The following equilibrium constants have been determined for hydrosulfuric acid at 25ºC:
H2S(aq) ⇌ H+(aq) + HS–(aq) K′c = 9.5 × 10^–8
HS–(aq) ⇌ H+(aq) + S2–(aq) K″c = 1.0 × 10^–19
Calculate the equilibrium constant for the following reaction at the same temperature: H2S(aq) ⇌ 2H+(aq) + S2–(aq).

Answer:

False

Explanation:

Regulation of blood coagulation by anticoagulant pathways Regulation of coagulation is exerted at each level of the pathway, either by enzyme inhibition or by modulation of the activity of the cofactors.

Answer: The molecular equation is given below.

Explanation:

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This means that total mass on the reactant side is equal to the total mass on the product side.

This also means that the total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side.

The balanced chemical equation for the reaction of ammonium bromide and lead (II) acetate follows:

[tex]2NH_4Br(aq.)+(CH_3COO)_2Pb(aq.)\rightarrow  2CH_3COONH_4(aq.)+PbBr_2(s)[/tex]

This is a type of double displacement reaction because here, exchange of ions takes place.

Hence, the molecular equation is given below.

The concentration will be equal to 0.899 mol/L.

The value of "A" will be found as follows:

[tex]6.28*10^-^3=\frac{1}{10} (0.962-[A])\\A= 0.899 mol/L[/tex]

It is important to remember that the concentration of a chemical solution refers to the amount of solute that exists within the solvent.

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The concentration of N2O after 10 seconds in a zero order reaction with initial concentration of 0.962 mol/L and rate constant of 6.28×10^-3 mol/L/s is 0.899 mol/L.

The question involves calculating the concentration of N2O after 10 seconds in a zero order reaction. The rate equation for a zero order reaction is given by rate = k, where the reaction rate is constant and does not depend on the concentration of the reactant. The formula to calculate the remaining concentration after a certain time in a zero order reaction is: [A] = [A]0 - kt, where [A] is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

In this case, the initial concentration of N2O is 0.962 mol/L, the rate constant k is 6.28×10−3 mol/L/s, and the time t is 10.0 seconds. Using the formula, we get [N2O] = 0.962 mol/L - (6.28×10−3 mol/L/s)(10.0 s) = 0.962 mol/L - 0.0628 mol/L = 0.899 mol/L.

Answer: The time required for the conversion of benzoyl chloride to benzoic acid is 390 seconds.

Explanation:

The expression for the rate law of first order kinetics is given as:

[tex]\ln \frac{N}{N_o}=e^{-kt}[/tex]      ......(1)

where,

[tex]N_o[/tex] = initial mass of isotope = 100 g

N = mass of the parent isotope left after the time = 100 - 25 = 75 g

t = time taken = 26 s

k = rate constant = ?

Putting values in above equation, we get:

[tex]\ln \frac{75}{100}=e^{-k\times 26}\\\\k=0.011065s^{-1}[/tex]

Now, we need to find the time when 99 % of the reaction is complete. Using equation 1 again, we get:

[tex]N_o[/tex] = initial mass of isotope = 100 g

N = mass of the parent isotope left after the time = 100 - 99 = 1 g

t = time taken = ? s

k = rate constant = [tex]0.011065s^{-1}[/tex]

Putting values in above equation, we get:

[tex]\ln \frac{1}{100}=e^{-0.011065\times t}\\\\t=416s[/tex]

Time required for the conversion of benzoyl chloride to benzoic acid = 416 - 26 = 390 s

To obtain 99% conversion of benzoyl chloride to benzoic acid, it would take approximately 420 seconds.

This is a first-order reaction, and the question asks for the time needed to obtain 99% conversion of benzoyl chloride to benzoic acid. The reaction is 25% complete after 26 seconds, so we can use the half-life formula to calculate the time required for complete conversion. In a first-order reaction, the half-life is given by t1/2 = (0.693/k), where k is the rate constant. Since the reaction is 25% complete after 26 seconds, we can find the rate constant by rearranging the formula to k = 0.693 / t1/2. The rate constant is approximately 0.0267 s-1. Now, we can find the time required for 99% conversion using the formula t = (ln(1-0.99)/-k), where k is the rate constant. Solving this equation gives approximately 420 seconds.

