Be sure to answer all parts and enter your answers in scientific notation. Write the 1 g equivalents using the following prefixes: (a) mega−, (b) kilo−, (c) deci−, (d) centi−, (e) milli−, (f) micro−, (g) nano−, (h) pico−. (a) 1 g = × 10 megagrams (b) 1 g = × 10 kilograms (c) 1 g = × 10 decigrams (d) 1 g = × 10 centigrams (e) 1 g = × 10 milligrams (f) 1 g = × 10 micrograms (g) 1 g = × 10 nanograms (h) 1 g = × 10 picograms

Answer: The percentage of calcium carbonate is 76.3% and magnesium carbonate is 23.7%.

Explanation:

[tex]CO_2[/tex] produced ca be calculated from ideal gas equation:

[tex]PV=nRT[/tex]

P= Pressure of the gas = 785 mmHg = 1.03 atm    (760 mmHg= 1 atm)

V= Volume of the gas = 1.94 L

T= Temperature of the gas = 25°C=(25+273)K= 298 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

[tex]n=\frac{PV}{RT}=\frac{1.03\times 1.94L}{0.0821 \times 298}=0.08moles[/tex]

Mass of [tex]CO_2={\text{Given moles}}\times {\text {Molar mass}}=0.08moles\times 44g/mol=3.6g[/tex]

If the mass of calcium carbonate is x g , (7.85-x) g of magnesium carbonate is there.

[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]

As, 100 g of [tex]CaCO_3[/tex] react to give 44 g of [tex]CO_2[/tex] gas

So, x g of [tex]CaCO_3[/tex] react to give [tex]\frac{44\times x}{100}g[/tex] of [tex]CO_2[/tex] gas

The balanced chemical reaction will be:

[tex]MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2[/tex]

As, 84 g of [tex]MgCO_3[/tex] react to give 44 g of [tex]CO_2[/tex] gas

So, [tex](7.85-x)g[/tex] of [tex]MgCO_3[/tex] react to give [tex]\frac{44\times (7.85-x)}{84}g[/tex] of [tex]CO_2[/tex] gas

[tex]\frac{44\times x}{100}+\frac{44\times (7.85-X)}{84}=3.6[/tex]

By solving the terms, we get the value of x

x = 5.99 g

The mass of [tex]CaCO_3[/tex] = x = 5.99 g

The mass of [tex]MgCO_3[/tex] = 7.85 - 5.99 = 1.86 g

Now we have to calculate the percentage of magnesium carbonate.

[tex]\%\text{ of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Total mass}}\times 100=\frac{5.99g}{7.85g}\times 100=76.3\%[/tex]

[tex]\%\text{ of }MgCO_3=\frac{\text{Mass of }MgCO_3}{\text{Total mass}}\times 100=\frac{1.86g}{7.85g}\times 100=23.7\%[/tex]

Therefore, the percentage of calcium carbonate is 76.3% and magnesium carbonate is 23.7%.

To determine the percentage of calcium carbonate and magnesium carbonate in the mixture, you need to calculate the moles of carbon dioxide produced from each carbonate and convert them to grams. Then, use the moles of CO2 to calculate the percentages of each carbonate in the sample.

To determine the percentage of calcium carbonate and magnesium carbonate in the given sample, we need to calculate the moles of carbon dioxide produced from each carbonate and then convert them to grams. First, we calculate the moles of CO2 produced using the ideal gas law equation, PV = nRT. We can assume the volume is at STP (standard temperature and pressure) conditions, so we substitute the given values (1.94L of CO2, 25 degrees C, 785mmHg) and solve for n. Next, we use the balanced chemical equation to relate the moles of CO2 produced to the moles of calcium carbonate and magnesium carbonate in the sample.

Let's assume the sample contains x grams of calcium carbonate and y grams of magnesium carbonate. To find the percentage of each carbonate in the sample, we calculate the moles of CO2 produced from each carbonate and then divide by the total moles of CO2 produced. Finally, we multiply these values by 100 to get the percentages.

Answer:

The concentration of the product at equilibrium when the flask is kept in the warm bath is more than the previous concentration of the product.

Explanation:

The formation of the ice crystals on the wall of the flask means that the several reagents react by absorbing some energy from the surroundings to form ice crystals. This means that the forward reaction of the equilibrium reaction is an endothermic reaction.

When the flask is placed in a warm bath, it means that the heat is being provided to the system which means to an endothermic reaction extra heat is being provided to the system. Also, on raising the temperature, the rate of the endothermic reaction increases. So, the new concentration of the product at equilibrium is more than the precious concentration.

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: 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 : 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:

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:

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 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 [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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hey there!:

Attached resolution.

