Question 8: A 150 mL of buffer contains 0.045 M carbonic acid (H2CO3) and 0.025 M sodium hydrogen carbonate (HCO3). If 3.0 mL of 0.15M HCl is added to the buffer, what is the new pH? pKa = 6.37

Answer: The density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL

Explanation:

To calculate the density of unknown metal, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]       ......(1)

Volume of unknown metal = 16.6 mL

Mass of unknown metal = 190.1 g

Putting values in equation 1, we get:

[tex]\text{Density of unknown metal}=\frac{190.1g}{16.6mL}\\\\\text{Density of unknown metal}=11.45g/mL[/tex]

The density of the metal remains the same.

Now, calculating the volume of unknown metal, using equation 1, we get:

Density of unknown metal = 11.45 /mL

Mass of unknown metal = 94.6 g

Putting values in above equation, we get:

[tex]11.45g/mL=\frac{94.6g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.26mL[/tex]

Hence, the density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

The concentration of the original potassium dichromate solution used in the mixture is 1.222 M. This was calculated by analyzing the total potassium concentration in the final solution, subtracting the contribution of potassium bromide, and then considering the volume of the potassium dichromate solution.

This is a problem related to the principle of conservation of matter focusing on concentration and volume in a chemical solution. As per the question, the potassium ion concentration ([K+]) in the final solution is 0.846 M. We know that the total amount of potassium (in moles) comes from both the potassium bromide and the potassium dichromate.

First, we calculate the moles of potassium from the potassium bromide: volume (L) x concentration (M) = 0.253 L x 0.19 M = 0.04807 moles. Now, consider that the total volume of the solution is 253 mL + 441 mL = 694 mL or 0.694 L. Since the given final concentration of the mixed solution is 0.846 M, the total moles of potassium in the solution would be: 0.694 L x 0.846 M = 0.587124 moles. We subtract the moles of potassium from the potassium bromide to find the moles contributed by potassium dichromate: 0.587124 moles - 0.04807 moles = 0.539054 moles.

This is the amount of potassium in the potassium dichromate solution. To find concentration, we divide this by the volume of the potassium dichromate solution: 0.539054 moles / 0.441 L = 1.222 M. So, the concentration of the original potassium dichromate solution is 1.222 M.

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Final answer:

The temperature of 304.89 Kelvin is equivalent to 31.74°C when converted using the formula C = K - 273.15.

Explanation:

The student has asked about converting the highest temperature recorded in a certain city from Kelvin to degrees Celsius. The formula for converting Kelvin to Celsius is: C = K - 273.15, where C is the temperature in Celsius and K is the temperature in Kelvin. Applying this formula to the given temperature (304.89 K), we get:

C = 304.89 K - 273.15

C = 31.74°C

Hence, the temperature of 304.89 K is equivalent to 31.74°C.

Answer:

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

Explanation:

The stoichiometry for this reaction is  

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

The rate for this reaction can be written as  

[tex]-r_{NO_2}=-\frac{d\left[NO_2\right]}{dt}=\frac{(0.01-0.008)M}{100s}=2\times{10}^{-5}\frac{M}{s}[/tex]

This rate of disappearence of [tex]NO_2[/tex] can be realated to the rate of appearence of [tex]O_2[/tex] as follows  (the coefficients of each compound are defined by the stoichiometry of the reaction)

[tex]-r_{O_2}=-r_{NO_2}\times\frac{coefficient\ O_2\ }{coefficient\ NO_2}=2\times{10}^{-5}\frac{M}{s}\times\frac{1\ mole\ O_2\ }{2\ mole\ NO_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

                                        w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International  System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s²   and 1 N =10^5 dynes

1 ton = 907,18 kg

1 k = 1000 gr

a) m = 30 lbm

[tex]w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf[/tex]

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

[tex]m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg[/tex]

c) m = 15 ton

[tex]m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes [/tex]

The weight of a 30 lbm object is 966 lbf. The mass of an object weighing 44N is 4.49 kg. And, the weight of a 15-ton object is 1.34 x 10¹² dynes.

