35S has a half-life of 88 days. You start with 128 micrograms of 3Ss. How much 35s is left after 264 days? O a. 32 micrograms b.8 micrograms O c. 64 micrograms O d. 16 micrograms O e. 4 micrograms

Answer: The values of 'x' are 0.074 and -0.132

Explanation:

The equation given to us is:

[tex]\frac{x^2}{0.160-x}=0.058[/tex]

Rearranging the above equation, we get a quadratic equation:

[tex]x^2+0.058x-0.009744=0[/tex]

To solve this equation, we use quadratic formula, which is:

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

where,

a = coefficient of [tex]x^2[/tex] = 1

b = coefficient of x = 0.058

c = constant = 0.009744

Putting values in above equation, we get:

[tex]x=\frac{-0.058\pm \sqrt{(0.058)^2-4(1)(0.009744)}}{2\times 1}\\\\x=0.074,-0.132[/tex]

Hence, the values of 'x' are 0.074 and -0.132

Answer:

The term half lime means: the time taken for the concentration of a reactant to decrease by a factor of 1/2.

Explanation:

In kinetics, the term half-life refers to the time that it takes to decrease the concentration of a reactant to half its initial concentration. Half-life depends on the reaction order, on the rate constant and, except for first-order kinetics, on the initial concentration of the reactant.

Answer:

The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.

Explanation:

Every weak acid or base is at equilibrium with its conjugate base or acid respectively when it is dissolved in water.

[tex]HA + H_{2}O[/tex] ⇄ [tex]A^{-} + H_{3}O^{+}[/tex]

This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,

[tex]pH = pKa + Log \frac{[A^{-}]}{[HA]}[/tex]

shows the dependency between the pH of the solution, the pKa and the concentration of the species. If the pH decreases the concentration of protons will increase and the ratio between A- and AH will decrease. Instead, if the pH increases the concentration of protons will decreases and the ratio between A- and AH will increase.

Final answer:

The ratio [A-]/[HA] changes with changing pH according to the Henderson-Hasselbalch equation.

Explanation:

The Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), shows how the ratio [A-]/[HA] changes with changing pH. In this equation, [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the weak acid. As the pH increases, the concentration of [A-] increases relative to [HA], resulting in a higher value for the ratio [A-]/[HA]. Conversely, as the pH decreases, the concentration of [A-] decreases relative to [HA], leading to a lower value for the ratio [A-]/[HA].

Answer:

Pipe racks are construction in chemical and other industries plants, that support the pipe line, electric cables and instrument cable.

Explanation:

The pipe racks also used to support mechanical equipment as valve and vessels. You can transfer material between equipment and sorage or utility areas. Pipe racks aren´t only non-building constructions that have similiraties to the Steel buildings but also have additional loads style. The requirements found in the building codes apply and dictate some of the design requirements.

Some industry references exist to help the designer apply the intent of the code and follow expected engineering practices.

Pipe racks have design criteria: In most of the United States, the governing building code is the International Building Code.

Includes:

*Dead Loads

*Live Loads

*Thermal Loads

*Earthquake Loads

*Wind Loads

*Rain Loads

*Snow Loads

*Ice Loads

*Load Combinations

Also have Design Considerations:

*Layout

*Seismic

*Seismic System Selection

*Period Calculations

*Analisys Procedure

*Selection

*Equivalent Lateral Force Method Analysis

*Modal Response Spectra Analysis

*Drift

*Seismic Detailing Requirements

*Wind

*Pressures and Forces

*Coatings

*Fire Protection

*Torsion on Support Beams  

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of [tex]O_2[/tex] = 8.00 g

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

Molar mass of CO = 28 g/mole

First we have to calculate the moles of [tex]O_2[/tex].

[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2C+O_2\rightarrow 2CO[/tex]

From the balanced reaction we conclude that

As, 2 mole of [tex]C[/tex] react with 1 mole of [tex]O_2[/tex]

So, 0.1 moles of [tex]C[/tex] react with [tex]\frac{0.1}{2}=0.05[/tex] moles of [tex]O_2[/tex]

From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]C[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]CO[/tex]

From the reaction, we conclude that

As, 2 mole of [tex]C[/tex] react to give 2 mole of [tex]CO[/tex]

So, 0.1 moles of [tex]C[/tex] react to give 0.1 moles of [tex]CO[/tex]

Now we have to calculate the mass of [tex]CO[/tex]

[tex]\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO[/tex]

[tex]\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g[/tex]

Therefore, the mass of carbon monoxide form can be 2.8 grams.

