When preparing 1-bromobutane from 1-butanol and HBr, why
mustit be washed with H2O, cold NaOH and NaCl?

Explanation:

Propanol is [tex]CH_3CH_2CH_2OH[/tex] , whereas propane is [tex]CH_3CH_2CH_3[/tex].

There is a presence of hydroxyl group in propanol which leads to the formation of hydrogen bonds with water and thus become soluble in water. Such bonding do not exists in propane.

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.

Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.

Final answer:

Propanol is more soluble in water than propane due to the presence of the hydroxyl (OH) group in propanol. The hydroxyl group allows propanol to engage in hydrogen bonding with water molecules, making it more soluble.

Explanation:

Propanol is more soluble in water than propane due to the presence of the hydroxyl (OH) group in propanol. The hydroxyl group allows propanol to engage in hydrogen bonding with water molecules, making it more soluble. On the other hand, propane lacks a hydroxyl group and therefore cannot form hydrogen bonds with water, resulting in lower solubility in water.

Answer:

a) 51.00000000 c) 51.32980000 d) 51.32975000

Explanation:

your welcome :)

Explanation:

Neutralization reaction -

The reaction of an acid and base to yield a salt and water , is a type of neutralization reaction .

The reaction of  potassium hydroxide and phosphoric acid  is a type of neutralization reaction ,

Hence , the reaction is as follows -

KOH (aq) + H₃PO₄ (aq) ----> K₃PO₄ (aq) + 3H₂O (l)

The reaction after balancing the atoms on the reactant side and on the product side is -

3 KOH (aq) + H₃PO₄ (aq) ----> K₃PO₄ (aq) + 3H₂O (l)

Answer: 15.850

Explanation:

The conversion used from liters to gallons is:

1 L = 0.264172 gallon

The conversion used from sec to min is:

60 sec = 1 min

1 sec =[tex]\frac{1}{60}\times 1=0.017min[/tex]

We are asked: liters/sec = gallons/min

[tex]liters/sec=\frac{0.264172}{0.017}=15.850gallons/min[/tex]

Therefore, to convert from liters/second to gallons/minute, multiply the number of liters/second by 15.850.

Answer:

The runner should run 51 minutes.

Explanation:

Distance wished by runner to cover = d = 10.4 km

Time taken by the runner to cover 10.4 km = T

Speed of the runner  = 7.6 mile/hour

1 mile = 1.60934 km

1 hour = 60 min

[tex]7.6 mile/Hour=\frac{7.6\times 1.60934 km}{1\times 60 min}=0.2038 km/min[/tex]

[tex]Speed=\frac{Distance}{Time}[/tex]

[tex]T=\frac{10.4 km}{0.2038 km/min}=51.0179 min\approx 51 minutes[/tex]

The runner should run 51 minutes.

Answer and Explanation:

Interstitial diffusion allows the purity atom like (C, O, H) to occupy the interstitial sites in the atom which are abundantly present as a result of weaker bond formation with the surrounding atom.

The interstitial diffusion of iron in Body Centered Cubic (BCC) crystal lattice is faster than Face Centered Cubic (FCC) and the reason being that BCC structure is more open with more interstitial spaces than FCC structure.

Also the packing fraction of BCC lattice is less than that of FCC lattice.

For BCC is 0.68 whereas for FCC, it is 0.74

Answer: mole fraction of methanol = 0.590

mole fraction of ethanol = 0.410

Explanation:

We are given:

Equal masses of methanol [tex]CH_4O[/tex] and ethanol [tex]C_2H_6O[/tex] are mixed.

let the mass be x g.

