What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A) 2.18x10 (B) 7.84*10J (C) 8.62x10 J (D) 5.34x10 3 Answer A

Final answer:

An interpretation of the results of many tests is called a theory. It is a well-substantiated explanation of an aspect of the natural world, based on extensive research and experimentation.

Explanation:

An interpretation of the results of many tests is called a theory. A theory is a well-substantiated explanation of an aspect of the natural world. It is a scientific explanation that has been repeatedly tested and supported by many experiments.

For example, the theory of evolution is a well-supported explanation of how species have evolved over time. It is based on extensive research and experimentation.

It is important to note that a theory in science is different from the everyday use of the word, which often refers to a guess or speculation. In science, a theory is a well-tested and supported explanation.

The distance travelled by charlotte is 7269ft/mi.

Given:

Charlotte's speed = 70.4 mi/h.

Time she takes her eyes off the road = 4.54 seconds.

To calculate the distance Charlotte has traveled in feet while looking at her phone, it is important to convert her speed from miles per hour (mi/h) to feet per second (ft/s), and then use that speed to calculate the distance.

Convertion of speed to feet per second:

1 mile = 5280 feet

1 hour = 3600 seconds

Speed in ft = (70.4 (mi/h) × 5280(ft/mi)) / (3600(s/h))

Speed in ft = 103.25ft.

Calculation of distance:

Distance =Speed / Time

Distance = 103.25 * 70.4

Distance = 7269 ft/mi

Therefore, the distance travelled by charlotte is 7269ft/mi.

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Charlotte traveled approximately 469.761 feet while looking at her phone for 4.54 seconds by driving at a speed of 70.4 mi/h.

Charlotte is driving at 70.4 mi/h. To find out how far she has traveled in feet while looking at her phone for 4.54 seconds, we need to convert the speed from miles per hour to feet per second and then multiply by the time in seconds.

First, we convert the speed from miles per hour to feet per second using the conversion factors 1 mile = 5280 feet and 1 hour = 3600 seconds:

70.4 mi/h × 5280 ft/mi × 1/3600 h/s = 103.46667 ft/s.

Now, we multiply the speed in feet per second by the time she is distracted:

103.46667 ft/s × 4.54 s = 469.761 ft.

Charlotte has therefore traveled approximately 469.761 feet while looking down at her phone for 4.54 second

Answer:

You need 8,53 L of ammonia

Explanation:

Global reaction of remotion of nitrogen oxide with ammonia is:

4 NH₃ + 6 NO ⇒ 5 N₂ + 6 H₂O

This balanced equation shows that 4 NH₃ moles reacts with 6 NO moles.

With 100% yield and temperature and pressure constants it is possible to apply Avogadro's law. This law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The formula is:

[tex]\frac{V_{1} }{n_{1} } = \frac{V_{2} }{n_{2} }[/tex]

Where:

V₁ is the NO volume =  12,8L

n₁ are NO moles = 6

n₂ are NH₃ moles = 4

V₂ is NH₃ volume, the unknown.

Thus, V₂ are 8,53 L of ammonia

I hope it helps!

Answer:

Total feed rate = 98.3 lbmol/h

methanol mole fraction = 0.315

water mole fraction = 0.685

Explanation:

First of all, it is needed to calculate the feed mass of methanol and water in kg/h.

For methanol:

[tex]m_{methanol} = m\%wt_{methanol}/100 = (1000kg/h)(45\%)/100[/tex]

[tex]m_{methanol} = 450kg/h[/tex]

For water:

[tex]m_{water} = m\%wt_{water}/100 = (1000kg/h)(55\%)/100[/tex]

[tex]m_{water} = 550 kg/h[/tex]

Now, change from mass units (kg/h) to moles units (kmol/h and lbmol/h) using simple conversion factors:

For methanol:

[tex]n_{methanol} = (450\frac{kg}{h})(\frac{1 kmol}{32 kg} )[/tex]

[tex]n_{methanol} = 14.1kmol/h[/tex]

For water:

[tex]n_{water} = (550\frac{kg}{h})(\frac{1 kmol}{18 kg} )[/tex]

[tex]n_{water} = 30.6kmol/h[/tex]

Change units from kmol/h to lbmol/h

For methanol:

[tex]n_{methanol} = (14.1\frac{kmol}{h})(\frac{1 lbmol}{0.454 kmol} )[/tex]

[tex]n_{methanol} = 31.0 lbmol/h[/tex]

For water:

[tex]n_{water} = (30.6\frac{kg}{h})(\frac{1 lbmol}{0.454 kmol} )[/tex]

[tex]n_{water} = 67.3 lbmol/h[/tex]

Sum moles of methanol and water in lbmol/h to compute the total feed rate:

[tex]n = 31.0 lbmol/h + 67.3 lbmol/h[/tex]

[tex]n = 98.3 lbmol/h[/tex]

Divide both methanol and water moles feed rates by total feed rate:

For methanol:

[tex]X_{methanol} = \frac{31.0 lbmol/h}{98.3 lbmol/h}[/tex]

[tex]X_{methanol} =  0.315[/tex]

For water:

[tex][X_{water} =  \frac{67.3 lbmol/h}{98.3 lbmol/h}[/tex]

[tex]X_{water} =  0.685[/tex]

End

Answer: The moles of carbon dioxide gas formed is 12.33 moles.

Explanation:

We are given:

Moles of oxygen gas = 12.33 moles

The chemical equation for the reaction of carbon and oxygen to produce carbon dioxide follows:

[tex]C+O_2\rightarrow CO_2[/tex]

By stoichiometry of the reaction;

1 mole of oxygen gas produces 1 mole of carbon dioxide gas.

So, 12.33 moles of oxygen gas will produce = [tex]\frac{1}{1}\times 12.33=12.33mol[/tex] of carbon dioxide gas.

Hence, the moles of carbon dioxide gas formed is 12.33 moles.

Answer:

At a pressure of I atm, the boiling point of a mixture of aniline and water is b) 100°C.

Explanation:

Assuming that aniline is insoluble in water there will be no interaction between liquids. There will be no positive interactions that can increase the boiling point or negative interactions that can decrease the boiling point. Thus the boiling points of each substance will not be affected. Since the boiling point of water is 100ºC, the mixture will start to evaporate at 100ºC.

Answer:

B.)

Explanation:

Answer:

Molarity of a solution that contains 3.11 mol of NaNO3 is 1,24 M

Explanation:

We understand molarity as the number of moles of solute that are contained in 1 L of solution, then if in a solution of 2.50 L we have 3.11 moles, it remains to calculate how many moles do we have in 1 liter.

2,50 L .......... 3,11 moles

1 L .................. x

X = ( 1 L x 3,11 moles) / 2,50 L = 1,24

Answer:

Part a:

Isobars: Ca-41, K-41 and Ar-41

Part b:

Number of proton

Part c:

Incorrect statement

Explanation:

Part a:

Nuclei with same mass number are called isobars.

Given:

Ca-40, Ca-41, K-41 and Ar-41

Ca-41, K-41 and Ar-41 have equal mass numbers, so these are isobars.

Part b:

Atomic number of Ca = 20

Atomic no. = Number of proton.

So, Ca-40 ans C-41 have same number of proton or in other words proton count is common in both.

Part c:

Number of neutrons in Ca-41

Atomic number = 20

Mass number = 41

Number of neutron = Mass number -atomic number

                                = 41 -20 = 21

Number of neutrons in K-41

Atomic number = 19

Mass number = 41

Number of neutron = Mass number -atomic number

                                = 41 -19 = 22

Number of neutrons in Ar-41

Atomic number = 18

Mass number = 41

Number of neutron = Mass number -atomic number

                                = 41 -18 = 23

So, the statement is incorrect.

The isobars among the examples given are Ca-41, K-41, and Ar-41. Ca-40 and Ca-41 are isotopes of calcium with different neutron counts. The statement about Ca-41, K-41, and Ar-41 having the same number of neutrons is incorrect because they have different atomic numbers.