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Answer:    The amount of energy needed to heat the body by 1 ° C (Q)

Explanation:  The heat capacity or as it is called the thermal mass is the amount of energy needed to heat the body by 1 ° C. When the body is said, it refers to any object in any aggregate state. The energy required for the body to warm up by 1 ° C is expressed in joules, and is obtained when the specific heat of the body multiplies with the body mass and with the change of temperature:

Q = m·c·ΔT

Q - heat capacity (J),

m - mass of the body (g),

c - specific heat of the body (J/g-°C),

ΔT - change in temperature (°C)

Answer : The correct option is, (a) 48 %

Explanation :

As we are given the efficiency of a spray chamber is, 95 %

Using unitary method :

If the efficiency of [tex]20\mu m[/tex] particles is 95 %

Then, the efficiency of [tex]10\mu m[/tex] particles will be = [tex]\frac{10\mu m}{20\mu m}\times 95\%=47.5\%\approx 48\%[/tex]

Therefore, the efficiency for [tex]10\mu m[/tex] particles is, 48 %

Answer:

D

Explanation:

The resolution of a light microscope is approximately 200 nanometers due to the wavelength of light. A eukaryotic cell is approximately 1 – 5 micrometers. This means that light can be used to view a eukaryotic cell because the wavelength of light is smaller than the size of the cell. The other options are of a smaller size than the light wavelength and hence lower wavelength beams (such as those of electron microscope) or laser can be used to view them.

Final answer:

The light microscope is best suited for viewing eukaryotic cells such as green algae, leveraging staining techniques to provide necessary contrast. It cannot resolve smaller entities like water molecules, DNA strands, or viruses, which are below the limit of light microscopy resolution.

Explanation:

A light microscope would be most advantageous for viewing eukaryotic cells such as green algae. Light microscopes can magnify cells up to approximately 400 times, which is suitable for viewing relatively large cellular structures. However, they are not powerful enough to resolve smaller structures such as water molecules, DNA strands, or viruses, which are below their limit of resolution. For improved observation, staining techniques are often used, as these colored chemicals make cellular components visible by providing the necessary contrast.

Other Microscope Types

A phase-contrast microscope is especially useful for viewing thick structures such as biofilms.

For very small surface structures of a cell, a scanning electron microscope (SEM) would be the best choice, as it provides a detailed three-dimensional image of the specimen's surface.

To see individual components of cells under a light microscope, scientists commonly use special stains.

Answer:

19.9 grams of [tex]CuSO_4.5H_2O[/tex] will be needed.

Explanation:

Required strength of the solution = 2% (w/v)

This means that 2 gram of solute in 100 ml of solution.

Mass of [tex]CuSO_4.5H_2O[/tex]  = 2 g

Moles of [tex]CuSO_4.5H_2O[/tex] =[tex]\frac{2 g}{249.68 g/mol}=0.008010 mol[/tex]

Volume of the solution = 100 mL = 0.1 L

Molarity of the solution:

[tex]M=\frac{0.008010 mol}{0.1 L}=0.08010 mol/L[/tex]

0.08010 moles of [tex]CuSO_4.5H_2O[/tex] are present 1 l of the solution.

Then mass of 0.08010 moles of [tex]CuSO_4.H_2O[/tex] will be:

0.08010 mol × 249.68 g/mol = 19.9993 g≈ 19.9 g

19.9 grams of [tex]CuSO_4.5H_2O[/tex] will be needed.

The isomers of C4H9Br are ranked in increasing reactivity towards SN2 and E2 reactions, taking into account steric hindrance and alkyl substitutions.

The constitutional isomers for the molecule C4H9Br are: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, and 2-bromo-2-methylpropane. (a) For SN2 reactions, reactivity increases with fewer alkyl substitutions (steric hindrance), thus the order is: 1-bromobutane > 2-bromobutane > 1-bromo-2-methylpropane > 2-bromo-2-methylpropane. (b) For E2 reactions, reactivity is greatest for bulkier bases and more alkyl substitutions, hence the order is: 2-bromo-2-methylpropane > 1-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane.

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Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

Answer:

b. Add a catalyst.

Explanation:

Only B is correct.

