briefly explain the difference between an isotope and two species that are isoelectronic provide an example for an isotope and isoelectronic species to argon-40

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: 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:

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.

Final answer:

The dipole moment points toward the more electronegative atom, Cl, in a C-Cl bond. The greater the electronegativity difference, the larger the dipole moment. Thus, (A) C←Cl with an electronegativity difference > 0.5 correctly predicts the dipole direction and suggests a significant dipole moment.

Explanation:

The direction and relative value of the dipole moment in a bond between two atoms depends on the difference in their electronegativities. Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. When two atoms form a bond, the more electronegative atom will attract the bonding electrons more strongly and will acquire a partial negative charge, while the less electronegative atom will have a partial positive charge.

Regarding the options given:
(A) C←Cl, with an electronegativity difference > 0.5 would have a dipole moment pointing towards the Cl, because Cl is more electronegative than C.
(B) C←Cl, with an electronegativity difference < 0.5 might suggest a relatively nonpolar bond, which is uncommon for C-Cl and might not be a realistic scenario.
(C) C→Cl, with an electronegativity difference > 0.5 would incorrectly suggest that C is more electronegative than Cl, which is not the case; thus this depiction of the dipole direction is incorrect.
(D) C→Cl, with an electronegativity difference < 0.5 would also be incorrect as it suggests the incorrect direction of the dipole moment.

The relative value of the dipole moment will be greater when the electronegativity difference is greater, leading to a stronger separation of charges, and hence a larger dipole moment.

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:

MPR=90,36%

Explanation:

The recrystallization is a purification process where the solid to purify is dissolved in an appropriated dissolvent and then, changing the conditions the solubility changes and that solid (that was in solution before) precipitates and form crystals.

In this case, for boiling water 5.50 g of  acetanilide could be dissolved and then cold the water, so the mass of crystals formed will be  

[tex]m_{crystal}=\left(5.5-0.53\right)g=4.97g[/tex]

The maximum percent recovery is then

[tex]MPR=\frac{mass\ of\ solid\ recovered}{mass\ of\ solid\ disolved}\times 100\% = \frac{4.97g}{5.5g}\times 100\%=90,36\%[/tex]

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

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:

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:

4714.950 kilograms  is the mass of a sample containing 45.0 kmol of methyl acetate.

Explanation:

Moles of methyl acetate =[tex]n_1[/tex]=45.0 kmol= 45000 mol

Mole percentage of methyl acetate = 60.0%

Total moles in the sample = n

[tex]60.0\%=\frac{45000 mol}{n}\times 100[/tex]

[tex]n=\frac{45000 mol}{60.0}\times 100=75000 mol[/tex]

Mole percentage of methyl alcohol = 20.0%

Moles of methyl alcohol = n_2

[tex]20.0\%=\frac{n_2}{75000 mol}\times 100[/tex]

[tex]n_2=15,000 mol[/tex]

Mass of methyl alcohol =  [tex]n_2\times 32.04 g/mol[/tex]

=[tex]15000 mol\times 32.04 g/mol=480,600 g[/tex]

Mole percentage of acetic acid  = 20.0%

Moles of acetic acid = n_3

[tex]20.0\%=\frac{n_3}{75000 mol}\times 100[/tex]

[tex]n_3=15,000 mol[/tex]

Mass of acetic acid= [tex]n_3\times 60.05 g/mol[/tex]

[tex]15000 mol\times 60.05 g/mol=900,750 g[/tex]

Mass of methyl methyl acetate= [tex]n_1\times 74.08 g/mol[/tex]

[tex]45000 mol\times 74.08 g/mol =3,333,600 g[/tex]

Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g

4714950 g = 4714.950 kg

(1 kg = 1000 g)

To find the mass of a sample with 45.0 kmol of methyl acetate, multiply the kmol amount by the molar mass of methyl acetate (74.08 g/mol), resulting in 3333.6 kg.