Final answer:

When 0.100 mol of carbon is burned with 8.00 g of oxygen, a maximum of 11.00 grams of carbon dioxide can be formed, assuming oxygen is the limiting reactant.

Explanation:

To determine how many grams of carbon dioxide (CO2) can form when burning carbon with oxygen, we look at the chemical equation for the combustion of carbon:

C(s) + O2(g) → CO2(g)

This reaction shows that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide. First, we need to find the number of moles of oxygen that 8.00 grams corresponds to:

Number of moles of O2 = mass (g) ÷ molar mass of O2 = 8.00 g ÷ 32.00 g/mol = 0.25 mol

Since the reaction requires equal moles of O2 and C to produce CO2, and we have 0.100 mol of carbon, we are limited by the amount of oxygen because it is less than the amount of carbon. Therefore, all of the oxygen will be used. To find the mass of CO2 that can be formed:

Mass of CO2 = moles of O2 × molar mass of CO2 = 0.25 mol × 44.01 g/mol = 11.00 g of CO2

So, a maximum of 11.00 grams of carbon dioxide can be formed from the combustion of 0.100 mol of carbon with 8.00 g of oxygen.

Answer:

Specific gravity of mercury is 13.56 and it is an unit-less quantity.

Explanation:

Mass of the mercury = m = 607.0 lb = 275330.344 g

1 lb = 453.592 g

Volume of the mercury  = v = [tex]0.717 ft^3=20,303.18 mL[/tex]

[tex]1 ft^3 = 28316.847 mL[/tex]

Density of the mercury = d=[tex]\frac{m}{v}=\frac{275330.344  g}{20,303.18 mL}[/tex]

d = 13.56 g/mL

Specific gravity of substance = Density of substance ÷ Density of water

[tex]S.G=\frac{d}{1 g/mL}[/tex]

Specific gravity of mercury :

[tex]S.G=\frac{13.56 g/mL}{1 g/mL}=13.56[/tex] (unit-less quantity)

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

First, we will convert 3.00 g of CaCl₂ to moles using its molar mass (110.98 g/mol).

[tex]3.00 g \times \frac{1mol}{110.98g} = 0.0270 mol[/tex]

The heat of solution (ΔHsoln) of CaCl₂ is −82.8 kJ/mol. The heat released by the solution of 0.0270 moles is:

[tex]0.0270 mol \times \frac{-82.8kJ}{mol} = -2.24 kJ[/tex]

According to the law of conservation of energy, the sum of the heat released by the solution (Qs) and the heat absorbed by the calorimeter (Qc) is zero.

[tex]Qs + Qc = 0\\\\Qc = -Qs = 2.24 kJ[/tex]

Assuming the density of water is 1 g/mL, we have 100 mL (100 g) of water and 3.00 g of CaCl₂. The mass of the solution (m) is:

[tex]m = 100g + 3.00 g = 103 g[/tex]

Finally, we can calculate the final temperature of the system using the following expression.

[tex]Qc = c \times m \times (T_2 - T_1)[/tex]

where,

c: specific heat of the solution (same as water 4.18 J/g.°C)

T₁ and T₂: initial and final temperature

[tex]T_2 = \frac{Qc}{c \times m} + T_1 = \frac{2.24 \times 10^{3}J }{(\frac{4.18J}{g.\° C} ) \times 103 g} + 23.0 \° C = 28.2 \° C[/tex]

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

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To determine the amount of heat involved in the dissolution of CaCl₂ and the final temperature of the solution, we can use the formula q = m * c * ΔT. First, we need to calculate the heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water. Assuming the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat.

To determine the amount of heat involved in the dissolution of CaCl₂ and the final temperature of the solution, we can use the formula q = m * c * ΔT, where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat of the solution, and ΔT is the change in temperature.

First, we need to calculate the heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water. The mass of the solution is 55.0 g (50.0 g + 5.00 g), and the change in temperature is 39.2°C - 23.0°C = 16.2°C. Assuming the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat using the formula: q = 55.0 g * 4.18 J/g°C * 16.2°C = 3660.36 J.

Since 1 kJ = 1000 J, we can convert the heat to kilojoules: 3660.36 J ÷ 1000 = 3.66 kJ. Therefore, the amount of heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water is 3.66 kJ.

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Explanation:

Ion-ion forces are defined as the forces that exist between oppositely charged ions.