To Calculate: (a) the weight (in lbf) of a 30.0 lbm object we need to use the fact that 1 lbm equals 32.2 lb force (lbf). Therefore, a 30.0 lbm object weight would be 30 lbm * 32.2 = 966 lbf.

In (b), the mass in kg of an object that weighs 44N can be calculated by dividing the weight by Earth's gravity (approximately 9.8m/s²). 44N / 9.8m/s² = 4.49 kg.

Lastly, in (c), to find the weight in dynes of a 15-ton object we first convert the weight to pounds since 1 ton equals 2000 lbs. Then we convert pounds to Newtons (1 lb = 4.44822 N) and finally Newtons to dynes (1 N = 1,000,000 dynes). So, 15 ton * 2000 = 30000lb * 4.44822 N/lb * 1,000,000 dynes/N = 1.34 x 10¹² dynes.

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Final answer:

The speed of light in a vacuum converted to kilometers per hour (km/h) is 1.079 x 10^12 km/hr, and in miles per minute (mi/min) is 11184.71 mi/min.

Explanation:

The speed of light in a vacuum is 2.998 x 108 m/s. To convert this speed into kilometers per hour (km/h), you multiply by the number of meters in a kilometer (1000) and the number of seconds in an hour (3600). This calculation gives us:

2.998 x 108 m/s x 1000 m/km x 3600 s/hr = 1.079 x 1012 km/hr.

Similarly, to find the speed in miles per minute (mi/min), you must use the conversion factor that 1 meter is approximately equal to 0.000621371 miles, and there are 60 seconds in a minute:

2.998 x 108 m/s x 0.000621371 mi/m x 60 s/min = 11184.71 mi/min.

Answer:

w >0, ΔE < 0

Explanation:

As per the first law of thermodynamics,

ΔE = Q - W

Where,

ΔE = Change in internal energy

Q = Heat receive or heat loss

W = work done

Work done by the system is always -ve.

Work done on the system is always +ve.

It is given that work done on the system.

W = +ve or W > 0

As, there is no heat receive or heat loss

So, Q = 0

Now, as per the first law of thermodynamics.

ΔE = Q - W

Q = 0

ΔE =  - W

or ΔE < 0

So, answer would be w > 0, ΔE < 0

Final answer:

To find the mass percent of chloride in the original dry sample, you can use the formula: Mass percent of chloride = (mass of chloride / mass of original sample) x 100%. Use the volume of AgNO3 solution used in the titration, the solution's molarity, and the molar mass of chloride to calculate the mass of chloride.

Explanation:

The percent by mass of chloride in the original dry sample can be calculated using the following formula:

Mass percent of chloride = (mass of chloride / mass of original sample) x 100%

In this case, the mass of chloride can be determined by multiplying the volume of AgNO3 solution used in the titration (28 mL) by the molarity of the solution (0.1 M) and the molar mass of chloride (35.453 g/mol).

Then, using the mass of chloride and the mass of the original sample (0.200 g), the percent by mass of chloride in the original dry sample can be calculated.

Answer:

Explanation has been given below.

Explanation:

Explanation:

The integrated rate law for the zeroth order reaction is:

[tex][A]=-kt+[A]_0[/tex]

The integrated rate law for the first order reaction is:

[tex][A]=[A]_0e^{-kt}[/tex]

The integrated rate law for the second order reaction is:

[tex]\frac{1}{[A]}=kt+\frac{1}{[A]_0}[/tex]

Where,

[tex][A][/tex] is the active concentration of A at time t

[tex][A]_0[/tex] is the active initial concentration of A

t is the time

k is the rate constant

Answer:

- 0th: [tex]C_A=C_{A0}-kt[/tex]

- 1st: [tex]C_A=C_{A0}exp(-kt)[/tex]

- 2nd: [tex]\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]

Explanation:

Hello,

For the ideal reaction A→B:

- Zeroth order rate law: in this case, we assume that the concentration of the reactants is not included in the rate law, therefore the integrated rate law is:

[tex]\frac{dC_A}{dt}=-k\\ \int\limits^{C_A}_{C_{A0}} {} \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\C_A-C_{A0}=-kt\\C_A=C_{A0}-kt[/tex]