The factors that affects the rate of chemical reactions are:

The rate of chemical reaction is a measure of the speed of the reaction.

When a foreign body is added to a reaction, the speed can be influenced. If the body increases the reaction rate, it is a positive catalyst. Catalysts are used to reduce the reaction time of many reactions.

To increase the rate of a chemical reaction, you can increase the temperature, increase the concentration of reactants, or add a catalyst.

To increase the rate of a chemical reaction, there are several factors to consider. One way is to increase the temperature of the reactants, as chemical reactions typically occur faster at higher temperatures. Another way is to increase the concentration of the reactants, which increases the likelihood of particles colliding. Additionally, adding a catalyst can provide an alternative pathway with a lower activation energy, thereby increasing the rate of the reaction.

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Answer:

Concentration of HCl is 1.2054 M

Explanation:

Acid-base titration is a type of qualitative analysis in which the concentration of the unknown acid/base is determined by treating it with a known concentration of base/acid. Indicators are used to determine the end point or the equivalence point (when all the acid/base is neutralized by base/acid).

At the end point,

Moles of acid = Moles of base

Since,

Molarity of a solution is given by the moles of solute per liter of the solution  

i.e.,

Molarity = moles of solute / volume of solution (in L)

moles = Molarity * Volume

moles = M * V  

M₁V₁ ( acid )= M₂V₂( base )

Where,

M₁ is the molarity of the acid

V₁ is the volume of the acid    

M₂ is the molarity of the base

V₂ is the total volume of the base

From the question,

M₁ =  ?

M₂ = 0.100 M

V₁ = 10.00 mL

V₂ = 120.54 mL

Using the above mentioned formula

M₁V₁ ( acid )= M₂V₂( base )

M₁ * 10.00mL = 0.100 M * 120.54 mL

M₁ = 1.2054 M

Hence, the concentration of of HCl is 1.2054 M

The concentration of the HCl solution is calculated by the formula c1v1 = c2v2, deriving from the one-to-one reaction between NaOH and HCl. The resulting concentration of the HCl solution is determined to be 1.21 M.

The subject of this question is a chemistry concept related to acid-base titration. In this context, you're given that 10.00 mL of an HCl solution is titrated with 120.54 mL of a 0.100 M NaOH solution to reach the equivalence point. From this, we can figure out the concentration of the unknown HCl solution using the formula c1v1 = c2v2, where c represents concentration and v represents volume. The reaction between NaOH and HCl is one-to-one, so the number of moles of NaOH that react will be equal to the number of moles of HCl initially present in the solution.

To calculate the number of moles of NaOH used in the reaction, you multiply the volume of NaOH solution used (in liters) by its molarity. Here, that is (120.54 mL / 1000 mL/L) * 0.100 mol/L = 0.012054 mol NaOH. Since the moles of NaOH must equal the moles of HCl, the molarity (M) of the HCl solution is the moles of HCl divided by the volume of HCl solution in liters. Therefore, the concentration of HCl is 0.012054 mol / (10.00 mL / 1000 mL/L) = 1.21 M.

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Answer : The molar volume of the solution is, [tex]24m^3/mole[/tex]

Explanation : Given,

Partial molar volumes of component A = [tex]30m^3/mole[/tex]

Partial molar volumes of component B = [tex]20m^3/mole[/tex]

Mole fraction of component A = 0.4

First we have to calculate the mole fraction of component B.

As we know that,

[tex]\text{Mole fraction of component A}+\text{Mole fraction of component B}=1[/tex]

[tex]\text{Mole fraction of component B}=1-0.4=0.6[/tex]

Now we have to calculate the molar volume of the solution.

Expression used :

[tex]V_s=X_A\times \bar V_A+X_B\times \bar V_B[/tex]

where,

[tex]\bar V_A[/tex] = partial molar volumes of component A

[tex]\bar V_B[/tex] = partial molar volumes of component B

[tex]V_s[/tex] =  molar volume of the solution

[tex]X_A[/tex] = mole fraction of of component A

[tex]X_B[/tex] = mole fraction of of component B

Now put all the give values in the above expression, we get:

[tex]V_s=0.4\times 30m^3/mole+0.6\times 20m^3/mole[/tex]

[tex]V_s=24m^3/mole[/tex]

Therefore, the molar volume of the solution is, [tex]24m^3/mole[/tex]

Answer:

Temperature of 100°C increase the reaction rate, due to the rise of the energetic collision between molecules that increase the collision, compared to the same reaction at 21 °C, who has less energy and for that reason will be more slow to react.