Nuclei with the same mass number but different atomic numbers are known as isobars. This means that while they have the same mass number (sum of protons and neutrons), they belong to different elements (different atomic numbers). Consequently, Ca-41, K-41, and Ar-41 are isobars since they all have a mass number of 41 but belong to different elements with atomic numbers 20 (calcium), 19 (potassium), and 18 (argon), respectively. Nevertheless, Ca-40 is not an isobar to these because it has a different mass number.

Isotopes are atoms with the same atomic number but differing numbers of neutrons, resulting in different mass numbers. In this case, Ca-40 and Ca-41 are isotopes because they are both calcium atoms (same atomic number of 20) with different mass numbers.

The commonality between Ca-40 and Ca-41 is they are isotopes of calcium and thus share the same atomic number and chemical properties, despite having different mass numbers due to different neutron counts.

The statement 'Atoms of Ca-41, K-41, and Ar-41 have the same number of neutrons' is incorrect because while these isotopes have the same mass number, they have different numbers of protons. To calculate the number of neutrons in these isotopes, you subtract the atomic number from the mass number and because the atomic numbers are different, the resulting neutron numbers will also differ.

Answer:

[tex]\Delta H_{rxn}[/tex] for the given reaction is -890.3 kJ

Explanation:

[tex]\Delta H_{rxn}=\sum n_{i}.\Delta H_{f}(product)_{i}-\sum n_{j}.\Delta H_{f}(reactant)_{j}[/tex]

where [tex]n_{i}[/tex] and [tex]n_{j}[/tex] represents number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence [tex]\Delta H_{rxn}=[1mol\times \Delta H_{f}(CO_{2})_{g}]+[2mol\times \Delta H_{f}(H_{2}O)_{l}]-[1mol\times \Delta H_{f}(CH_{4})_{g}]-[2mol\times \Delta H_{f}(O_{2})_{g}][/tex]

so, [tex]\Delta H_{rxn}=[1mol\times -393.5kJ/mol]+[2mol\times -285.8kJ/mol]-[1mol\times -74.8kJ/mol]+[2mol\times 0kJ/mol]=-890.3 kJ[/tex]

So, Correct answer is -890.3 kJ

Answer:

E) More neutrons in the nucleus

Explanation:

Isotope -

The atoms which have same number of protons but different number of neutrons.

And since , atomic number = protons number

and , mass number = proton + neutrons .

Hence ,

The atomic number will not change , but the mass number will change .

Hence , the correct option is more neutrons in the nucleus .

Answer:

OH-

Explanation:

H2O can act as base or acid.

When H2O acts as a base the conjugate acid is H3O+

When H2O acts as an acid the conjugate base is OH-

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

[tex]r = k [A]^{x} [B]^{y}[/tex]

where:

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

[tex]v(t) = -\frac{d[A]}{dt} = k [A]^{n}[/tex]

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law:                                    rate = k

Concentration-time Equation:   [A]=[A]o - kt

where

For first-order kinetics, we have:

Rate Law:                                        rate= k[A]

Concentration -Time Equation:      ln[A]=ln[A]o - kt

where:

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

Explanation:

A mixture is material which is composed up of two or more substances and these substances are physically combined. The identities of each specie in the mixture are retained.

A solution is a type of the homogeneous mixture which is composed of two or more substances. The specie which is dissolved into the another substance is known as solute and in which it is dissolved is known as a solvent.

A compound is the substance which is formed when two or more elements are bonded together chemically.

A molecules are electrically neutral and is a combination of of two or more atoms which are held together by means of chemical bonds.

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species.

The reduction of permanganate ion to manganese ion in acidic solution is represented by the balanced half-reaction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l). It shows the permanganate ion gaining 5 electrons to become a manganese(II) ion, with the protons reacting with permanganate to form water.

The reduction of the permanganate ion (MnO⁴⁻) to the manganese(II) ion (Mn²⁺) in acidic solution can be illustrated with the following balanced half-reaction:

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

This equation indicates that the permanganate ion gains 5 electrons (is reduced) in an acidic solution to result in a manganese(II) ion. The 8 protons react with permanganate to form 4 molecules of water, which is also reflected in the equation.