The Lewis Dot structure for diprotic acid HOOC-(CH2)4-COOH connects H to O, O to C, and C to C, repeating twice. The balanced chemical equation to titrate this diprotic acid with KOH would be: HOOC-(CH2)4-COOH + 2 KOH -> KOO-(CH2)4-COOK + H2O. This formula along with the volumes and concentration given can be used to find the unknown concentration of the adipic acid, 0.37 M.

The Lewis Dot structure for diprotic acid HOOC-(CH2)4-COOH, also known as adipic acid, is not easy to represent in textual format, but each H atom is connected to an O atom, which is connected to a C atom. The O atom is also connected to another C atom, and this structure repeats itself twice. Each C in the middle CH2 sequence is bonded to two H atoms.

The balanced chemical equation would be:

HOOC-(CH2)4-COOH + 2 KOH -> KOO-(CH2)4-COOK + H2O.

The titration involves using KOH (potassium hydroxide) to neutralize the adipic acid until it reaches an endpoint. To work out the unknown concentration of the adipic acid, we can use the formula c1v1 = c2v2, where c1 and v1 are the concentration and volume of the KOH solution, and c2 and v2 are the concentration and volume of the adipic acid solution.

In this case: c1 = 0.970M (concentration of the KOH), v1 = 27.1mL/1000 = 0.0271L (volume of the KOH solution), v2 = 35.5mL/1000 = 0.0355L (volume of the adipic acid). Re-arranging and substituting the numbers we will find the concentration of the adipic acid: c2 = (c1*v1)/(v2*2) = (0.970 * 0.0271) / (0.0355 * 2) = 0.37 M.

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Answer:

True

Explanation:

The percentage yield of a reaction , is given by the formula -

Percentage yield = (actual yield)/(theoretical yield)  * 100

Actual yield = the exact amount of yield obtained from an experiment

Theoretical yield = the calculated yield of the experiment.

The denominator value i.e. theoretical yield can never exceed the numerator value , i.e. the actual yield ,

Hence,

percentage yield can never be more than 100% .

The statement is true; percent yield is always between 0% and 100%, with yields above 100% indicating impurities or measurement errors.

True, the percent yield can never be greater than 100%. The calculation of percent yield is done by dividing the actual yield by the theoretical yield and then multiplying by 100%. Cases where a yield appears to be greater than 100% typically indicate an impure product or an error in weighing the reactants or products. The law of conservation of mass holds true and ensures that percent yields must logically be between 0% and 100%. A percent yield of 100% indicates a perfect conversion of reactants to products, which is rare in practice due to factors like reaction inefficiencies and measurement errors. Most laboratory or industrial procedures aim for a percent yield that is as close to 100% as practical while addressing waste disposal concerns.

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:

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:

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:

0.291

Explanation:

Given data

1070 torr × (1 atm/ 760 torr) = 1.41 atm

According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present.

Psolution = Psolvent × Χsolvent

Χsolvent = Psolution/Psolvent

Χsolvent = 1.00 atm/1.41 atm

Χsolvent = 0.709

The sum of the mole fraction of the solvent (water) and the solute (ethylene glycol) is 1.

Χsolvent + Χsolute = 1

Χsolute = 1 - Χsolvent = 1 - 0.709

Χsolute = 0.291

Considering the Roult's law, the mole fraction of ethylene glycol in the solution is 0.291

In first place, you know:

On the other side, if a solute has a measurable vapor pressure, the vapor pressure of its solution is always less than that of the pure solvent. Thus, the relationship between the vapor pressure of the solution and the vapor pressure of the solvent depends on the concentration of the solute in the solution.

In other words, Raoult's Law states that the relationship between the vapor pressure of each component in an ideal solution is dependent on the vapor pressure of each individual component and the mole fraction of each component in the solution.

Mathematically, this law expresses that in an ideal solution, the partial pressures of each component in the vapor are directly proportional to their respective molar fractions in the solution. That is, the partial pressure of a solvent on a solution P is given by the vapor pressure of the pure solvent, multiplied by the mole fraction of the solvent in the solution:

P1=x1P1°

where P1 is the partial pressure of a solvent over a solution, x1 is the mole fraction of the solution component, and P1 ° is the vapor pressure of the pure solvent.

In this case, for the solvent, you can calculate the mole fraction as:

Psolvent=xsolventPsolvent°

1 atm=xsolvent1.41 atm

1 atm÷1.41 atm= xsolvent

0.709 atm= xsolvent

The sum of the mole fraction of the solvent (water) and the solute (ethylene glycol) is 1.

xsolvent + xsolute = 1

xsolute = 1 - xsolvent

xsolvent= 1 - 0.709

xsolute = 0.291

Finally, the mole fraction of ethylene glycol in the solution is 0.291

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Answer:

maybe blower motor , but im not really sure .

Explanation:

Answer:

The Correct Answer is A

Explanation:

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.