Explanation:

The enzymatic reactions are reactions between organic molecules named as substrate and some proteic biological structures called enzymes. These reactions can be simplified as a regular chemistry process between reagents to obtain a product, that in this case is the transformations of the substrate.

So we can use the following process:

Enzyme (E) + substrate (S) = product (P)

Of course, this an uncomplex way to see this process, just to understand this example. In reality, the product or these reactions involves a transformation of the substrate and the enzyme. But for now, let's just use this equation

Using just the letters:

E + S = P

Now, we can use the concept of rate or velocity on chemical equations to analyze the effect of temperature in the enzymatic reactions:

Remember that for any chemistry reactions, the rate depends on the capacity of the molecules to collide, these collisions will be major when the reagents have enough kinetic energy to move around and interact between them. The frequency of these collisions is affected by different variables such as temperature.

Temperature is equal to energy, so if to reactions are supplied by external energy like thermic energy, the molecules, in this case, enzyme and substrate can move faster, and the collides can be more frequent when the temperature increases.

In conclusion, the increases in temperature to 100°C, increase the reaction rate, due to the rise of the energetic collision between molecules that increase the collision, and are these who result in the product of a enzymatic reactions, compared to the same reaction at 21 °C, who has less energy and for that reason will be more slow to react.

Answer:

maybe blower motor , but im not really sure .

Explanation:

Answer:

The Correct Answer is A

Explanation:

Explanation:

When a non-volatile solute is added in a solvent then decrease in its freezing point is known as freezing point depression.

Mathematically,     [tex]\Delta T = K_{f}m[/tex]

where      [tex]\Delta T[/tex] = change in freezing point

                     [tex]K_{f}[/tex] = freezing point depression constant [tex]\text({in ^{o}C/mol/kg})[/tex]

                           m = molality

First, calculate the number of moles as follows.

                       No. of moles = [tex]\frac{\text{mass of solute}}{\text{molar mass of solute}}[/tex]

                                             = [tex]\frac{22.78g}{62.07 g/mol}[/tex]

                                             = 0.367 mol

Now, it is given that mass of solvent is 87.95 g. As there are 1000 grams in 1 kg.

So,          [tex]87.95g \times \frac{1 kg}{1000 g}[/tex] = 0.08795 kg

Hence, molality of the given solution is as follows.

                   Molality = [tex]\frac{\text{no. of moles}}{\text{mass in kg}}[/tex]

                                 = [tex]\frac{0.367 mol}{0.08795}[/tex]

                                 = 4.172 mol/kg

Therefore, depression in freezing point will be as follows.

                          [tex]\Delta T = K_{f}m[/tex]

            [tex]T_{solvent} - T_{mixture}[/tex] = [tex]K_{f}m[/tex]

Since, freezing point of pure water is [tex]0^{o}C[/tex]. Now, putting the given values as follows.

                [tex]0^{o}C - T_{mixture}[/tex] = [tex]1.86^{o}C/m \times 4.172mol/kg[/tex]

                                         = [tex]-7.759 ^{o}C[/tex]

Thus, we can conclude that the freezing point of water in the given mixture is [tex]-7.759 ^{o}C[/tex].

Answer: 71.65 L

Explanation:

Decomposition of sodium azide is shown by equation below:

[tex]2NaN_3\rightarrow 2Na+3N_2[/tex]

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

[tex]\text{Number of moles of sodium azide}=\frac{125g}{65g/mol}=1.92moles[/tex]

According to stoichiometry:

2 moles of [tex]NaN_3[/tex] produce 3 moles of [tex]N_2[/tex]

Thus 1.92 moles of [tex]NaN_3[/tex] will produce=[tex]\frac{3}{2}\times 1.92=2.88[/tex] moles of [tex]N_2[/tex]

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = 756 torr = 0.99 atm    (1 torr= 0.0013 atm)

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K       (0°C = 273 K)

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 2.88  

[tex]V=\frac{nRT}{P}=\frac{2.88\times 0.0821\times 300}{0.99}=71.65L[/tex]

Thus the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide is 71.65 L

Answer:

glycerol-3-phosphate, ADP, H⁺

Explanation:

The reaction of converting glycerol to glycerol-3-phosphate which makes is unfavorable and is coupled with the second reaction which involves conversion of ATP to ADP which is high energetically favorable.