Calculating the moles of methanol in the solution, by using the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{32.04g/mol}[/tex]

Calculating the moles of ethanol in the solution, by using the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}[/tex]

To calculate the mole fraction of methanol, we use the equation:

[tex]\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}[/tex]

[tex]\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590[/tex]

To calculate the mole fraction of ethanol, we use the equation:

[tex]\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}[/tex]

[tex]\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410[/tex]

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

To compute the mole fractions, first calculate the number of moles of each component by dividing mass by molar mass. The mole fraction is then calculated as each component's moles divided by the total moles. The mole fractions of methanol and ethanol are approximately 0.591 and 0.409 respectively.

To determine the mole fraction of methanol (CH4O) and ethanol (C2H6O) in a solution where equal masses of both are mixed, we first need to calculate the number of moles of each component. The number of moles is determined by the formula: moles = mass/molar mass. The molar mass of methanol is approximately 32.04 g/mol, and the molar mass of ethanol is approximately 46.07 g/mol.

Let's say that we've mixed 50g of each component. The moles of methanol would be 50g/32.04 g/mol which is approximately 1.56 moles. For ethanol, it would be 50g/46.07g/mol, approximately 1.08 moles.

Next, we calculate the total moles in the solution, which is the sum of moles of methanol and ethanol, or 1.56 moles methanol + 1.08 moles ethanol = 2.64 moles.

The mole fraction is then calculated as the number of moles of each component divided by the total moles. So, the mole fraction of methanol would be 1.56 moles methanol / 2.64 total moles = 0.591 to three significant figures. For ethanol, the mole fraction would be 1.08 moles ethanol / 2.64 total moles = 0.409 to three significant figures.

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Answer: The mass of alcohol is 790 grams.

Explanation:

To calculate the mass of alcohol, we use the equation:

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

Volume of alcohol = 1.0 L = 1000 mL    (Conversion factor:  1 L = 1000 mL)

Density of alcohol = 0.79 g/mL

Putting values in above equation, we get:

[tex]0.79g/mL=\frac{\text{Mass of alcohol}}{1000mL}\\\\\text{Mass of alcohol}=790g[/tex]

Hence, the mass of alcohol is 790 grams.

Final answer:

The mass of 1.0 L of alcohol is 790 grams.

Explanation:

To calculate the mass in grams of 1.0 L of alcohol, you can use the density of alcohol.

The given density is 0.79 g/mL.

This means that for every 1 mL of alcohol, there is a mass of 0.79 grams.

Since 1 L is equal to 1000 mL, you can multiply the density by the volume to find the mass: 0.79 g/mL × 1000 mL = 790 grams of alcohol.

The concept at hand is Dalton's Law of Partial Pressures, stating that the total pressure of a mixture of gases is the sum of their individual partial pressures. Knowing the total pressure and the partial pressure of He, you can calculate the partial pressure of Ne as 0.19 atm.

The question revolves around the concept known as Dalton's Law of Partial Pressures. In a mixture of non-reactive gasses, Dalton's Law states that the total pressure exerted by the mixture of gases is the sum of the partial pressures of each component gas. In simpler terms, total pressure is the sum of the pressures of each individual gas.

If the total pressure of a gas mixture consisting of He (Helium) and Ne (Neon) is given as 0.51 atm, and we know the partial pressure of He is 0.32 atm, then we can calculate the partial pressure of Ne. Since the partial pressure of any given gas in a mixture is part of the total pressure, subtracting the known He partial pressure from the total will provide the Ne partial pressure:

Total pressure - He partial pressure = Ne partial pressure

So, 0.51 atm (total) - 0.32 atm (He) = 0.19 atm (Ne).

Hence, the partial pressure of the Ne in the mixture is 0.19 atm.

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Answer:

The correct option is: conjugate acid

Explanation:

Bases are the chemical substances that are proton acceptors and electron-pair donors. Therefore, when a base accepts a hydrogen proton from an acid it forms a conjugate acid.

Therefore, conjugate acids are formed when a base accepts a hydrogen proton from an acid. Also, if the base is a weak base then its conjugate acid is strong. Whereas, if the base is a strong base then the conjugate acid is weak.