Dipole-dipole interactions are defined as the forces which exist between positive end of polar molecule and negative end of another polar molecule.

Ion-dipole forces are defined as the forces that exist between a charged ion and a polar molecule.

London dispersion forces are defined as the forces that arise due to the development of temporary charges on the combining atoms of a molecule.

Hence, the given substances are classified as follows.

(a) HF - It is a covalent compound but due to the difference in electronegativity of hydrogen and fluorine there will be development of partial charges on both of them.

Hence, in a HF molecule there will be dipole-dipole forces.

(b) [tex]CH_{3}OH[/tex] - There will also be development of partial charges due to the difference in electronegativity of oxygen and hydrogen atoms.

Hence, in a [tex]CH_{3}OH[/tex] there will be dipole-dipole forces.

(c) [tex]CaCl_{2}[/tex] - It is an ionic compound. Hence, there will be partial positive charge on calcium and partial negative charge on chlorine atom.

Hence, in a [tex]CaCl_{2}[/tex] molecule there will exist ion-ion forces.

(d) [tex]FeBr_{3}[/tex] - It is an ionic compound. Hence, there will also exist ion-ion forces.

Answer:

a) (2)

b) (2)

c) (3)

d) (3)

Explanation:

The intermolecular forces are the forces that make molecules to be bond in a substance. When a solvent dissolves a solute, the molecules of the solvent and the solute will be attached by the forces. The types of forces are:

The bonds between substances can mix these forces. So if one is polar and the other is nonpolar, the bond will be London dipole; if both are polar, dipole dipole, if one is polar and the other is ionic ion dipole; and if one is nonpolar and the other is ionic, ion London.

Water (H2O) is a polar molecule, so the dipole must happen in all of the dissolutions.

a) HF is a polar molecule, so the bond of it and water will be dipole dipole. In both substances the hydrogen is bonded to high electronegativity elements, so its hydrogen bond! But, because there's no answer to it, we can call it dipole dipole. (2)

b) CH3OH is a polar compound, and have hydrogen bonds, but, as explained above, it'll be dipole dipole forces. (2)

c) CaCl2 is an ionic compound (cation Ca+2 and anion Cl-), thus the force will be ion dipole. (3)

d) FeBr3 is an ionic compound (cation Fe+3 and anion Br-), thus the force will be ion dipole. (3)

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

        [tex]C_{p}_{liquid}[/tex] = 4.19 [tex]kJ/kg ^{o}C[/tex]

        [tex]C_{p}_{vaporization}[/tex] = 1.9 [tex]kJ/kg ^{o}C[/tex]

Heat of vaporization ([tex]\DeltaH^{o}_{vap}[/tex]) at 1 atm and [tex]100^{o}C[/tex] is 2259 kJ/kg

        [tex]H^{o}_{liquid}[/tex] = 0

Therefore, calculate the enthalpy of water vapor at 1 atm and [tex]100^{o}C[/tex] as follows.

            [tex]H^{o}_{vap}[/tex] = [tex]H^{o}_{liquid}[/tex] + [tex]\DeltaH^{o}_{vap}[/tex]        

                                   = 0 + 2259 kJ/kg

                                   = 2259 kJ/kg

As the desired temperature is given [tex]179.9^{o}C[/tex] and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.

             [tex]H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})[/tex]

                             = [tex]0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)[/tex]

                              = 334.781 kJ/kg

Hence, enthalpy of water vapor at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.

               [tex]H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})[/tex]

             [tex]H^{D}_{vap}[/tex] = [tex]2259 kJ/kg + 1.9 \times (179.9 - 100)[/tex]            

                              = 2410.81 kJ/kg

Therefore, calculate the latent heat of vaporization at 10 bar and [tex]179.9^{o}C[/tex] as follows.

       [tex]\Delta H^{D}_{vap}[/tex] = [tex]H^{D}_{vap} - H^{D}_{liq}[/tex]              

                         = 2410.81 kJ/kg - 334.781 kJ/kg

                         = 2076.029 kJ/kg

or,                      = 2076 kJ/kg

Thus, we can conclude that at 10 bar and [tex]179.9^{o}C[/tex] latent heat of vaporization is 2076 kJ/kg.

Answer:

K₅ is smaller because the reaction is slower due to steric and electronic effects.

Explanation:

K is the rate constant of the reaction. The higher the value of log K, the higher the value of K and the faster is the reaction.

