvin, calculate the number of moles using the ideal gas law equation, and then convert moles to grams using the molar mass of water. Finally, convert grams to milligrams. The mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

To find the mass of water vapor, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. Then, we can rearrange the equation to solve for n, the number of moles. Once we have the moles, we can use the molar mass of water to convert it to mass in grams. Finally, we can convert grams to milligrams by multiplying by 1000.

Given:

Vapor pressure of water = 28.3 mm Hg
Temperature = 28 °C
Vapor volume = 600 mL

Converting the temperature to Kelvin:

T = 28 °C + 273.15 = 301.15 K

Using the ideal gas law equation to find the number of moles (n):

n = PV / RT

Substituting the values:

n = (28.3 mm Hg * 600 mL) / (62.36 L mm/mol K * 301.15 K)

n = 0.2829 mol

Using the molar mass of water (18.015 g/mol) to find the mass:

mass = n * molar mass

mass = 0.2829 mol * 18.015 g/mol

mass = 5.097 g

Converting grams to milligrams:

mass = 5.097 g * 1000 mg/g

mass = 5097 mg

Therefore, the mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

Answer:

30.73 Doses and 39.42 g

Explanation:

It is necessary to know the conversion factors in that manner change the initial 500 mL to the number of doses as follow:

[tex]500 ml * \frac{1tsp}{4.93mL} * \frac{1TBS}{3tsp} * \frac{1 Dose }{1.1 TBS} = 30.7 doses\\[/tex]

And the for b. it is such necessary to take in account that 1 mL = 1 [tex]cm^{3}[/tex]

35.2 mL * 1.12 [tex]\frac{g}{cm^{3} }[/tex] = 39.42 g

Answer:

Part A

[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]

Part B

[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]

Explanation:

Part A

The rate law is the equation that relates the rate of the reaction, the kinetic constant and the concentration of the reactant or reactants.

For the given chemical reaction we can write a general expression for the rate law as follows:

[tex]rate= k * [N2O5]^{x}[/tex]

where k is the rate constant and x is the order of the reaction with respect of N2O5 concentration. Particularly, a first order reaction kinetics indicate that the rate of the reaction is directly proportional to the concentration of only one reactant. Then x must be 1.

Replacing the value of the rate constant given in the text we can arrive to the following expression for the rate law:

[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]

Part B

Replacing the value of the concentration of N2O5 given, we can get the rate of reaction:

[tex]rate= 4.82*10^{-3}s^{-1} *2.80*10^{-2}M[/tex]

[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]

Answer:

There will be 13107200 number of bacteria after 12 hours

Explanation:

1 hour = 60 minutes

So, 12 hours = ([tex]12\times 60[/tex]) minutes = 720 minutes

Initially there are 50 bacteria.

As number of bacteria doubles in every 40 minutes therefore rate of increase in number of bacteria will be similar to a first order reaction.

Hence, number of bacteria after 720 minutes = [tex]50\times (2)^{\frac{720}{40}}[/tex] = 13107200

Answer:

0.002 %

Explanation:

Given that:

[tex]K_{a}=4.0\times 10^{-10}[/tex]

Concentration = 1.0 M

Consider the ICE take for the dissociation of Hydrocyanic acid as:

                                      HCN    ⇄     H⁺ +        CN⁻

At t=0                            1.0                -              -

At t =equilibrium        (1.0-x)                x           x            

The expression for dissociation constant of Hydrocyanic acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}[/tex]

[tex]4.0\times 10^{-10}=\frac {x^2}{1.0-x}[/tex]

x is very small, so (1.0 - x) ≅ 1.0

Solving for x, we get:

x = 2×10⁻⁵  M

Percentage ionization = [tex]\frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%[/tex]

Answer:

The mole fraction composition of the liquid is :

Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.