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The reduction half-reaction for permanganate to manganese ion in an acidic solution is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O(l), where oxygen and hydrogen are balanced by adding water and hydrogen ions, respectively, and the charge is balanced by adding electrons.

To write a balanced half-reaction for the reduction of the permanganate ion (MnO4-) to manganese ion (Mn2+) in an acidic aqueous solution, we start with the half-reaction:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O(l)

In this balanced equation, the oxygen atoms are balanced by adding water molecules (H2O) to the right side, and the hydrogen atoms are balanced by adding hydrogen ions (H+) to the left side. Additionally, 5 electrons (5e^-) are added to the left side to ensure that the overall charge is balanced between the reactant and product sides of the equation.

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Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii)  truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna:  number of trucks needed/week= 134400 / 30000  = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks  

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

Answer: [tex]41.4\times 10^{6}[/tex]

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s preceding the first integers are never significant.

All zero’s after the decimal point are always significant.

Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power.  It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.

The engineering notation written in the form:

[tex]a\times 10^b[/tex]

where,

a = the number which is greater than 0 and less than 999

b = an integer multiple of 3

Now converting the given value of 41,405,000 into engineering notation, we get [tex]41.4\times 10^{6}[/tex]

Therefore the scientific notation with 3 significant figures is [tex]41.4\times 10^{6}[/tex]

Based on conversion ratios, 47 mg of sodium would be:

1g is 1,000 grams.

47 mg would be:

= 0.47 / 1,000

= 0.047 grams

16 ounces is 453.6 grams so 0.047 grams would be:

= (0.047 x 16) / 453.6

= 0.00166 ounces.

1 pound is 453.6 grams so 0.047 grams would be:
= 0.047 / 453.6

= 0.000104 pounds

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To convert 47 mg of sodium to grams, divide by 1000. To convert grams to ounces, use the conversion factor 1 oz = 28.35 g. To convert grams to pounds, use the conversion factor 1 lb = 453.6 g.

To convert 47 mg to grams, divide by 1000:

47 mg = 47/1000 = 0.047 grams

To convert grams to ounces, use the conversion factor:

0.047 g x (1 oz/28.35 g) = 0.00166 oz

To convert grams to pounds, use the conversion factor:

0.047 g x (1 lb/453.6 g) = 0.0001035 lb

The rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g) based on the input data is rate = k[BF3][NH3], where k is the rate constant, [BF3] and [NH3] denote the molar concentrations of BF3 and NH3, indicating a first order relationship for both reactants.

In order to determine the rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g), we have to look at how the initial rate changes with respect to the change in concentration of the reactants. Using the given data, we can compare experiments where only one reactant's concentration is changed while the concentration of the other reactant remains constant. Upon analysis, we see that when the concentration of BF3 doubles, the rate also doubles, suggesting a first order relationship. Similarly, when the concentration of NH3 doubles, the rate doubles, indicating a first order relationship for NH3 as well. Hence, given that both BF3 and NH3 are first order, the rate law for the reaction should be: rate = k[BF3][NH3].

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The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) is rate = k[BF3][NH3].

The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) can be determined by analyzing the effect of changing concentrations on the initial rate of the reaction. By comparing the rates of reaction for different experiments, we can observe how changing the concentrations of the reactants affects the rate. In this case, it is clear that the rate of reaction is directly proportional to the concentration of BF3 and NH3, so the rate law for the reaction is rate = k[BF3][NH3].

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Infrared spectroscopy is a technique used to identify functional groups in molecules by measuring absorption frequencies. It helps determine the presence or absence of specific groups in a molecule.

Infrared spectroscopy is a technique used to identify different functional groups in molecules based on their characteristic absorption frequencies. It involves passing infrared light through a sample and measuring the frequencies that are absorbed. Different functional groups have different absorption frequencies, allowing scientists to determine the presence or absence of specific groups in a molecule. For example, if a sample absorbs in the carbonyl frequency range but not in the alkyne range, it can be inferred that the molecule contains a carbonyl group but not an alkyne.