- First order rate law: in this case, we assume that the concentration of the reactant is included lineally in the rate law, therefore the integrated rate law is:

[tex]\frac{dC_A}{dt}=-kC_A\\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A} } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\ln(\frac{C_{A}}{C_{A0}} )=-kt\\C_A=C_{A0}exp(-kt)[/tex]

- Second order rate law: in this case, we assume that the concentration of the reactant is squared in the rate law, therefore the integrated rate law is

[tex]\frac{dC_A}{dt}=-kC_A^{2} \\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A^{2} } } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\-\frac{1}{C_A}+\frac{1}{C_{A0}}=-kt\\\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]

Best regards.

Answer:

FeCl: Ferric Chloride (also called iron chloride), comes from Fe (ferrum, or iron), and Cl (Chlorine)

HNO: Nitroxyl, from N (Nitrogen), and the acidic nature of a radical ending in -yl.

NaSO:  Sodium sulfate, Na (Sodium), S (Sulfur), O (Oxygen).

SO: Sulfur monoxide (Mono-One), O (Oxygen) and S (Sulfur).

Answer:

a) b.i.d: twice daily.

b) 12 hour between each administration.

c) mg amoxicilin/administration = 413.9 mg/administration.

d) should be stored in the refrigerator

Explanation:

∴ b.i.d. : refer to twice a day; so 12 hours will pass between each administration of the medication.

⇒ mg amoxicilin/administration = 25 mg/ kg * 16.556 Kg = 413.9 mg amoxicilin.

Amoxicillin should be stored in the refrigerator, since in this section the temperature is kept within the storage range

Answer:

i) pH = 0.6990

ii) pH = 2.389

Explanation:

i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:

HCl + H₂O ⇒ H₃O⁺ + Cl⁻

The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.

The pH is related to the hydronium ion concentration as follows:

pH = -log([H₃O⁺]) = -log(0.2000) = 0.699

ii) Addition of NaOH causes the following reaction:

H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺

The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:

n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH

Thus, 4.800 mmol of H₃O⁺ were neutralized.

The initial amount of H₃O⁺ present was:

n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺

The amount of H₃O⁺ that remains after addition of NaOH is:

(5.000 mmol) - (4.800 mmol) = 0.2000 mmol

The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL

C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M

The pH is finally calculated:

pH = -log([H₃O⁺]) = -log(0.004082) = 2.389

To find the molar concentration of NaCl, subtract the tare weight from the total weight to get the mass of NaCl, calculate moles of NaCl, and divide by the solution volume. To estimate the standard deviation, propagate the errors from the mass and volume measurements according to the rules of error propagation. Specific numerical values for the standard deviation cannot be provided without exact formulas.

The question pertains to calculating the molar concentration of sodium chloride (NaCl) in a solution, and its associated standard deviation, given certain experimental measurements and potential error margins. First, the mass of NaCl added to the solution is found by subtracting the tare weight of the volumetric flask from the total weight after NaCl was added, yielding 8.84 g of NaCl. The molecular weight of NaCl is 58.44 g/mol, which allows determination of the moles of NaCl present.

To find the molar concentration, divide the moles of NaCl by the volume of the solution in liters (assuming the 100.0 mL flask volume as ideal, the error in volume would be considered in calculating the standard deviation, not the concentration itself). Then, to address the error margins, propagate the errors from the mass and volume measurements to estimate the standard deviation of the calculated concentration.

Note: Without specific formulas for error propagation and the exact calculation method for standard deviation provided in the question, a detailed numerical solution including the standard deviation calculation cannot be accurately presented. However, this process typically involves the square root of the sum of squared fractional uncertainties of the measurements involved.

Answer:

Stereochemistry is a branch of chemistry that studies the spatial arrangement of atoms or groups in a molecule.

The molecules with the same molecular formula, bond connectivity, and reactivity but a different arrangement of atoms in the space are known as stereoisomers. These molecules interact differently in a chiral environment or optical light.

Answer:

esters

Explanation:

The -OR group of the alcohol replaces the -OH of the carboxylic acid, forming an ester. See attachment for condensed mechanism.