A species that forms when a base gains a proton is called a conjugate acid. This concept is a part of acid-base reactions involving conjugate acid-base pairs, where the strength of conjugates is inversely related to their parent compounds.

When a base accepts a proton (H+), it becomes a conjugate acid. This process can be observed in acid-base reactions where a conjugate acid-base pair is formed; the base gains a proton to become the conjugate acid. Conversely, when an acid donates a proton, what remains is known as the conjugate base of that acid. The strength of these conjugate species is inversely related to their parent compounds. Notably, a conjugate base arises from a weak acid, and as such, may exhibit significant strength as a base itself, such as the acetate ion which is capable of accepting a hydrogen ion from water to produce acetic acid and a hydroxide ion.

Explanation:

Fischer Projections allow to represent the three dimensional molecular structures in two dimensional environment without the change in the properties or the structural integrity of the compound. It consists of horizontal as well as vertical lines both, where horizontal lines represent atoms which are pointed toward viewer while vertical line represents atoms which are pointed away from viewer. The point of the intersection between horizontal and vertical lines represents central carbon.

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day [tex]1 MGD = 3785411.8 litre/day = 3785411.8 L/d[/tex]

[tex]F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d [/tex]

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

[tex]C_f = \frac{F1*C1 +F2*C2}{F1 +F2}[/tex]

Replacing every value in L/d and mg/L

[tex]C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L[/tex]

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration [tex]C_f[/tex]. It is required to calculate per day, let's take a time of t = 1 day.  

[tex]F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg[/tex]

After that, mg are converted to pounds.  

[tex]M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb[/tex]

b) A total of 137.6 pounds pass a given spot downstream per day.

Answer:

PART A 1st order in A and 0th order in B

Part B The reaction rate increases

Explanation:

PART A

The rate law of the arbitrary chemical reaction is given by

[tex]-r_A=k\times\left[A\right]^\alpha\times\left[B\right]^\beta\bigm[/tex]

Replacing for the data

Expression 1 [tex]0.00639=k\times{0.12}^\alpha\times{0.22}^\beta[/tex]

Expression 2 [tex]0.01280=k\times{0.24}^\alpha\times{0.22}^\beta[/tex]

Expression 3 [tex]0.00639=k\times{0.12}^\alpha\times{0.11}^\beta[/tex]

Making the quotient between the fist two expressions

[tex]\frac{0.00639}{0.01280}=\left(\frac{0.12}{0.24}\right)^\alpha[/tex]

Then the expression for [tex]\alpha[/tex]

[tex]\alpha=\frac{ln\frac{0.00639}{0.01280}}{ln\frac{0.12}{0.24}}=1\bigm[/tex]

Doing the same between the expressions 1 and 3  

[tex]\frac{0.00639}{0.00639}=\left(\frac{0.22}{0.11}\right)^\beta[/tex]

Then

[tex]\beta=\frac{ln\frac{0.00639}{0.00639}}{ln\frac{0.22}{0.11}}=0\bigm[/tex]

This means that the reaction is 1st order respect to A and 0th order respect to B .

PART B

By the molecular kinetics theory, if an increment in the temperature occurs, the molecules will have greater kinetic energy and, consequently, will move faster. Thus, the possibility of colliding with another molecule increases. These collisions are necessary for the reaction. Therefore, an increase in temperature necessarily produces an increase in the reaction rate.

To determine the amount of sodium ions [tex]\rm (Na^+)[/tex]in the final solution, we need to take into account the contribution of sodium ions from both sodium chloride (NaCl) and sodium sulfate ([tex]\rm Na_2SO_4[/tex]). The volumes can be considered as individual contributions and then added together because they are additive.

Sodium chloride (NaCl):

Given volume of NaCl solution = 3.58 ml

Molarity of NaCl = 0.288 M

Number of moles of NaCl = Molarity × Volume (in liters)

So, the number of moles of NaCl = 0.288 M × (3.58 / 1000) L = 0.00082704 moles

The amount of sodium ions from NaCl is also 0.00082704 moles because each molecule of sodium chloride dissociates into one sodium ion ([tex]\rm Na^+[/tex]).