To calculate the mass of a sample containing 45.0 kmol (kilo moles) of methyl acetate, we first need to know the molar mass of methyl acetate. Methyl acetate (C3H6O2) has a molar mass of approximately 74.08 g/mol. Knowing this, we can calculate the mass of the methyl acetate in the sample.

The calculation is as follows:

This calculation reveals that a sample containing 45.0 kmol of methyl acetate has a mass of 3333.6 kilograms.

Answer:

Explanation has been given below.

Explanation:

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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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:

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

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:

[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:

You need to add 203 mL of 1,00M KH₂PO₄ and 147 mL of 1,00M K₂HPO₄.

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for 7,07 phosphate buffer is:

7,07 = 7,21 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]

0,7244 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex] (1)

As the buffer concentration must be 1,00 M:

1,00 = [H₂PO₄⁻] + [HPO4²] (2)

Replacing (2) in (1):

[H₂PO₄⁻] = 0,5799 M

Thus:

[HPO4²] = 0,4201 M

To obtain these concentrations you need to add:

0,5799 M × 0,350 L × [tex]\frac{1L}{1mol}[/tex] = 0,203 L ≡ 203 mL of 1,00M KH₂PO₄

And:

0,4201 M × 0,350 L × [tex]\frac{1L}{1mol}[/tex] = 0,203 L ≡ 147 mL of 1,00M K₂HPO₄

I hope it helps!

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:

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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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:

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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Explanation:

The given data is as follows.

       Mass of a lead atom = [tex]3.14 \times 10^{-22}[/tex]

       Volume = 2.00 [tex]cm^{3}[/tex]

        Density = 11.3 [tex]g/cm^{3}[/tex]

As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.

So, in 2 cubic centimeter there will be [tex]2 \times 11.3 g = 22.6 g[/tex] of lead atoms.

One lead atom has a mass of [tex]3.14 \times 10^{-22}[/tex]. Therefore, number of atoms present in 22.6 g of lead will be as follows.

                [tex]\frac{22.6 g}{3.14 \times 10^{-22}}[/tex]

                  = [tex]7.197 \times 10^{22} atoms[/tex]

Thus, we can conclude that there are [tex]7.197 \times 10^{22} atoms[/tex] of lead are present.

Answer:

(B) F⁻, HCOOH

Explanation:

(A) CH₄, HCOOH

(B) F⁻, HCOOH

(C) F⁻, CH₃-O-CH₃

The hydrogen bonds are formed when the hydrogen is found between two electronegative atoms such as oxygen (O), nitrogen (N) or florine (F).

O····H-O, F····H-O, O····H-N

(A) CH₄, HCOOH

- here methane CH₄ is not capable to form hydrogen bond with water

- formic acid HCOOH can form hydrogen bonds with water

H-C(=O)-O-H····OH₂

(B) F⁻, HCOOH

-both floride (F⁻) and formic acid can form hydrogen bonds with water

F····OH₂

H-C(=O)-O-H····OH₂

(C) F⁻, CH₃-O-CH₃

-  dimethyl-ether CH₃-O-CH₃ is not capable to form hydrogen bond with water

- floride (F⁻) can form hydrogen bonds with water

F····OH₂

The correct pairs of species that can form hydrogen bonds with water are option (B) F, HCOOH and option (C) F.CH, OCH. This is because hydrogen bonds are formed between hydrogen and a highly electronegative atom such as Oxygen, Nitrogen, or Fluorine.

The correct pairs of species that can form hydrogen bonds with water are option (B) F, HCOOH and option (C) F .CH, OCH. Hydrogen bonds are primarily formed between hydrogen and a highly electronegative atom such as Oxygen, Nitrogen, or Fluorine, which are present in both HCOOH (formic acid) and OCH (a group from a larger molecule such as methanol).