The reaction can be represented by the following reaction:

6 NH₃ + [Cu(OH₂)₆]²⁺ → [Cu(NH₃)₆]²⁺ + 6 H₂O

This means that the reaction is exchanging the H₂O ligands by NH₃ ligands.

The more NH₃ ligands we add to the complex, the more difficult (slower) is the substitution. This happend because the addition of NH₃ ligands promotes a steric hindrance and electronic repulsion, which makes it harder for the next NH₃ to approach the complex and substitute the H₂O ligand.

This is the reason why K₅ is negative. The rate of this substitution is extremelhy low.

Answer:

3.3 %

Explanation:

According to the question , the following reaction takes place -

 Ba(OH)₂ + 2 HBr   →   BaBr₂  +  2 H₂O

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

According to the question ,

V = volume of Ba(OH)₂ = 19.5 mL = 0.0195 L    ( since , 1 ml = 10 ⁻³ L )

M = Molarity of Ba(OH)₂ = 0.374 M

The moles of Ba(OH)₂ can be calculated by using the above equation ,

M = n / V  

n = M * V = 0.374 M  *  0.0195 L  =  0.0072 mol

From the above balanced reaction ,

2 mol of HBr reacts with 1 mol  Ba(OH)₂

1 mol of HBr reacts with 1 / 2 mol  Ba(OH)₂

From the above data ,

1 mol HBr reacts with = 1 / 2 * 0.0072 mol  = 0.0036 mol

Hence , number of moles of HBr = 0.0036 mol

Now,

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

As calculated above ,

n = 0.0036 mol

As we know , the m = molecular mass of HBr =  81 g/mol

n = w / m  

w = n * m =  0.0036 mol  *  81 g/mol  = 0.2916 g

Now ,

mass % = mass of HBr /  mass of solution   * 100

mass % = 0.2916 g / 8.72 g * 100 = 3.3 %

To find the percent by mass of hydrobromic acid in the mixture, we need to use the concept of titration. We can calculate the moles of hydrobromic acid using the volume and concentration of the barium hydroxide solution used to neutralize the acid. Then, we can determine the percent by mass.

To find the percent by mass of hydrobromic acid in the mixture, we need to use the concept of titration. In a neutralization reaction, the moles of acid can be determined by multiplying the volume of the base solution with its concentration. This can be expressed using the formula:

Moles of acid = Volume of base solution (L) x Concentration of base solution (M)

In this case, we are given the volume and concentration of the barium hydroxide solution used to neutralize the hydrobromic acid. So, we can calculate the moles of hydrobromic acid and then determine the percent by mass.

Answer:

When you start to make this operations, you will find out that the correct answer is, NaCl 5.56 mM glucose, 16.7 mM.

Explanation:

First of all you should need to find, how many mols are in the first solutions you add: In glucose you have 0.100m, so as you know they are in 1000ml,  how many, in 25 ml? this is 2,5 *10^-3 moles. In NaCl, you should do the same, 1000 ml has 0.5 mols, so how many are, in 15ml?. The answer is 7.5 *10^-3. Now, that you have your mols you have to take account the water which is in 450 ml. So, let's go again, in 450ml you have 2,5 *10^-3 moles of glucose and 7.5 *10^-3 moles of NaCl, how many moles of them, are in 1000 ml. You will get that concentrations are 0,0167 M in NaCl and 5,56 *10^-3 M. Let's see that this numbers are in M, so if u want to get mM, just *1000.

The correct concentrations in the student's mixed solution are 5.10 mM glucose and 15.3 mM NaCl, calculated by using the dilution formula and considering the total volume of the final solution.

To calculate the concentrations of glucose and NaCl in the final solution after mixing 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl, and 450.0 mL water, one would use the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the initial solutions, and M2 and V2 are the molarity and volume of the final solution respectively.

For glucose: moles of glucose = M1V1 = 0.100 mol/L × 0.025 L = 0.0025 mol.

For NaCl: moles of NaCl = M1V1 = 0.500 mol/L × 0.015 L = 0.0075 mol.

Total volume of the final solution = 25.0 mL + 15.0 mL + 450.0 mL = 490.0 mL = 0.490 L (to convert mL to L, divide by 1000).

Concentration of glucose in the final solution = moles of glucose / total volume = 0.0025 mol / 0.490 L = 5.10 mM (since 1 mM = 0.001 M).

Concentration of NaCl in the final solution = moles of NaCl / total volume = 0.0075 mol / 0.490 L = 15.3 mM.