Sodium sulfate (Na₂SO₄):

Given volume of Na₂SO₄ solution = 500 ml

Molarity of Na₂SO₄ = 6.51 * 1/1000 M = 0.00651 M

Number of moles of Na₂SO₄ = Molarity × Volume (in liters)

So, the number of moles of Na₂SO₄ = 0.00651 M × (500 / 1000) L = 0.003255 moles

Since [tex]\rm Na_2SO_4[/tex] gives 2 sodium ions ([tex]\rm Na^+[/tex]) per formula unit, the amount of sodium ions from [tex]\rm Na_2SO_4[/tex] is 2 * 0.003255 = 0.00651 moles.

Adding the moles of sodium ions from both sources:

Total moles of sodium ions = Moles from NaCl + Moles from Na₂SO₄

Total moles of sodium ions = 0.00082704 + 0.00651 = 0.00733704 moles

So, the molarity of sodium ions in the final solution:

Molarity of sodium ions = Total moles of sodium ions / Total volume (in liters)

Molarity of sodium ions = 0.00733704 moles / (500 / 1000 + 3.58 / 1000) L ≈ 0.0162 M

Therefore, the molarity of sodium ions in the final solution is approximately 0.0162 M.

Learn more about molarity, here:

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The molarity of sodium ions in the solution made by mixing 3.58 mL of 0.288 M sodium chloride and 500 mL of 6.51/1000 M sodium sulfate is 0.00851 M.

To calculate the molarity of sodium ions in the mixture, we need first to calculate the moles of sodium in each solution and then sum those up. This can then be divided by the total volume of the mixture in liters.

For sodium chloride:

moles = Molarity * volume = 0.288 mol/L * 3.58 mL * (1 L/1000 mL) = 0.00103 mol

For sodium sulfate:

moles = Molarity * volume = 6.51/1000 mol/L * 500 mL * (1 L/1000 mL) = 0.00326 mol

Total moles of sodium ions = moles from sodium chloride + moles from sodium sulfate = 0.00103 mol + 0.00326 mol = 0.00429 mol

Total volume = volume of sodium chloride + volume of sodium sulfate = 3.58 mL + 500 mL = 503.58 mL = 0.50358 L

Therefore, molarity = moles / volume = 0.00429 mol / 0.50358 L = 0.00851 M

So, the molarity of sodium ions in the solution is 0.00851 M.

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Answer:

9.23 μm = 0.000363 in

Explanation:

In order to convert 9.23 μm into inches, we need to keep in mind two conversion factors:

Now we proceed to calculate, keeping in the denominator the unit we want to convert, and in the numerator the unit that we wish to convert to:

[tex]9.23um*\frac{1cm}{10000um}*\frac{1in}{2.54cm}  =0.0363 in[/tex]

Thus 9.23 μm are equal to 0.000363 inches.

Answer:

Degree of ionization = 0.0377

pH of the solution = 1.769

Explanation:

Initial concentration of HF = 0.45 M

[tex]K_a = 6.8 \times 10^{-4}[/tex]

                     [tex]HF \leftrightharpoons  H^+ + F^-[/tex]

Initial        0.45                             0          0

At equi      0.45 - x                      x           x

Equilibrium constant = [tex]\frac{[H^+][F^-]}{HF}[/tex]

                   [tex]6.8 \times 10^{-4}= \frac{[x][x]}{0.45 - x}[/tex]

           [tex]x^2 + 6.8 \times 10^{-4} x -  6.8 \times 10^{-4} \times 4.5 = 0[/tex]

x = 0.017 M

x = Cα

α = Degree of ionization

C = Concentration

Degree of ionization = [tex]\frac{0.017}{0.45} = 0.0377[/tex]

[tex]pH = -log[H^+][/tex]

[H^+]=0.017 M

[tex]pH = -log[0.017][/tex]

             = 1.769

Final answer:

The acid dissociation reaction for hydrofluoric acid in water is [tex]HF (aq) + H_2O (l)[/tex] ⇌ [tex]H_3O^+ (aq) + F- (aq)[/tex]. The concentration of a 0.1 L solution containing 0.05 g of HF is 0.025 M. To find the pH for such a solution using the given Ka, the ICE table method can be utilized.