Answer:

Because for dilute and aqueous solutions the mass of solvent will be a very close value to the volume of solution.

Explanation:

Molar concentration is defined as:

[tex][M]=\frac{molessolute}{volumesolution}[/tex]

And molal concentration is defined as:

[tex][m]=\frac{molessolute}{kgsolvent}[/tex]

And:

Msolution=Msolute+Msolvent

For dilute solutions, we have small amounts of solute, then we have:

Msolution=Msolute+Msolvent, and as the mass of  solute is very small: Msolution≅Msolvent

If the solution is also aqueous (water as solvent), and considering that the density of water is around 1 gm/cm3 or 1 kg/m3:

Msolvent≅Msolution≅Vsolution

Therefore, if we look to the molar and molal equations, we have the same numerator in both (moles of solute) and nearby numbers for the denominator, giving to the molar and molal concentration close values.  

Final answer:

To prepare a 0.05 M magnesium sulfate heptahydrate solution with a volume of 200 mL, you will need 2.46 grams of magnesium sulfate heptahydrate.

Explanation:

To calculate the amount of magnesium sulfate heptahydrate required to prepare 200 mL of 0.05 M solution, we need to use the formula:

moles = molarity x volume

First, we convert the volume from milliliters to liters:

200 mL x (1 L/1000 mL) = 0.2 L

Next, we substitute the given values into the formula:

moles = 0.05 mol/L x 0.2 L = 0.01 mol

Finally, we calculate the mass of magnesium sulfate heptahydrate using its molar mass:

moles = mass (g) / molar mass (g/mol)

0.01 mol = mass (g) / 246.48 g/mol

mass (g) = 0.01 mol x 246.48 g/mol = 2.46 g

The mass of 11 nanomoles of the substance is 616 nanograms.

Given that one mole of a substance has a mass of 56.0 g, we can calculate the mass of 11 nanomoles of the substance as follows:

1 mole = 56.0 g

1 nanomole = 56.0 g / 1,000,000,000

=[tex]5.60 * 10^{-8} g[/tex]

Now, to find the mass of 11 nanomoles:

Mass = [tex]11 nanomoles * 5.60 * 10^{-8}g/nanomole[/tex]

Calculating this gives us:

Mass =[tex]6.16 * 10^{-7} g[/tex]

To express the answer in nanograms, we need to convert grams to nanograms:

1 g = 1,000,000,000 ng

So, [tex]6.16 * 10^{-7} g = 6.16 * 10^2 ng = 616 ng[/tex]

Therefore, the mass of 11 nanomoles of the substance is 616 nanograms.

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To find the mass of 11 nanomoles of a substance, multiply the number of nanomoles by the molar mass of the substance. In this case, the mass is 616.0 ng.

To calculate the mass of 11 nanomoles of a substance, we need to know the molar mass of the substance. If one mole of the substance has a mass of 56.0 g, then the molar mass is 56.0 g/mol. To find the mass of 11 nanomoles, we can use the conversion factor:

11 nmol * 56.0 g/mol = 616.0 ng

Therefore, the mass of 11 nanomoles of the substance is 616.0 nanograms (ng).

To round 129.752416 to the requested number of significant figures:
- For 3 significant figures, round down to 129.7.
- For 4 significant figures, round up to 129.8.
- For 6 and 7 significant figures, keep the given digits.

To round 129.752416 to the requested number of significant figures:

Answer:

Mass of sea food = 30.98 Kg

Mass of sea food in pound = 68.31 lbs

Explanation:

Salmon, crab and oysters all are sea food.

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

Mass of salmon = 22 kg

Mass of crab = 5.5 kg

Mass of oysters = 3.48 kg

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

                             = 22 + 5.5 + 3.48

                             = 30.98 Kg

1 Kg = 2.205 lbs

Therefore, 30.98 kg = 30.98 × 2.205

                                 = 68.31 lbs

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

Final answer:

The rate of disappearance of NO2 in the experiment at 300 °C is approximately 1.32 × 10^-5 M/s, calculated by the change in concentration over time.