Explanation:

Mass of the liquid mixture = 200 g

Percentage of butane = 30%

Mass of butane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]

Moles of butane = [tex]n_1=\frac{60 g}{58 g/mol}=1.0345 mol[/tex]

Percentage of pentane= 40%

Mass of pentane= [tex]\frac{40}{100}\times 200 g=80 g[/tex]

Moles of pentane= [tex]n_2=\frac{80 g}{58 g/mol}=1.1111 mol[/tex]

Percentage of hexane = 100% - 30% - 40% = 30%

Mass of hexane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]

Moles of hexane = [tex]n_2=\frac{60 g}{86 g/mol}=0.6977 mol[/tex]

Mole fraction of butane, pentane and hexane : [tex]\chi_1, \chi_2 \& \chi_3[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908[/tex]

[tex]\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454[/tex]

Answer:

[tex]-r_{O_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

Explanation:

The decomposition of [tex]NO_2[/tex] follows the equation  

[tex]2NO_2\rightarrow2NO+O_2[/tex]

By definition, the rate of a chemical reaction can be expressed by  

[tex]-r_{NO_2}=\frac{d\left[NO_2\right]}{dt}=\frac{0.010-0.008\ M}{100\ s}=2\times{10}^{-5}\frac{M}{s}[/tex]

The rate of appearance of [tex]O_2[/tex] is related to the rate of disappearance of [tex]NO_2[/tex] by the stoichiometry. This means that, for each mole of [tex]O_2[/tex] that appears 2 moles of [tex]NO_2[/tex]  are consumed. So  

[tex]-r_{O_2}=-r_{NO_2}\times\frac{1\ mole\ \ O_2}{2\ mole\ \ {\rm NO}_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

The rate of appearance of O2 for this period is -1.0x10^-5 M/s.

The rate of appearance of O2 for a given period can be determined using the change in concentration of NO2 over that period. In this case, the concentration of NO2 drops from 0.0100 to 0.00800 M in 100 s. First, calculate the change in concentration of NO2:

Δ[NO2] = [NO2]final - [NO2]initial = 0.00800 M - 0.0100 M = -0.00200 M

Since the reaction is 2NO2 → 2NO + O2, the stoichiometric ratio between NO2 and O2 is 2:1. Therefore, the change in concentration of O2 is half of the change in concentration of NO2:

Δ[O2] = -0.00200 M ÷ 2 = -0.00100 M

Finally, divide the change in concentration of O2 by the time period to find the rate of appearance of O2:

Rate = Δ[O2] ÷ time = -0.00100 M ÷ 100 s = -1.0x10-5 M/s

Explanation:

Atomic number of P = 15

Common oxidation state of P = -3, +3 and +5

Electronic configuration of P^3+: [tex]1s^2 2s^22p^63s^2[/tex]

Electronic configuration of P^3-: [tex]1s^2 2s^22p^63s^23p^6[/tex]

Electronic configuration of P^5+: [tex]1s^2 2s^22p^6[/tex]

Atomic number of Ar = 18

Argon has stable octet, so it generally does not exist as ions.

Electronic configuration of Ar: [tex]1s^2 2s^22p^63s^23p^6[/tex]

Atomic number of Te = 52

Common oxidation states of Te = -2, +2, +4, +6

Electronic configuration of Te^2-: [tex][Kr]4d^10 5s^2 5p^6[/tex]

Electronic configuration of Te^2+: [tex][Kr]4d^10 5s^2 5p^2[/tex]

Electronic configuration of Te^4+: [tex][Kr]4d^10 5s^2[/tex]

Electronic configuration of Te^6+: [tex][Kr]4d^10[/tex]

Boiling point of HF is more as compared to HCl

Reason:

Molecular weight of HF is low, but is boiling point is high because of presence of hydrogen bonding. Whereas in case of HCl, its molecular weight is high but has only weak van der Waals intercations. As hydrogen bonding is stronger than van der Waals interactions, therefore, boiling point of HF is more.

Answer:

specific volume O2 = 0.5 Kg/m³

molar volume O2 = 8 E-3 m³/mol

Explanation:

specific volume (Sv):

∴ Sv = 1 / ρ

∴ ρ = mass / volume

∴ V = 500 L * ( m³/1000 L) = 0.5 m³

∴ ρ = 1 Kg / 0.5m³ = 2 Kg/m³

⇒ Sv = 1 / 2 Kg/m³ = 0.5 m³/Kg

molar volume ( Vm ):

∴ Vm = volume/mol

∴ Mw O2 = 1000g O2 * ( mol/16g O2) = 62.5 mol O2

⇒ Vm = 0.5 m³ / 62.5 mol

⇒ Vm = 8 E-3 m³/mol