What volume of phenytoin suspension 30 mg/5 mL is required to be added to a suitable diluent to obtain 150 mL phenytoin suspension 20 mg/5 mL? 100

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:

HCI (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

Explanation:

A double displacement reaction is a type of reaction where two reactants exchange ions to form two new compounds. Here we can see that both the acid and the base exchanged an ion each to for salt and water, 2 new products.

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:

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

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:

(a) 17,178 mg/m3

(b) 11,625 mg/m3

Explanation:

The concentration of CO in mg/m3 can be calculated as

[tex]C (mg/m3) =(P/RT)*MW*C_{ppm}[/tex]

For standard conditions (1 atm and 25°C), P/RT is 0.0409.

Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.

The molecular weigth of CO is 28 g/mol.

(1) For 25°C and 1 atm conditions

[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0409*28*15,000=17,178[/tex]

(b) For 200°C and 1.1 atm,

[tex]P/RT=0.0409*(P/P_{std})*(T_{std}/T)\\P/RT=0.0409*(1.1atm/1atm)*(273+15K/273+200K)=0.0277[/tex]

Then the concentration in mg/m3 is

[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0277*28*15,000=11,625[/tex]

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: The wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]

Explanation:

The relationship between wavelength of light and frequency of light is given by the equation:

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency of radiation = [tex]6.912\times 10^{-14}s^{-1}[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] = wavelength of radiation = ?

Putting values in above equation, we get:

[tex]6.912\times 10^{-14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}\\\\\lambda=4.34\times 10^{21}m[/tex]

Hence, the wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]

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:

[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

Answer:

0.1933 M

Explanation:

Data provided in the question:

Number of Moles of Na₂SO₄ initially taken = 0.145 mol

Volume of solution, V = 750 mL = 0.750 L

Now, the molarity is given as:

Molarity = [tex]\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}[/tex]

or

Molarity = [tex]\frac{0.145}{0.750}[/tex]

or

Molarity = 0.1933 M

The molarity of the solution is 0.1933 M.

To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution. In this case, we have 0.145 mol of Na2SO4 and a volume of 750 mL. First, convert the volume to liters by dividing by 1000: 750 mL = 0.75 L. Now, calculate the molarity using the formula: Molarity (M) = moles of solute / volume of solution in liters.

So, Molarity = 0.145 mol / 0.75 L = 0.1933 M

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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:

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 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]

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: 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:

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: The cell potential of the cell is 0.31 V

Explanation:

We know that:

[tex]E^o_{(Br_2/2Br^-)}=1.07V\\E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, [tex]MnO_4^-[/tex] will undergo reduction reaction will get reduced. And, bromine will get oxidized.

Oxidation half reaction: [tex]2Br^-(aq.)+2e^-\rightarrow Br_2(l);E^o_{(Br_2^/2Br^-)}=1.07V[/tex]      ( ×  5)

Reduction half reaction: [tex]MnO_4^-(aq.)+8H^+(aq.)+5e^-\rightarrow Mn^{2+}(aq.)+4H_2O(l);E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]  ( ×  2)

The net reaction follows:

[tex]2MnO_4^-(aq.)+10Br^-(aq.)+16H^+(aq.)\rightarrow 2Mn^{2+}(aq.)+5Br_2(l)+8H_2O(l)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=1.51-(1.07)=0.44V[/tex]

To calculate the EMF of the cell, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]^2}{[MnO_4^{-}]^2\times [Br^-]^{10}\times [H^+]^{16}}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = ?V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.44 V

n = number of electrons exchanged = 10

[tex][H^{+}]=1M[/tex]

[tex][Mn^{2+}]=0.15M[/tex]

[tex][MnO_4^{-}]=0.01M[/tex]

[tex][Br^{-}]=0.01M[/tex]

Putting values in above equation, we get:

[tex]E_{cell}=0.44-\frac{0.059}{10}\times \log(\frac{(0.15)^2}{(0.01)^2\times (0.01)^{10}\times (1)^{16}})\\\\E_{cell}=0.31V[/tex]

Hence, the cell potential of the cell is 0.31 V

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.

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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Explanation:

Actual value = 102.0 mg/dL

Calculation of absolute error and relative error for the measured value, 98.4 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 98.4 | = 3.6

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{3.6}{102.0} = 0.03529[/tex]

Calculation of absolute error and relative error for the measured value, 104.3 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 104.3 | = 2.3

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{2.3}{102.0} = 0.02255[/tex]

Calculation of absolute error and relative error for the measured value, 97.4 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 97.4 | = 4.6

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{4.6}{102.0} = 0.04509[/tex]

Calculation of absolute error and relative error for the measured value, 106.7 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 106.7 | = 4.7

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{4.7}{102.0} = 0.04607[/tex]

Calculation of absolute error and relative error for the measured value, 93.0 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 93 | = 9

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{9}{102.0} = 0.08823[/tex]

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:

The answer is the option "A"

Explanation:

The reaction rate is the amount of substance formed or transformed per unit of time. The catalytic action allows to increase the reaction rate or the selectivity. The units are moles / time (mol / s)

mol = moles

s = seconds

So the correct answer is "A"

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:

No, it doesn't.

Explanation:

To convert ppb to mg/m³ we first need to convert to ppm, by just divide the amount by 1,000, so the concentration in the sample is 4.8 ppm.

mg/m³ = (ppm x molar mass)/molar volume

Using the molar mass in gram and the molar volume in liters, multiplying by the parts per million, we will get the concentration in mg/m³.

Molar mass of C = 12 g/mol; molar mass of Cl = 35.5 g/mol

Molar mass of CCl4 = 12 + 4x35.5 = 154 g/mol

Assuming, 25ºC and 1 atm, the molar volume of an ideal gas is 24.45 L, so:

mg/m³ = (4.8 x 154)/24.45

mg/m³ = 30.2

Which is higher than the limit of 12.6 mg/m³

Answer:

Sample is greater than the standard (32.96 mg/m³ > 12.6 mg/m³)

Explanation:

Please look at the solution in the attached Word file

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 final pH is 7

Explanation:

KOH + HCl ------> KCl + H2O

KOH ----> K+  + OH-

0,3 M        0,3 M  +  0,3 M

HCl -----> H+  +  Cl-

0,2 M         0,2M   +  0,2M

In both cases they are strong acid and base because they dissociate completely; The initial molarity is the same, at the end of the dissociation.

As we have an acid next to a strong base, the final solution will have a neutralization reaction where the OH- of the base react with the H + of the acid to give water. The excess of OH- or H + defines whether the solution will have an acid or alkaline pH.

Molarity = moles / volume (L)

Moles of OH- = 0,3 M x 0,020 L = 0,006 moles

(remember volume in L, so 20mL = 0,020 L)

Moles of H+ = 0,2 M x 0,030 L = 0,006 moles

As we have the same quantity for OH- and H+, the solution will have neutral pH.

Answer:

a) HCN  - hybridization sp

b) C(CH₃)₄ - hybridization sp³

c) H₃O⁺  - hybridization sp³

d) - CH₃ - hybridization sp³

Explanation:

Hybridization occurs to allow an atom to make more covalent bonds than the original electronic distribution would allow or to allocate ligands in an energetically stable geometry.

Carbon can have thre hybridization states: sp³ , sp² and sp.

Oxygen usualluy has an sp³ hybridization.

In order to determine the hybridization, we need to consider the number of atoms attached to the central atom and the number of lone pairs.

The figure attached shows the species and the hybridization of their central atoms.

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:

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]

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