Therefore, the correct concentrations in the solution are 5.10 mM glucose and 15.3 mM NaCl.

Answer:

The correct option is: Activity coefficient            

Explanation:

Ideal solution is a solution that has thermodynamic properties similar to the ideal gases. The pressure of an ideal solution obeys the Raoult's law. An ideal solution has zero enthalpy of mixing and the activity coefficient of all the components of the solution is unity.

Example: The solution of 1-butanol and 2-butanol, is nearly ideal as the two chemical compounds or molecules are chemically similar.        

Answer:

4.88 g / cm³

Explanation:

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass of the metal = 24.54 g

The volume of the metal = 5.02 mL

Hence , by using the above formula ,and putting the corresponding values , the density is calculated as -

d = m / V

d = 24.54 g / 5.02 mL

d = 4.88 g /mL

The unit 1mL = 1 cm³

Hence ,

d = 4.88 g / cm³

Answer:

Explanation:

Hello,

At first, consider the balanced reaction for the combustion of the isooctane:

[tex]C_8H_{18}+\frac{25}{2} O_2-->8CO_2+9H_2O[/tex]

Now, the stoichiometric relationship between the hydrocarbon and the oxygen leads to:

[tex]molO_2=230molC_8H_{18}(\frac{\frac{25}{2}mol O_2 }{1mol C_8H_{18}} )=2875molO_2[/tex]

Best regards.

The molar concentration of O2 in the surface water of a mountain lake at 20 °C and an atmospheric pressure of 665 torr is approximately 1.21×10-3 M.

To calculate the molar concentration of O2 in the surface water of a mountain lake using Henry's law, we first need to understand how pressure affects the solubility of gases and vice versa.

As per Henry's law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.  The partial pressure of O2 in air at sea level is 0.21 atm. This means that when the atmospheric pressure is 1 atm, the molar concentration of O2 is 1.38×10−3 M. At higher altitudes, the atmospheric pressure reduces. The given atmospheric pressure at the mountain lake is 665 torr, which is approximately 0.875 atm.

Using these values in Henry's law, the molar concentration of O2 can be calculated as:

C = P * x

where C is molar concentration, P is atmospheric pressure, and x is given solubility at 1 atm. Substituting the values:

C = (0.875 atm) * (1.38×10−3 M) = 1.21x10-3 M approximately

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Using Henry's law, the molar concentration of O₂ in water at a 20°C mountain lake with an atmospheric pressure of 665 torr is approximately 2.5×10⁻⁴ M. This is calculated by converting the pressure to atm and then using the given solubility at standard conditions. The relationship is direct proportionality between solubility and partial pressure.

We will use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid. The formula is:

Concentration = kH × P

Given:

The solubility of O₂ in water at 20°C with 1 atm O₂ pressure (P₀) is 1.38×10⁻³ M (C₀).

Partial pressure of O₂ in air at sea level (P₀) is 0.21 atm.

Atmospheric pressure at the mountain lake is 665 torr.

First, convert 665 torr to atm:

665 torr × (1 atm / 760 torr) = 0.875 atm

Next, calculate the partial pressure of O₂ in air at the mountain lake:

Partial Pressure of O₂ = 0.21 atm × 0.875 atm = 0.184 atm

Using Henry's law, we can find the new concentration (C):

C = kH × P = 1.38×10⁻³ M/atm × 0.184 atm = 2.5×10⁻⁴ M

Therefore, the molar concentration of O₂ in the mountain lake's surface water at 20°C is approximately 2.5×10⁻⁴ M.

Answer:

6,45mmol/L of NaOH you need to add to reach this pH.

Explanation:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,74

Henderson-Hasselbalch equation for acetate buffer is:

5,0 = 4,74 + log₁₀[tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex]

Solving:

1,82 = [tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex] (1)

As total concentration of acetate buffer is:

10 mM = [CH₃COOH] + [CH₃COO⁻] (2)

Replacing (2) in (1)

[CH₃COOH] = 3,55 mM

And

[CH₃COO⁻] = 6,45 mM

Knowing that:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Having in the first 10mmol/L of CH₃COOH, you need to add 6,45 mmol/L of NaOH. to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃COOH .

I hope it helps!

Answer:

The correct option is: a) saturated

Explanation:

Relative humidity is a primary measurement of humidity. At a given temperature, relative humidity describes the present state of absolute humidity relative to the maximum humidity.