Hydrogen bonds form due to the attraction between the slightly positive Hydrogen of one molecule and the slightly negative Oxygen, Nitrogen, or Fluorine of another. For example, in a water molecule, the oxygen atom carries a slight negative charge due to its higher electronegativity while hydrogen atoms carry a slight positive charge.

Option (A) CH cannot form a hydrogen bond with water as it is a nonpolar molecule and lacks an electronegative atom. Also, individual fluorine atoms as given in option (B) do not form hydrogen bonds as they lack the H-F bond necessary for hydrogen bonding.

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Answer:

The correct option is: c) The formation of an oxide layer inhibits further diffusion and corrosion

Explanation:

Aluminium is a chemical element which belongs to the group 13 of the periodic table and has atomic number 13. It is a soft metal and a member of the p-block.

When aluminium is exposed to the normal atmosphere, the top layer of the metal gets oxidized to form a thin protective layer of aluminium oxide. The thin and hard protective aluminium oxide layer then protects the metal from getting further corroded. This process is called passivation.

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]

where M is magnification, n is grain size

therefore we have

[tex]64 = 2^{7-1}(\frac{100}{M})^2[/tex]

solving for M we get

M = 100 X

B)  total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]

where M is magnification, n is grain size

therefore we have[tex]500 = 2^{7-1}(\frac{100}{M})^2[/tex]

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

[tex]Nm = 2^{n-1} (\frac{100}{M})^2[/tex]

where M is magnification, n is grain size

therefore we have[tex]20 = 2^{7-1}(\frac{100}{M})^2[/tex]

solving for M we get

M = 178.88 X

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

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 : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = [tex]\frac{x}{100}[/tex]

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = [tex]\frac{10}{100}[/tex]

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = [tex]\frac{(90-x)}{100}[/tex]

Now put all the given values in above formula, we get:

[tex]48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})][/tex]

[tex]x=78\%[/tex]

Therefore, the percent abundance of the heaviest isotope is, 78 %

Answer:

78 %

Explanation:

The atomic mass is the weighted average of the atomic masses of each isotope.

In a weighted average, we multiply each value by a number representing its relative importance.

In this problem, the percent abundance represents the relative importance of each isotope.

Data:

X-47: mass = 47.00 u; abundance = 10.0 % = 0.100  

X-48: mass = 48.00 u  

X-49: mass = 49.00 u

Calculations:

                 Let x = abundance of X-49

Then 0.900 - x = abundance of X-48

[tex]\begin{array}{cccc}\\\textbf{Isotope} & \textbf{Mass/u} & \textbf{Abundance} & \textbf{Contribution/u}\\\text{X-47} & 47.00 & 0.100 & 4.700\\\text{X-48} & 48.00 & 0.900 - x & 48.00(0.900 - x)\\\text{X-49} & 49.00 & x & 49.00x\\& \text{TOTAL} & = & \mathbf{48.68}\\\end{array}[/tex]

[tex]\begin{array}{rcr}4.700 + 48.00(0.900 - x) + 49.00x & = & 48.68\\4.700 + 43.20 - 48.00x + 49.00x & = & 48.68\\47.90 +x & = & 48.68\\x & = & \mathbf{0.78}\\\end{array}[/tex]

The heaviest isotope has an abundance of 78 %.

Answer:

The density of the solid is 1,316 g/mL

Explanation:

The weight of both Toluene and the solid insoluble is 58,68 g. And the weight of the solid is 32,50 g. Thus, weight of toluene is:

58,68 g - 32,50 g = 26,18 g of Toluene

To know how much volume that toluene occupy you must use density thus:

26,18 g of toluene × ( 1 mL / 0,864 g) = 30,30 mL of toluene

The volume of both Toluene and the solid is 55,00 mL and the volume of toluene is 30,30 mL. Thus, the volume of the solid is:

55,00 mL - 30,30 mL = 24,70 mL

Knowing both volume and weight it is possible to know the density thus:

32,50 g / 24,70 mL = 1,316 g/mL

I hope it helps!

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.