To calculate the mass we use the following formulas:

PV=nRT     (1)

and

n = m / M    (2)

where:

P - pressure (atm)

V - volume (L)

n - moles

R - gas constant = 0.082 (L × atm) / (mol × K)

T - temperature (°K) (25°C + 273 = 298°K)

m - mass (g)

M - molecular mass (g/mole)

Now we rewrite equation (1):

n = PV / RT

And replace n with m / M from equation (2):

m / M = PV / RT

m = (P × V × M) / (R ×T)

1 L of He will have a mass of:

m = (1 × 1 × 4) / (0.082 × 298) = 0.1637 g

1 L of Cl₂ will have a mass of:

m = (1 × 1 × 71) / (0.082 × 298) = 2.9055 g

1.0 L of air will contain 0.79 L of N₂ and 0.21 L of O₂

0.79 L of N₂ will have a mass of:

m = (1 × 0.79 × 28) / (0.082 × 298) = 0.9052 g

0.21 L of O₂ will have a mass of:

m = (1 × 0.21 × 32) / (0.082 × 298) = 0.2750 g

mass of air = mass of N₂ + mass of O₂

mass of air = 0.9052 + 0.2750 = 1.1802 g

A balloon filed with helium will rise because as you see 1 L of helium is lighter than 1 L of air.

Chlorine gas is dangerous because chlorine is very toxic for human life and more of that is heavier than the air so will diffuse very hard from the area where the leak appeared.

Explanation:

The given data is as follows.

Boiling point of water ([tex]T^{o}_{b}) = 100^{o}C[/tex] = (100 + 273) K = 323 K,

Boiling point of solution ([tex]T_{b}) = 101.24^{o}C[/tex] = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

              [tex]\Delta T_{b} = (T_{b} - T^{o}_{b})[/tex]

                           = 374.24 K - 323 K

                           = 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

            Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]

Let molar mass of the solute is x grams.

Therefore,   Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]

                        m = [tex]\frac{288 g \times 1000}{x g \times 90}[/tex]              

                          = [tex]\frac{3200}{x}[/tex]

As,    [tex]\Delta T_{b} = k_{b} \times molality[/tex]

                 [tex]1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}[/tex]

                       x = [tex]\frac{0.512 ^{o}C/m \times 3200}{1.24}[/tex]

                          = 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is [tex]C_{n}H_{2n}O_{n}[/tex].

As, its empirical formula is [tex]CH_{2}O[/tex] and mass is 30 g/mol. Hence, calculate the value of n as follows.

                n = [tex]\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

                   = [tex]\frac{1321.29 g}{30 g/mol}[/tex]

                   = 44 mol

Thus, we can conclude that the formula of given material is [tex]C_{44}H_{88}O_{44}[/tex].

Answer:

Examples of storage polysaccharides - starch and glycogen and structural polysaccharides - cellulose and chitin

Explanation:

Polysaccharides are the complex carbohydrate polymers, composed of monosaccharide units that are joined together by glycosidic bond.

In other words, polysaccharides are the carbohydrate molecules that give monosaccharides or oligosaccharides on hydrolysis.

The examples of storage polysaccharides are starch and glycogen. The examples of structural polysaccharides are cellulose and chitin.

Answer:

0.8 pg/ml

Explanation:

To make the dilutions, you will take 1 ml of the original solution (tube 1) and add 4 ml of solvent. You will now have 20 pg per 5 ml of solution, so your new concentration will be 4 pg/ml (tube 2). Then you will repeat the process, so you will have 4 pg per 5 ml of solution, resulting in a concentration of 0.8 pg/ml (tube 3). The same process will be repeated for tubes 4 and 5.

Answer:

The concentration of the copper sulfate solution is 83 mM.

Explanation:

The absorbance of a copper sulfate solution can be calculated using Beer-Lambert Law:

A = ε . c . l

where

ε is the extinction coefficient of copper sulfate (ε = 12 M⁻¹.cm⁻¹)

c is its molar concentration (what we are looking for)

l is the pathlength (0.50 cm)

We can use this expression to find the molarity of this solution:

[tex]c=\frac{A}{\epsilon.l } =\frac{0.5}{12M^{-1}cm^{-1}0.50cm  } =0.083M=83mM[/tex]

The concentration of copper sulfate in the solution is calculated using Beer's law and the given values, resulting in a concentration of 0.042 M.