It is generally expressed as percentage. Therefore, 100% relative humidity means that the air is entirely saturated.

Answer:

The answers is the option a: 46.062

Explanation:

First, you must determine the limiting reagent, that is the reagent that is consumed first during the reaction.

So, in first place, you must determine the molecular weight of the molecules CH4, O2, CO2 and H20. You know that

So, to determine the mass of a molecule, you must multiply the individual masses of each atom by the amount present in the molecule. This would be:

In the same way, you can determinate the mass of all reagents and products involved in the reaction.

Now you can apply stoichiometry to determine the limiting reagent  using these numbers and observing how many molecules react.

On one side, it is known that, by stoichiometry, 1 mol of CH4 and 2 moles of O2 react. This means that 16 kg/mol of CH4 and 64 kg/kmol ( 2moles* 32 kg/kmol) of O2 react

And it is known that there are 71 kg/hr of CH4 reacting with 67 kg/hr of 02

So, using the stoichiometric information (16 kg/mol of CH4 and 64 kg/kmol of O2), 71 kg/hr of CH4 and The Rule of Three, you can determine the limiting reagent:

16 kg/kmol CH4 ⇒ 64 kg/kmol O2 (stoichiometry)

71 kg/kmol CH4 ⇒ x

So [tex]x=\frac{71*64}{16}[/tex]

x=284 kg/kmol

This means that to react 71 kg/kmol of CH4, 284 kg/kmol of O2 are needed. But you only have 67 kg/mol that can react. That is why O2 is the limiting reagent, because it is consumed first.

Now, you can calculate the rate of CO2 that is generated, using the data of the amount of limiting reagent and stoichiometry. This is:

64 kg/kmol O2 ⇒ 44 kg/kmol CO2 (stoichiometry)

67 kg/kmol O2 ⇒ x

So [tex]x=\frac{67*44}{64}[/tex]

x=46.0625 kg/kmol

This means that 46.0625 kg/kmol of CO2 are generated.

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically,         Molarity = [tex]\frac{\text{no. of moles}}{\text{Volume in liter}}[/tex]

Also, when number of moles are equal in a solution then the formula will be as follows.

                     [tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]

It is given that [tex]M_{1}[/tex] is 8.00 M, [tex]V_{1}[/tex] is 7.00 mL, and [tex]M_{2}[/tex] is 0.80 M.

Hence, calculate the value of [tex]V_{2}[/tex] using above formula as follows.

                    [tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]

                 [tex]8.00 M \times 7.00 mL = 0.80 M \times V_{2}[/tex]

                      [tex]V_{2} = \frac{56 M. mL}{0.80 M}[/tex]

                                  = 70 ml

Thus, we can conclude that the volume after dilution is 70 ml.

Answer:

As can be seen in the explanation, there will be a greater increase in the enthalpy of the working fluid in the adiabatic process than in the isothermal process, since in the adiabatic process the change of internal energy is added; In the isothermal process there is no change in internal energy that is added to the value of enthalpy.

Explanation:

∴ Q = 0  ⇒ ΔU = W......first law

⇒ ΔU = Cv*ΔT

⇒ W = P*dV  ∴ P = nRT/V

⇒ ΔH = ΔU + P*ΔV = Cv*ΔT + nRT/V*ΔV

∴ ΔT = 0 ⇒ ΔU = 0

⇒ Q = - W = nRTLn(V2/V1)

⇒ ΔH = PΔV = nRT/V*ΔV

Final answer:

For a greater increase in the enthalpy of the working fluid in a reversible flow compressor, the isothermal process is more favorable than the adiabatic process. This is because the isothermal process allows for heat exchange with the surroundings, directly increasing the enthalpy, unlike the adiabatic process where there's no heat exchange.

Explanation:

To determine which condition between adiabatic and isothermal will lead to a greater increase in the enthalpy of the working fluid, it is essential to understand the concepts of adiabatic and isothermal processes in the context of a reversible flow compressor. In an adiabatic process, the system is thermally insulated, leading to no heat exchange with the surroundings, which in turn means that any work done on the system results directly in changes to the system's internal energy and hence, its temperature. On the other hand, during an isothermal process, the system's temperature is kept constant by allowing heat exchange with the surroundings, effectively absorbing or releasing heat as work is done on or by the system.