Explanation:

To answer your questions regarding hydrofluoric acid (HF) and its properties, we can proceed as follows:

a) Write out the acid dissociation reaction for hydrofluoric acid. Label the conjugate acid/base pairs.

Hydrofluoric acid dissociates in water as follows:

[tex]HF (aq) + H_2O (l)[/tex] ⇌ [tex]H_3O^+ (aq) + F- (aq)[/tex]

In this reaction, HF is the conjugate acid and F- is the conjugate base.

b) What is the concentration (M) of a solution containing 0.05 g of HF in 0.1 L H2O?

The molecular weight of HF is approximately 20.01 g/mol. To find the molarity, first convert grams to moles:

0.05 g HF × (1 mol HF/20.01 g HF) = 0.0025 mol HF

Then, divide the moles of HF by the volume of the solution in liters:

0.0025 mol HF / 0.1 L = 0.025 M

c) Using the given Ka value, calculate the pH of the solution from part b

Since HF is a weak acid, and given that Ka = 7.2 × [tex]10^-^4[/tex], you can use the ICE table method to find the concentration of H3O+ and then calculate the pH.

Answer:

All description is given in explanation.

Explanation:

Van der Waals forces:

It is the general term used to describe the attraction or repulsion between the molecules. Vander waals force consist of two types of forces:

1.  London dispersion forces

2.  Dipole-dipole forces

1. London dispersion forces:

These are the weakest intermolecular forces. These are the temporary because when the electrons of atoms come close together they create temporary dipole, one end of an atom where the electronic density is high is create negative pole while the other becomes positive . These forces are also called induce dipole- induce dipole interaction.

2.  Dipole-dipole forces:

These are attractive forces , present between the molecules that are permanently polar. They are present between the positive end of one polar molecules and the negative end of the other polar molecule.

Hydrogen bonding:

It is the electrostatic attraction present between the atoms which are chemically bonded. The one atom is hydrogen while the other electronegative atoms are oxygen, nitrogen or flourine. This is weaker than covalent and ionic bond.

Ionic bond or electrostatic attraction:

It is the electrostatic attraction present between the oppositely charged ions. This is formed when an atom loses its electron and create positive charge and other atom accept its electron and create negative charge.

Hydrophobic interaction:

It is the interaction between the water and hydrophobic material. The hydrophobic materials are long chain carbon containing compound. These or insoluble in water.

Covalent bond:

These compounds are formed by the sharing of electrons between the atoms of same elements are between the different element's atoms. The covalent bond is less stronger than ionic bond so require less energy to break as compared to the energy require to break the ionic bond.

Answer:

mass flow = 64.9086 Lbm/min

Explanation:

SG = 0.700 = ρ gas / ρ H2O

∴ ρ H2O = 1000 Kg/m³

⇒ ρ gas = 700 Kg/m³

∴ F auto tank = 20.0 gal / 1.8 min = 11.11 gal/min * ( m³/264.172 gal)

⇒ F auto = 0.042 m³/min

⇒ mass flow = 0.042 m³/min * (700 Kg/m³) * ( 2.20462 Lbm/ Kg )

⇒ mass flow =  64.9086 Lbm/min

Answer:

3,8×10⁻⁵ mol/L of potassium permanganate solution

Explanation:

To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.

3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.

100 mL × ( 1L / 1000 mL) = 0,100 L

Thus, concentration in mol/L is:

3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution

I hope it helps!