The concentration of copper sulfate in solution can be calculated using Beer's law, which is the relation A = εbc where A is absorbance, ε is the extinction coefficient, b is the path length, and c is the concentration. Given the question, the extinction coefficient for copper sulfate is 12 M-1.cm-1, the absorbance is 0.50, and the cuvette path length is 0.50 cm. To find the concentration c, rearrange the equation to c = A / (εb) and substitute into that the values provided, resulting in a concentration of 0.042 M for copper sulfate in the solution.

Answer:

The answers are:

a) 30°C; 303.15K

b) -30°C; 243.15K

c) 50°C; 323.15K

d) -40°C; 233.15K

e) 0°C; 273.15K

f) -273.15 °C ; 0K

Explanation:

To convert the temperature from ° F to ° C we use the following expression:

[tex]C=(F-32)\frac{5}{9}[/tex]

where C es temperature en °C and F is temperature in °F

To obtain the temperature in K we need to add 273.15 to each Celcius temperature

[tex]K=C+273.15[/tex]

Answer:

You need 0.8 ml of 5M stock solution and you have to add 9.2 ml of water.

Explanation:

Protocol solution (1X): 100 mM=0.1M

4X: 0.4M

The concentration of a solution is inversely proportional to the volume of a solution, so:

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

where:

M1= 5M stock solution

V1= amount of solution we need to collect

M2=4X solution

V2= 10 ml (volumen of 4X solution)

Therefore:

5M×V1=0.4M×10ml

V1={0.4M}{5M}10ml=0.8ml

[tex]5M*V_{1}=0.4M*10ml\\ V_{1}=\frac{0.4M}{5M}10ml=0.8ml[/tex]

To make a 10 ml solution we have to add 9.2 ml of water because V2 es 10 ml.

Answer:

H₂ gas

Explanation:

The reaction between nitrogen gas and hydrogen gas forms ammonia (the Haber-Bosch process):

N₂ + 3H₂ ⇒ 2NH₃

The excess reactant can be found by comparing the moles of nitrogen and hydrogen. The molar mass of N₂ is 28.00 g/mol and the molar mass of H₂ is 2.02 g/mol.

(100 kg N₂)(1000g/kg)(mol/28.00g) = 3570 mol

(100 kg H₂)(1000g/kg)(mol/2.02g) = 49500 mol

The molar ratio between the reactant N₂ and H₂ is 1N₂:3H₂. The moles of nitrogen required to react with H₂ is:

(49500 mol H₂)(1N₂ / 3H₂) = 16500 mol

The amount of nitrogen required is more than what is available, so nitrogen is the limiting reagent and hydrogen is the excess reagent.

Explanation:

It is known that density is the amount of mass present in liter of solution or substance.

Mathematically,      Density = [tex]\frac{mass}{volume}[/tex]

It is given that volume is 3.25 L and mass is [tex]4.10 \times 10^{3} g[/tex]. Hence, calculate the density of glycerol as follows.

                     Density = [tex]\frac{mass}{volume}[/tex]

                                   = [tex]\frac{4.10 \times 10^{3} g}{3.25 L}[/tex]

                                    = [tex]1.26 \times 10^{3} g/L[/tex]

As, 1 L = 1000 [tex]cm^{3}[/tex].

So,           [tex]1.26 \times 10^{3} g/L \times \frac{1000 cm^{3}}{1 L}[/tex]

                 = [tex]1260 \times 10^{6} g/cm^{3}[/tex]    

Thus, we can conclude that the density of glycerol is [tex]1260 \times 10^{6} g/cm^{3}[/tex].