By comparing the expressions dU = Tds - PdV and dH = TdS + VdP, and knowing that dH represents the increase in enthalpy (the heat added at constant pressure), it's evident that during an isothermal process, where temperature is constant, the system can absorb or release heat without changing temperature, ideally through reversible compressions or expansions that allow it to maintain equilibrium with a thermal reservoir. This contrasts with an adiabatic process where any work done on the system directly influences its internal energy and, consequently, temperature, without the exchange of heat with its surroundings.

Therefore, for a greater increase in the enthalpy of the working fluid, the isothermal process is favorable. This is because it allows for heat exchange, facilitating a direct increase in enthalpy, as opposed to the adiabatic process, where the increase in internal energy does not directly translate to an increase in enthalpy due to the lack of heat exchange.

Answer:

The velocity of flow in 10in pipe is 4.16 ft/s.

Explanation:

Given that

Specific gravity = 0.8

Viscosity =0.00042[tex]lbf/ft^2[/tex]

For pipe 1

[tex]V_1=6 ft/s,d_1=12\ in[/tex]

For pipe 1

[tex]V_2,d_1=10\ in[/tex]

If we assume that flow in the both pipe is laminar

For laminar flow through circular pipe

[tex]\dfrac{\Delta P}{L}=\dfrac{32V\mu }{d^2}[/tex]

So same pressure drop we can say that

[tex]\dfrac{V_1 }{d^2_1}=\dfrac{V_2}{d^2_2}[/tex]

[tex]\dfrac{6}{12^2}=\dfrac{V_2}{10^2}[/tex]

[tex]V_2=4.16 ft/s[/tex]

So the velocity of flow in 10in pipe is 4.16 ft/s.

Explanation:

It is known that [tex]mv_{r} = \frac{nh}{2p}[/tex]

where,      m = mass of the electron

                 r = radius of the orbit

                 [tex]v_{r}[/tex] = orbital speed of the electron

Equation when the electron is experiencing uniform circular motion is as follows.

            [tex]\frac{Kze^{2}}{r^{2}} = \frac{mv^{2}}{r}[/tex] ........ (1)

Rearranging above equation, we get the following.

                        [tex]mv^{2} = \frac{Kze^{2}}{r}[/tex]

Also,         v = [tex]\frac{nh}{2pmr}[/tex] .......... (2)

Putting equation (2) in equation (1) we get the following.

                [tex]\frac{mn^{2}h^{2}}{4p^{2}m^{2}r^{2}} = \frac{Kze^{2}}{r}[/tex]

Hence, formula for radius of the nth orbital is as follows.

                 [tex]r_{n} = [\frac{h^{2}}{4p^{2}mKze^{2}}]n^{2}[/tex]

                   [tex]r_{n} = [5.29 \times 10^{-11}m] \times (6)^{2}[/tex]

                             = [tex]19.044 \times 10^{-10} m[/tex]

                             = [tex]19.044 A^{o}[/tex]

Thus, we can conclude that the value for the radius r for a n= 6 Bohr orbit is [tex]19.044 A^{o}[/tex].

Answer:

A) [tex]2.7\frac{km}{day}[/tex]

B) [tex]1.03*10^{2}\frac{ft}{s}[/tex]

C) [tex]31\frac{m}{s}[/tex]

D) [tex]2.05*10^{3}\frac{yd}{min}[/tex]

Explanation:

A) Convert [tex]70\frac{mi}{hr} to \frac{km}{day}[/tex]

[tex]70\frac{mi}{hr}*\frac{24hr}{1day}*\frac{1.60934km}{1mi}=2.7\frac{km}{day}[/tex]

B) Convert  [tex]70\frac{mi}{hr} to \frac{ft}{s}[/tex]

[tex]70\frac{mi}{hr}*\frac{1hr}{3600s}*\frac{5280ft}{1mi}=1.03*10^{2}\frac{ft}{s}[/tex]

C) Convert [tex]70\frac{mi}{hr} to \frac{m}{s}[/tex]

[tex]70\frac{mi}{hr}*\frac{1hr}{3600s}*\frac{1609.34m}{1mi}=31\frac{m}{s}[/tex]

D) Convert [tex]70\frac{mi}{hr} to \frac{yd}{min}[/tex]

[tex]70\frac{mi}{hr}*\frac{1hr}{60min}*\frac{1760yd}{1mi}=2.05*10^{3}\frac{yd}{min}[/tex]

To convert the speed limit of 70 mi/hr into alternative units, multiply the speed by the appropriate conversion factors.