Final answer:

The concentration of the potassium permanganate solution is 3.8 x 10⁻⁵ M when rounded to two significant digits.

Explanation:

The concentration of a solution is calculated by dividing the number of moles of the solute by the volume of the solution in liters. To calculate the concentration of potassium permanganate (KMnO₄) in the chemist's solution, you need to use the equation:

C = n / V

where C is the molarity (concentration) in moles per liter (mol/L), n is the number of moles of KMnO₄, and V is the volume of the solution in liters.

In this case, the student already has 3.8 μmol (or 3.8 x 10⁻⁶ mol) of KMnO₄ and the total volume is 100 mL, which is equivalent to 0.1 L. Therefore, the molarity (C) of the solution is:

C = 3.8 x 10⁻⁶ mol / 0.1 L = 3.8 x 10⁻⁵ M

Thus, the concentration of the potassium permanganate solution is 3.8 x 10⁻⁵ M, which can be rounded to two significant digits as 3.8 x 10⁻⁵ M.

Burning 1 kg of coal with a 50% carbon content produces 1.833 kg of CO2 after performing a stoichiometric calculation based on the molar masses of carbon and CO2.

To calculate the amount of CO2 released when 1 kg of coal is burned, we first consider that the coal is 50% carbon by mass. This means that 0.5 kg (or 500 g) of that coal is carbon. Using the stoichiometry of the combustion reaction (C + O2 -> CO2), we find that each 12 grams of carbon (C) will produce 44 grams of CO2 (since the molar mass of carbon is 12 g/mol, and the molar mass of CO2 is 44 g/mol).

Therefore, to find out how much CO2 is produced from the carbon in coal, you can use the ratio (44 g CO2 / 12 g C). Calculating the amount of CO2 produced from 500 g of carbon:

500 g C x (44 g CO2 / 12 g C) = 1833.33 g CO2

Converting this to kilograms:

1833.33 g CO2 x (1 kg / 1000 g) = 1.833 kg CO2

Answer: The chemical formula for magnesium sulfite is [tex]MgSO_3[/tex]

Explanation:

The given compound is formed by the combination of magnesium and sulfite ions. This is an ionic compound.

Magnesium is the 12th element of periodic table having electronic configuration of [tex][Ne]3s^2[/tex].

To form [tex]Mg^{2+}[/tex] ion, this element will loose 2 electrons.

Sulfite ion is a polyatomic ion having chemical formula of [tex]SO_3^{2-}[/tex]

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium sulfite is [tex]MgSO_3[/tex]

Answer:

MgSO3

Explanation:

Answer:

Percentage of the isotope left is 75.87 %.

Explanation:

Initial mass of the isotope = 1 mg

Time taken by the sample, t = [tex]t_{\frac{1}{2}}=123 years[/tex]

Formula used :

[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]

where,

[tex]N_o[/tex] = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope

[tex]\lambda[/tex] = rate constant

[tex]\lambda =\frac{0.693}{123 year}=0.005635 year^{-1}[/tex]

[tex]N=N_o\times e^{-\lambda \times t}[/tex]

Now put all the given values in this formula, we get

[tex]N=1 mg\times e^{-0.005634 year^{-1}\times 49 years}[/tex]

[tex]N=0.7587 mg[/tex]

Percentage of the isotope left:

[tex]\frac{N}{N_o}\times 100[/tex]

=[tex]\frac{0.7587 mg}{1 mg}\times 100[/tex]

Percentage of the isotope left is 75.87 %.

Answer: The number of moles of gas present is 0.276 moles

Explanation:

To calculate the number of moles of gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of the gas = 725 mm Hg

V = Volume of the gas = 7.55 L

n = number of moles of gas = ?