Answer:

q = 2.343 W/m^2

Explanation:

Given data:

Ambient temperature =  0°C

Normal Body temperature = 37 °C

Thermal conductivity of tissue is K 0.3W/m °K

Heat transfer coefficient 20 W/m2 °K

Heat flux can be determined by using following formula

[tex]q = \frac{\Delta T}{\frac{c}{K} + \frac{1}{h}}[/tex]

putting all value to get flux value

[tex]q = \frac{37 -0}{\frac{4\times 10^{-3}}{0.3} + \frac{1}{20}}[/tex]

q = 2.343 W/m^2

Answer:

(a) The consistency as a function of time is C=0.15*t.

(b) The tank will become clogged in 24 minutes.

Explanation:

The rate of accumulation of the pulp stock can be defined as

[tex]\frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*C_{o}[/tex]

In this case, Co is 0, because the exit flow is only water and 0% fiber.

[tex]frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*0=Q_{i}*C_{i}[/tex]

Rearranging adn integrating

[tex]dC = (Q_{i}*C_{i})dt\\\int dC = \int (Q_{i}*C_{i})dt\\C=(Q_{i}*C_{i})*t+constant[/tex]

At t=0, C=0,

[tex]C=(Q_{i}*C_{i})*t+constant\\0=(Q_{i}*C_{i})*0+constant\\0=constant\\\\C=(Q_{i}*C_{i})*t[/tex]

[tex]C=(15*0.01)*t=0.15*t[/tex]

(b) The time at when the concentration reaches 6% is 0.4 hours or 24 minutes.

[tex]C=0.15*t\\0.06=0.15*t\\t=0.06/0.15=0.4[/tex]

The consistency of the pulp stock in the tank can be calculated using the equation of continuity. The consistency as a function of time can be determined by calculating the flow rate, density, mass flow rate, and the ratio of fiber mass to pulp stock mass. The graph of consistency as a function of time can be plotted using time intervals and consistency values.

The equation of continuity states that the mass flow rate into a volume has to equal the mass flow rate out of the volume. In this case, the tank is well-mixed and completely full, so the flow in equals the flow out. We can use this equation to calculate the consistency of the pulp stock in the tank as a function of time.

The flow rate can be calculated by multiplying the input flow rate (15 ft/hr) by the cross-sectional area of the tank (10 ft2). This gives us a flow rate of 150 ft3/hr.

The density of the pulp stock is given as 1% consistency. To convert this to density, we need to know the density of water, which is 62.4 lb/ft3. The density of the pulp stock can then be calculated as (0.01)(62.4 lb/ft3).

The mass flow rate can be calculated by multiplying the flow rate by the density. This gives us a mass flow rate of (150 ft3/hr)(0.01)(62.4 lb/ft3).

The consistency is the mass of fiber in the tank divided by the mass of the pulp stock in the tank. Since all of the fiber is kept in the tank, the consistency is equal to the mass of fiber divided by the mass flow rate. This can be calculated as 100 ft (the height of the tank) divided by the mass flow rate calculated in step 3.

To plot the consistency as a function of time, you can create a table with time intervals and calculate the consistency at each interval using the formula calculated in step 4. Then, plot the time intervals on the x-axis and the consistency values on the y-axis.

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Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

[tex]N=\frac{eq_{solute}}{V_{solution}}[/tex]

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

[tex]eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq[/tex]

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

[tex]m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4[/tex]

Best regards.

Answer:

For 0.1 M sodium acetate solution

if concentration of acid is 0.0025 then pH will  6.075

if concentration of acid is 0.005 then pH will  5.775

if concentration of acid is 0.01 then pH will  5.475

if concentration of acid is 0.05 then pH will  4.775

For 0.01 M sodium acetate solution

if concentration of acid is 0.0025 then pH will  5.075

if concentration of acid is 0.005 then pH will  4.775

if concentration of acid is 0.01 then pH will  4.475

if concentration of acid is 0.05 then pH will  3.775

Explanation:

to calculate the pH of a buffer solution we use the following formula

pH = pKa + log [B]/[A] ------------- eq (1)

[B] = concentration of base

[A] = concentration of acid

Given data

[B] = 0.1 M , 0.01M

[A] = 0.0025 M , 0.005 M, 0.01 M, 0.05 M

pKa value for sodium acetate is 4.75

1. First we will calculate the pH values for 0.1 M acetate solution.

If the concentration of acid is 0.0025, then:

[B] = 0.1 M

[A] = 0.0025 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.0025]

pH = 4.75 + log [40]

pH = 4.475 + 1.6

pH = 6.075

If the concentration of acid is 0.005 M, then:

[B] = 0.1 M

[A] = 0.005 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.005]

pH = 4.75 + log [20]

pH = 4.475 + 1.3

pH = 5.775

If the concentration of acid is 0.01, then:

[B] = 0.1 M

[A] = 0.01 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.01]

pH = 4.75 + log [10]

pH = 4.475 + 1

pH = 5.475

If the concentration of acid is 0.05, then:

[B] = 0.1 M

[A] = 0.05 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.05]

pH = 4.75 + log [2]

pH = 4.475 + 0.3

pH = 4.775

2. Now we will calculate the pH values for 0.01 M acetate solution.

If the concentration of acid is 0.0025, then:

[B] = 0.01 M

[A] = 0.0025 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.0025]

pH = 4.75 + log [4]

pH = 4.475 + 0.6

pH = 5.075

If the concentration of acid is 0.005 M, then:

[B] = 0.01 M

[A] = 0.005 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.005]

pH = 4.75 + log [2]

pH = 4.475 + 0.3

pH = 4.775

If the concentration of acid is 0.01 M, then:

[B] = 0.01 M

[A] = 0.01 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.01]

pH = 4.75 + log [1]

pH = 4.475 + 0

pH = 4.475

If the concentration of acid is 0.05 M, then:

[B] = 0.01 M

[A] = 0.05 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.05]

pH = 4.75 + log [0.2]

pH = 4.475 + (-0.7)

pH = 4.475 - 0.7

pH = 3.775

Answer:

a) pH = 4.71

b) pH = 4.704

c) pH = 4.57

d) No buffer here, the pH will be between 2-3

Explanation:

Applying Henderson Hasselbach equation:

pH = pKa + log([A]/[HA])

a) For 0.0025 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.0025 M of acid:

[A] = 0.05 - 0.0025 = 0.0475 M

[HA] = 0.05 + 0.0025 = 0.0525 M

[tex]pH=4.75+log(\frac{0.0475}{0.0525} )=4.71[/tex]

b) For 0.005 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.005 M of acid:

[A] = 0.05 - 0.005 = 0.0495 M

[HA] = 0.05 + 0.005 = 0.055 M

[tex]pH=4.75+log(\frac{0.0495}{0.055} )=4.704[/tex]

c) For 0.01 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.01 M of acid:

[A] = 0.05 - 0.01 = 0.04 M

[HA] = 0.05 + 0.01 = 0.06 M

[tex]pH=4.75+log(\frac{0.04}{0.06} )=4.57[/tex]

d) For 0.05 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.05 M of acid:

[A] = 0.05 - 0.05 = 0

[HA] = 0.05 + 0.05 = 0.1 M

No buffer here, the pH will be between 2-3

Final answer:

The D and L stereochemical descriptors are used to represent the configuration of stereoisomers in monosaccharides. The D- or L- designation is based on the position of the -OH group on the penultimate carbon in the Fisher projection. The D-configuration is commonly found in nature and only dextrorotary amino acids are used by cells to build proteins.

Explanation:

The D and L stereochemical descriptors are used to represent the configuration of stereoisomers, specifically in the context of monosaccharides or sugars. The designation of D or L is based on the position of the -OH group on the second-last carbon (penultimate C) in the Fisher projection. If the -OH group is on the right side, it is assigned D-configuration, and if it is on the left side, it is assigned L-configuration.

These descriptors do not indicate the rotation of plane polarized light, but purely define the configuration. Enantiomers that are D- and L- pairs have the same common name, with the D- or L- designation indicating their configuration. It's important to note that the D- and L- designation does not always correlate with the dextro/levo rotatory nature of the enantiomers in a polarimeter.

For example, D-glucose and L-glucose are enantiomers with the penultimate C defining D- or L-configuration. The D-configuration is commonly found in nature, and only dextrorotary d amino acids (L amino acids) are used by cells to build polypeptides and proteins.