Explanation:

The rate of disappearance of NO2 in the experiment at 300 °C is calculated by determining the change in concentration over the change in time. The initial concentration of NO2 is 0.0138 M and it drops to 0.00886 M over 374 s. The change in concentration (Δ[NO2]) is 0.0138 M - 0.00886 M = 0.00494 M. The rate of disappearance is then Δ[NO2]/Δt = 0.00494 M/374 s.

After calculating this, we get the rate of disappearance of NO2 to be approximately 1.32 × 10-5 M/s, which is one of the possible choices provided in the question.

Answer:

The mechanisms for the elimination reactions between NaOH in ethanol and the halogenoalkanes are demonstrated in the figure attached.

Explanation:

(i) 1-bromobutane will suffer elimination to for an alkene. The mechanism will be E2, which means that the attack and the elimination will occur simultaneously. This is the preferred mechanism because the bromine is in a primary carbon.

(ii) 2-bromo-2-methylpentane will suffer elimination to for an alkene. The mechanism will be E1, which means that the attack and the elimination will occur in two different steps. The bromine will be eliminated in the first step with the formation of a carbocation and in a second step the double bond will be formed after the anionic attack. This is the preferred mechanism because the bromine is in a terciary carbon which is able to stabilize the carbocation formed.

Answer:

Kc = 168.0749

Explanation:

initial mol:   0.822               0          0

equil. mol: 2(0.822 - x)         x           x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

⇒ Kc = 168.0749

The equilibrium constant (Kc) for the reaction where 2HI decomposes into H2 and I2 at a given temperature can be calculated using the ICE method. After setting up the Kc expression and solving for x, the concentration changes of HI, H2, and I2 can be used to find Kc, which is approximately 0.1176.

To calculate the equilibrium constant (Kc) for the decomposition of hydrogen iodide (HI) into hydrogen gas (H₂) and iodine gas (I₂), we must first write the Kc expression for the reaction:

2HI (g) ⇌ H₂ (g) + I₂ (g)

The Kc expression is Kc = [H₂][I₂] / [HI]^2.

Next, we use an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations throughout the reaction:

To calculate x, solve the equation 0.7409 - 2x = 0.055, yielding x = 0.34295 M. Now we know at equilibrium:

Inserting these values into the Kc expression we get:Kc = (0.34295)(0.34295) / (0.055)^2

Simplifying gives us:Kc = 0.1176 to 4 decimal places.

Answer:

Carbonic acid/bicarbonate; bicarbonate  

Explanation:

H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻; pKₐ₁ =   6.35

HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃²⁻;  pKₐ₂ = 10.33

[tex]\text{pH}& = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\7.4 & = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}[/tex]

The best buffer is one for which pKₐ ≈ pH.  

6.35 is closer to 7.4, so the carbonic acid/bicarbonate form of the buffer predominates

The pH of the blood is higher (more basic) than the pKₐ of carbonic acid, so its basic form (bicarbonate, HCO₃⁻) predominates.

Final answer:

The bicarbonate form of carbonic acid predominates in the buffer system at pH 7.4.

Explanation:

The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution. In this case, we are trying to determine which form of the carbonic acid buffer predominates at pH 7.4, which is the homeostasis pH of the blood.

The Henderson-Hasselbalch equation is:

pH = pKa + log ([base] / [acid])

Given that the pKa of carbonic acid is 6.35, we can plug in the values to calculate:

pH = 6.35 + log (0.024 / 0.0012)

Simplifying the equation, we get:

pH = 6.35 + log (20)

pH = 6.35 + 1.3010

pH = 7.65

So, at pH 7.4, the bicarbonate (HCO3-) form of carbonic acid predominates in the buffer system.

Answer:

Volume of cube = [tex]L^3[/tex] = [tex]1 cm^3[/tex]

Density = [tex] \frac{m}{V} = 3.48 \frac{g}{cm^3}[/tex]