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

T = Temperature of the gas = [tex]45^oC=(45+273)K=318K[/tex]

Putting values in above equation, we get:

[tex]725mmHg\times 7.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 318K\\\\n=0.276mol[/tex]

Hence, the number of moles of gas present is 0.276 moles

Answer:

(b) amino group

e) carboxyl group

Explanation:

The functional group in an organic compound distinguishes it from other compounds. It is usually the site where chemical reactions takes place.

       CH₃-HC(NH₂)-COOH

There are two functional groups in this organic compoud:

Answer: The molar solubility of carbon dioxide gas is 0.003 M

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.7\times 10^{-2}mol/L.atm[/tex]

[tex]p_{CO_2}[/tex] = partial pressure of carbonated drink = 0.71 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 0.71atm\\\\C_{CO_2}=2.637\times 10^{-2}mol/L=0.003M[/tex]

Hence, the molar solubility of carbon dioxide gas is 0.003 M

Answer: The molar solubility of carbon dioxide is [tex]2.63\times 10^{-2}M[/tex]

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.7\times 10^{-2}mol/L.atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas = ?

[tex]p_{CO_2}[/tex]  = partial pressure of carbon dioxide gas = 0.71 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 0.71atm\\\\C_{CO_2}=2.63\times 10^{-2}M[/tex]

Hence, the molar solubility of carbon dioxide is [tex]2.63\times 10^{-2}M[/tex]

Answer:

Mass of 246 lb object is 7.646 slugs.

Explanation:

Mass of the object = m

Weight of the object = W = 246 lb

Acceleration due to gravity= g = [tex]32.174 ft/s^2[/tex]

[tex] Weight=mass \times g[/tex]

[tex]W=mg[/tex]

[tex]1 lb=slugs\times 1 ft/s^2[/tex]

[tex]1 slugs=\frac{lb\times s^2}{ft}[/tex]

[tex]m=\frac{W}{g}=\frac{246 lb}{32.174 ft/s^2}=7.646 lb s^2/ft=7.646 slugs[/tex]

Mass of 246 lb object is 7.646 slugs.

Answer: The density of substance is [tex]3.38g/cm^3[/tex]

Explanation:

To calculate density of a substance, we use the equation:

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

We are given:

Mass of substance = 61.6 g

Volume of substance = [tex]18.2cm^3[/tex]

Putting values in above equation, we get:

[tex]\text{Density of substance}=\frac{61.6g}{18.2cm^3}\\\\\text{Density of substance}=3.38g/cm^3[/tex]

Hence, the density of substance is [tex]3.38g/cm^3[/tex]

Final answer:

The density of the substance with a mass of 61.6 g and a volume of 18.2 [tex]cm^3[/tex] is 3.385 g/[tex]cm^3[/tex], calculated by dividing mass by volume using the density formula.

Explanation:

The density of a substance is calculated using the formula d = m/V, where d is the density, m is the mass, and V is the volume.

To find the density of a substance with a given mass and volume, you simply divide the mass by the volume. For a substance with a mass of m = 61.6 g and a volume of V = 18.2 [tex]cm^3[/tex], the density d would be calculated as follows:

d = m/V
d = 61.6 g / 18.2 [tex]cm^3[/tex]
d \3.385 g/[tex]cm^3[/tex]

This means that the density measurement of the substance is 3.385 grams per cubic centimeter, expressed as g/[tex]cm^3[/tex].

Answer:

The sum of the coefficient is: 1 + 2 +2 + 1 = 6 ( option c)

Explanation:

First we will balance on both sides Na

On the right side we have 2x Na but on the left side we have only 1x Na. So we have to multiply NaOH on the left side by 2.

This will give us:

H2SO4 + 2 NaOH → H2O + Na2SO4

Now we have on both sides 2x Na

We see that on the left side we have 4x H ( 2x H of H2SO4 and 2x H of NaOH), but on the right side we only have 2x H. So, we have to multiply H2O on the right side by 2.

This will give us:

H2SO4 + 2 NaOH → 2 H2O + Na2SO4

Now we have on both sides 2x Na and 4x H.