Reaction 1:  Glycerol + HPO₄²⁻ ⇒ Glycerol-3-phosphate + water

Reaction 2:             ATP  + H₂O ⇒ ADP + HPO₄²⁻ + H⁺

The coupled reaction of both the reactions become favorable. Thus, the overall coupled reaction is:

Glycerol + ATP ⇒ Glycerol-3-phosphate + ADP + H⁺

The net products are = glycerol-3-phosphate, ADP, H⁺

The net products of the coupled reactions as described are; glycerol-3-phosphate, ADP and H+.

According to the reaction;

The reactions given is as follows;

glycerol + (HPO4)²− + ATP+H2O ⟶⟶ glycerol-3-phosphate+H2O ADP+(HPO4)²- + H+

Since, the (HPO4)²- cancels out on both sides of the equation; the net products of the coupled reactions above are; glycerol-3-phosphate, ADP and H+.

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Answer : The [tex]C_p[/tex] of the the calorimeter is, [tex]98.35J/^oC[/tex]

Explanation :

First we have to calculate the moles of NaOH.

[tex]\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }NaOH=5.6mole/L\times 0.0044L=0.02464mole[/tex]

conversion used for volume : (1 L = 1000 ml)

As we know that, HCl and NaOH are the strong acid and bases. So, the heat of neutralisation is -55.9 kJ/mole.

Now we have to calculate the heat released by neutralization.

As, 1 mole of heat released by neutralization = 55.9 kJ

So, 0.02464 mole of heat released by neutralization = [tex]0.02464\times 55.9kJ=1.377kJ[/tex]

Now we have to calculate the [tex]C_p[/tex] of the the calorimeter.

[tex]Q=C_p\times \Delta T[/tex]

where,

Q = heat of reaction = 1.377 kJ

[tex]C_p[/tex] = specific heat capacity of calorimeter =?

[tex]\Delta T[/tex] = change in temperature = [tex]14^oC[/tex]

Now put all the given values in this expression, we get:

[tex]1.377kJ=C_p\times 14^oC[/tex]

[tex]C_p=0.0983kJ/^oC=98.35J/^oC[/tex]

Therefore, the [tex]C_p[/tex] of the the calorimeter is, [tex]98.35J/^oC[/tex]

The Cp of the calorimeter is equal to 98.35J/ºC

[tex]Mole_N_a_O_H= 5.6 mol/L*0.0044L= 0.02464 mole[/tex]

[tex]\left \ {{1mole=-55.9 kJ} \atop {0.02464 mole=x}} \right.\\x= 55.9*0.02464 mole\\x= 1.377kJ[/tex]

[tex]Q=C_p*\Delta T\\1,377= C_p*14 \textdegree C\\C_p= \frac{1.377}{14} = 98.35J\textdegree C[/tex]

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Answer: 61 grams

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles of octane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{97g}{114g/mol}=0.85moles[/tex]

[tex]\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{150g}{32g/mol}=4.69moles[/tex]

The chemical equation for the combustion of octane in oxygen follows the equation:

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

By stoichiometry of the reaction;

25 moles of oxygen react with 2 moles of octane

4.69 moles of oxygen react with=[tex]\frac{2}{25}\times 4.69=0.37[/tex]  moles of octane

Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.

25 moles of oxygen produce 18 moles of water

4.69 moles of oxygen produce=[tex]\frac{18}{25}\times 4.69=3.38[/tex]  moles of water.

Mass of water produced=[tex]moles\times {\text{Molar mass}}=3.38\times 18g/mol=61g[/tex]

The maximum mass of water that could be produced by the chemical reaction is 61 grams.