To convert the speed limit of 70 mi/hr into alternative units:

Answer:

Magnitude of the aftershock = 0.3

Explanation:

Based on the information in the problem, an expression can be written that relates the magnitude of the earthquake (M₁) with the magnitude of the aftershock (M₂):

M₁ = 25M₂

We can then solve for M₂ and substitute in the values in the problem:

M₂ = M₁ / 25 = 6.3 / 25 = 0.3

Answer:

0.4429 m

Explanation:

Given that mass % of the ethanol in water = 2%

This means that 2 g of ethanol present in 100 g of ethanol solution.

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of [tex]CH_3CH_2OH[/tex] = 2 g

Molar mass of [tex]CH_3CH_2OH[/tex] = 46.07 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{2\ g}{46.07\ g/mol}[/tex]

[tex]Moles\ of\ CH_3CH_2OH= 0.0434\ moles[/tex]

Mass of water = 100 - 2  g = 98 g = 0.098 kg ( 1 g = 0.001 kg )

So, molality is:

[tex]m=\frac {0.0434\ moles}{0.098\ kg}[/tex]

Molality = 0.4429 m

Answer:

Vinegar has 4.09% of acetic acid.

Explanation:

The neutralization reaction is:

[tex]CH_{3}COOH + KOH=>CH_{3}COOK + H_{2}O[/tex]

Each mol of KOH reacts with each mole of acetic acid so the quantity of moles of acetic acid is:

[tex]M= 0.46 \frac{mol}{l}*37ml*\frac{1l}{1000ml} =0.01702 mol[/tex]

The mass of acetic acid is:

0.01702 mol×60.02g/mol=1.0215 g Acetic Acid

Finally, the percentage is:

%=1.0215 g Acetic Acid÷25g vinegar(solution)=4.09%

Answer:

 Chemical equivalence:

The chemical equivalence is defined as, the point at which multiple protons are under same electronic condition, at that point they are artificially equal.

In the chemical equivalence, the weight of the substance in gram consolidates with or dislodges one gram of hydrogen. Substance counter parts as a rule are found by partitioning the equation weight by the valence.

The law of chemical equivalence is basically define as, the point at which  two substances respond, the reciprocals of one will be equivalent to the counterparts of other and the reciprocals of any item will likewise be equivalent to that of the reactant.

Answer:

Force of attraction = 35.96 [tex]\times 10^{27}[/tex]N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = [tex]10^{-9}[/tex] m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

[tex]F = \dfrac{k \times q_1 q_2}{ \r^2}[/tex] ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = [tex]8.99 \times 10^9 \ Nm^{2}C^{-2}[/tex]

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

[tex]F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }[/tex]

Force of attraction = 35.96 [tex]\times 10^{27}[/tex]N

The mass of Benadryl that should be given to a dog weighing 26.6 kg is indeed 65 milligrams.

When calculating the appropriate dosage of Benadryl for a dog weighing 29.5 kg, it's important to consider that the recommended dosage is given per pound. Since the dog's weight is provided in kilograms, it's necessary to convert it to pounds for accurate dosage determination. Converting 29.5 kg to pounds results in approximately 64.7 pounds.

Convert the weight from kilograms to pounds:

29.5 kg × 1 lb / 0.454 kg ≈ 64.7 lbs

Calculate the mass of Benadryl in milligrams using the dog's weight in pounds: (1 mg/lb) × 64.7 lbs ≈ 64.7 mg

Rounding this to 65 mg ensures practicality. This process accounts for the difference in units (kilograms to pounds) and utilizes the given dosage information to arrive at the correct amount of Benadryl needed to treat the dog's itchy skin.

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To determine the appropriate Benadryl dosage for a 26.6 kg dog, you first convert the dog's weight to pounds (26.6 kg = 58.6 lbs). You then multiply the weight in pounds by the recommended dosage, which leads to a recommended dosage of 58.6 mg.

In order to calculate the appropriate dosage of Benadryl for dogs, you first transform the dog's weight from kilograms to pounds, as the given dosage is in milligrams per pound. Given that 1 kilogram is roughly equal to 2.20462 pounds,  you can find the weight of a 26.6 kg dog in pounds as follows:

26.6 kg * 2.20462 lbs/kg = 58.6 lbs

Next, you multiply the weight of the dog in pounds by the recommended dosage:

58.6 lbs * 1 mg/lb = 58.6 mg

So, a dog that weighs 26.6 kg should be administered a 58.6mg dose of Benadryl.

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