Also the number of O is on both sides equal, due to this. ( Both sides have 6x O).

Finally, we have this reaction:   H2SO4 + 2 NaOH → 2 H2O + Na2SO4

The sum of the coefficient is: 1 + 2 +2 + 1 = 6 ( option c)

Final answer:

The sum of the coefficients when the equation H2SO4 + NaOH → H2O + Na2SO4 is balanced is 6. The balanced equation is 1 H2SO4 + 2 NaOH → 2 H2O + 1 Na2SO4.

Explanation:

The sum of the coefficients when the chemical equation H2SO4 + NaOH → H2O + Na2SO4 is balanced is the total of the numbers that are used to balance the equation.

To balance the equation, we need to ensure that there is the same number of each type of atom on both the reactant and product sides of the equation. In this case, we balance the equation as follows: 1 H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + 1 Na2SO4 (aq). So, the coefficients in the balanced equation are 1, 2, 2, and 1 respectively.

Adding these coefficients, we get:

1 (for H2SO4) + 2 (for NaOH) + 2 (for H2O) + 1 (for Na2SO4) = 6.

The answer is option c: 6.

Answer:

The correct option is: B. 33%

Explanation:

Orbital hybridisation refers to the mixing of atomic orbitals of the atoms in order to form new hybrid orbitals. The concept of orbital hybridization is used to explain the structure of a molecule.

The sp² hybrid orbitals are formed by the hybridization of one 2s orbital and two 2p orbitals. The three sp² hybrid orbitals formed have 33% s character and 67% p character.

Answer:

It's B. 33%

Explanation:

Because it refers to the mixing of atomic orbitals of the atoms in order to form new hybrid orbitals. The concept of orbital hybridization is used to explain the structure of a molecule.

Also, the sp² hybrid orbitals are formed by the hybridization of one 2s orbital and two 2p orbitals. Then the three sp² hybrid orbitals formed have 33% s character and 67% p character.

Answer:

Pressure in duct = 799.75 mmHg

Atmospheric pressure = 774.75 mmHg

Gauge pressure = 24.99 mmHg

Explanation:

First of all, it is needed to set a pressure balance (taking in account that diameter of manometer is constant) in the interface between the air of the duct and the fluid mercury.

From the balance in the sealed-end manometer, we have the pressure of air duct as:

[tex]P = \rho g h_1[/tex]

We have that ρ is density of mercury and g is the gravity

[tex]\rho = 13600 kg/m^{3}[/tex]

[tex]g = 9.8 m/s^{2}[/tex]

So, replace in the equation:

[tex]P = (13600 kg/m^{3} )(9.8 m/s^{2})(800 mmHg)(\frac{1 mHg}{1000 mmHg})[/tex]

[tex]P = 106624.0 \frac{kg}{s^{2}} = 106624.0 Pa[/tex]

Transforming from Pa to mmHg

[tex]P =  106624.0 Pa (\frac{760 mmHg}{101325 Pa}) = 799.7 mmHg[/tex]

From the balance in the open-end manometer, we have the pressure of air duct as:

[tex]P = \rho g h_2 + P_atm[/tex]

Isolate [tex]P_atm[/tex]:

[tex]P_atm = P - \rho g h_2[/tex]

Calculating:

[tex]P_atm = 799.75 mmHg - (13600 kg/m^{3} )(9.8 m/s^{2})(25 mmHg)(\frac{1 mHg}{1000 mmHg})(\frac{760 mmHg}{101325 Pa} )[/tex]

[tex]P_atm = 774.75 mmHg[/tex]

Finally, gauge pressure is the difference between duct pressure and atmospheric pressure, so:

[tex]P_gau = P - Patm[/tex]

[tex]P_gau = 799.75 mmHg - 774.75 mmHg[/tex]

[tex]P_gau = 24.99 mmHg[/tex]

End.