To determine the maximum mass of water produced from the combustion of octane, one must first calculate the moles of octane and water using the balanced chemical equation. The stoichiometry shows that for every 2 moles of octane, 18 moles of water are produced. The final mass of water is calculated by considering the limiting reagent, which could be either octane or oxygen.

To calculate the maximum mass of water that could be produced from the combustion of octane, we should start with the balanced chemical equation for octane (C8H18):

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

First, we determine the moles of octane based on the given mass:

Next, we calculate the moles of water produced per mole of octane using the stoichiometry of the reaction:

Now, we find the mass of water produced using the moles of octane we calculated:

Finally, we check if oxygen is the limiting reagent by comparing moles of oxygen provided with the moles required based on the moles of octane we have. To calculate this, use the molar mass of oxygen (O2) which is 32.00 g/mol and the stoichiometry of the equation. If the available moles of oxygen are less than required, we must recalculate the maximum mass of water based on the moles of the limiting reagent (oxygen).

Note that we haven't provided actual numerical results, and you will need to perform the calculations to find the maximum mass of water that can be produced. Remember to round your final answer to the correct number of significant digits based on the given data in the question.

A generator is a device that converts mechanical energy into electrical energy. They are usually referred to as electric generators. They usually have a dynamo that utilizes the Flemings left- and the right-hand principle of interaction between electric energy fields and magnetic energy fields to produce motion. A motor generator does the reverse by producing mechanical energy from electric energy.

Answer:

In electricity generation, a generator is a device that converts motive power (mechanical energy) into electrical power for use in an external circuit. Sources of mechanical energy include steam turbines, gas turbines, water turbines, internal combustion engines, wind turbines and even hand cranks. The first electromagnetic generator, the Faraday disk, was invented in 1831 by British scientist Michael Faraday. Generators provide nearly all of the power for electric power grids.

Explanation:

Answer: 1. [tex]2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O[/tex]

2. 100.8 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Balanced equation for the decomposition of the compound sodium hydrogen carbonate is:

[tex]2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O[/tex]

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

moles of [tex]CO_2=\frac{\text {given mass}}{\text {molar mass}}=\frac{27.0g}{44g/mol}=0.6moles[/tex]

According to stoichiometry:

1 mole of [tex]CO_2[/tex] is produced from 2 moles of [tex]NaHCO_3[/tex]

Thus 0.6 moles of [tex]CO_2[/tex] is produced from =[tex]\frac{2}{1}\times 0.6=1.2[/tex] moles of [tex]NaHCO_3[/tex]

Mass of  [tex]NaHCO_3=moles\times {\text {Molar mass}}=1.2\times 84=100.8g[/tex]

Thus  the mass of [tex]NaHCO_3[/tex] required to produce 27.0 g of  [tex]CO_2[/tex] is 100.78 grams.

The balanced equation for the decomposition of sodium bicarbonate is 2 NaHCO3 → Na2CO3 + CO2 + H2O. To produce 27.0 g of CO2, you would need approximately 51.0 g of NaHCO3.

The balanced chemical equation for the decomposition of sodium bicarbonate or baking soda is 2 NaHCO3 → Na2CO3 + CO2 + H2O. When heated, sodium bicarbonate decomposes into sodium carbonate, carbon dioxide, and water. This CO2 is what makes baked goods rise.

To calculate the mass of NaHCO3 required to produce 27.0 g of CO2, we use molar mass and stoichiometry. The molar mass of NaHCO3 is about 84 g/mol, and that of CO2 is about 44 g/mol. Since the reaction produces one mole of CO2 for every two moles of NaHCO3, we can set up the equation: (27g CO2 / 44 g/mol CO2) * (2 mol NaHCO3 / 1 mol CO2) * (84 g NaHCO3 / 1 mol NaHCO3) = 51.0 g NaHCO3

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Compound A will be 2-chloro-butane which in the presence of sodium methoxide will undergo the elimination of the halogen with a hydrogen from the neighboring atom carbon (not the one with which the halogen in bound),  which have less hydrogen atoms, forming the trans-2-butene.

Answer:

2-chlorobutane.

Explanation:

Hello,

In this case, the treatment of the substance A with sodium methoxide to yield trans-2-butene stands for a dehydrohalogenation in the presence of a strong base (the sodium methoxide), such chemical reaction is undergone by alkyl halides, in this case with chlorine, based on its molecular formula. In such a way, on the attached picture, you will find both the structure of the 2-chlorobutane and its chemical reaction to yield the trans-2-butene throughout the usage of sodium methoxide.

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Answer:

D. CH₃CH₂C(CH₃)₂C≡CCH(CH₃)₂  

Explanation:

You start numbering from the end closest to the triple bond (on the right). The triple bond is between C3 and C4, and there is one methyl group on C3 and two on C5.

A. CH₃CH₂CH(CH₃)C≡CCH₂CH(CH₃)₂ is wrong. The longest chain has eight C atoms, so the compound is an octyne.

B. CH₃CH₂CH(CH₃)C≡CC(CH₃)₃ is wrong. This is a molecule of 2,2,5-trimethylhept-3-yne.

C. (CH₃CH₂)₂C≡CCH₂CH₃ is wrong. This is a molecule of 6-ethyl-5-methylhept-3-yne.

E. CH₃CH₂CH₂CH(CH₃)C≡CC(CH₃)₃ is wrong. The longest chain has eight C atoms, so the compound is an octyne.

Final answer:

The correct structure for 2,5,5-trimethylhept-3-yne, according to IUPAC nomenclature, is the one that features a seven-carbon backbone with a triple bond beginning at the third carbon and contains three methyl groups precisely at the second, fifth, and fifth carbon positions. The only structure that fits these requirements is (CH₃CH₂)C(CH₃)₂C≡CCH(CH₃)₂.

Explanation:

To determine the acceptable structure for 2,5,5-trimethylhept-3-yne, it is essential to understand the basics of organic chemistry nomenclature. This name indicates a seven-carbon chain (hept-) with a triple bond starting from the third carbon (3-yne) and three methyl groups (CH3) attached to the second (2-), fifth (5-), and fifth (5-) carbons. Thus, the structure should broadly fit this description to be correct.

Reviewing the options provided, we look for the structure that accurately aligns with the IUPAC naming rules which prioritize the length of the carbon chain, the position of the triple bond, and the placement of the methyl groups.

Answer:   k [[tex]F_{2}[/tex]] [tex][Cl_{2}O]^{2}[/tex]

Explanation:  The given reaction is -

[tex]F_{2}(g) + 2Cl_{2}O(g) \rightarrow  2ClO_{2}(g) + Cl_{2}[/tex]

The rate law of the reaction is the written as the concentration of the reactant species rest to the power of its stiochiomeric coefficient.

Thus the rate law of the given reaction can be written as -

Rate = k [[tex]F_{2}[/tex]] [tex][Cl_{2}O]^{2}[/tex]

Rate law usually is determined from the slowest step of the reaction.

Answer:

The w/w 5 of the solution is 10 %.

Explanation:

w/w % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

[tex]w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]

w/v %: The percentage of mass of the of solute present in total volume of the solution.

[tex]w/v\%=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100[/tex]

v/v % : The percentage volume of the of solute present in total volumeof the solution.

[tex]v/v\%=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100[/tex]

Mass of solution =  Mass of solute + Mass of solvent

Mass of solute= 10 g of NaOH

Mass of solvent = 90 g

Mass of solution = 10 g + 90 g = 100 g

[tex]w/w\%=\frac{10 g}{100 g}\times 100 =10\%[/tex]

The w/w 5 of the solution is 10 %.

Answer: The number of moles of nitrogen gas is 9.9 moles.

Explanation:

To calculate the mass of bromine gas, we use the ideal gas equation, which is:

PV = nRT

where,

P = Pressure of nitrogen gas = 2.30 atm

V = Volume of nitrogen gas = 120.0 L

n = Number of moles of nitrogen gas = ? mol

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = Temperature of nitrogen gas = 340 K

Putting values in above equation, we get:

[tex]2.30atm\times 120.0L=n\times 0.0821\text{L atm }mol^{-1}K^{-1}\times 340K\\\\n=9.88mol\approx 9.9mol[/tex]

Rule of significant figures in case of multiplication and division:

The least number of significant figures in any number of the problem will determine the number of significant figures in the solution.

Here, the least precise number of significant figures are 2. Thus, the number of moles of nitrogen gas is 9.9 moles.