The reaction given is best described as an aqueous reaction where hydrochloric acid and potassium hydroxide react to produce aqueous potassium chloride and water. This is a type of neutralization reaction which is common in Chemistry where an acid and a base react to produce a salt and water.

The statement that best describes the given reaction, HCl(aq) + KOH(aq) -> KCl(aq) + H2O(l), is 'Aqueous solutions of hydrochloric acid and potassium hydroxide react to produce aqueous potassium chloride and water'. This type of reaction is a neutralization reaction, which is a type of chemical reaction where an acid and a base react quantitatively with each other. In this reaction, hydrochloric acid, which is the acid, reacts with potassium hydroxide, the base, to produce potassium chloride, the salt, and water.

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"The correct option is d. An acid + a base react to produce a salt and water. Option d is the best description of the reaction as it encapsulates the nature of the reactants (an acid and a base) and the products (a salt and water) in the most general sense.

The given reaction is a classic example of a neutralization reaction, where an acid reacts with a base to form salt and water. In this case, hydrochloric acid (HCl), which is an acid, reacts with potassium hydroxide (KOH), which is a base, to produce potassium chloride (KCl), which is a salt, and water (H2O).

Let's analyze the options:

a. Hydrochloric acid and potassium hydroxide solutions react to produce potassium chloride solution and water.

- This option is not incorrect, but it is not as general as option d. It describes the reaction accurately but does not explicitly state that it is a reaction between an acid and a base.

b. Hydrochloric acid and potassium hydroxide solutions react to produce potassium chloride and water.

- This option is similar to option a but omits the word solution, which is not a significant issue since potassium chloride is soluble in water and will exist as a solution in the given reaction context.

c. Aqueous solutions of hydrochloric acid and potassium hydroxide react to produce aqueous potassium chloride and water.

- This option is correct in specifying that the reactants and products are aqueous. However, it does not generalize the reaction as being between any acid and base.

d. An acid + a base react to produce a salt and water.

- This option is the most general and correct description of the reaction. It identifies the reactants as an acid and a base and the products as a salt and water, which is the essence of a neutralization reaction.

e. Hydrochloric acid and potassium hydroxide produce potassium chloride and water.

- While this statement is true, it does not explicitly state that the reaction is between an acid and a base, nor does it mention the physical state of the reactants and products.

Therefore, option d is the best description of the reaction as it encapsulates the nature of the reactants (an acid and a base) and the products (a salt and water) in the most general sense."

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:   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:

Molarity of the base solution was 0.1830 M

Explanation:

Total volume of NaOH added for complete standardization = (final reading-initial reading) = (30.89-1.98) mL = 28.91 mL

Neutralization reaction between NaOH and HCl :

                                 [tex]NaOH+HCl\rightarrow NaCl+H_{2}O[/tex]

So, 1 mol of HCl neutralizes 1 mol of NaOH

Moles of HCl added = [tex]\frac{28.58\times 0.1851}{1000}[/tex] moles

If molarity of NaOH was C (M) then moles of NaOH added is [tex]\frac{C\times 28.91}{1000}[/tex]

[tex]\frac{28.58\times 0.1851}{1000}[/tex] = [tex]\frac{C\times 28.91}{1000}[/tex]            

or, C = 0.1830

So, molarity of NaOH was 0.1830 M            

Final answer:

By calculating the moles of HCl reacted and knowing the 1:1 stoichiometry of the reaction with NaOH, the molarity of NaOH is found to be approximately 0.183 M.

Explanation:

To determine the molarity of the sodium hydroxide (NaOH) solution used in the titration, we will use the data provided from the titration of hydrochloric acid (HCl) with NaOH. The reaction between a strong acid and strong base like HCl and NaOH is one to one:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The amount of NaOH used is the final buret reading minus the initial buret reading:

VNaOH = 30.89 mL - 1.98 mL = 28.91 mL

Convert this volume to liters (L) by dividing by 1000:

VNaOH in liters = 28.91 mL ÷ 1000 = 0.02891 L

Next, calculate the number of moles of HCl since it is directly proportional to the number of moles of NaOH:

nHCl = MHCl × VHCl

Substitute given values to find the number of moles of HCl:

nHCl = 0.1851 M × 0.02858 L = 0.005290138 moles

Since the reaction ratio is 1:1, the moles of NaOH will be the same as the moles of HCl:

nNaOH = nHCl = 0.005290138 moles

Finally, determine the molarity of NaOH using the formula:

MNaOH = nNaOH ÷ VNaOH

Substitute the values:

MNaOH = 0.005290138 moles ÷ 0.02891 L = 0.183 M (rounded to three significant figures)

The molarity of the sodium hydroxide solution used in the titration is approximately 0.183 M.

Hey there!:

More soluble in acidic solution than in neutral solution :

*  Here the conjugate acids of AgCl , HgBr2 , and Pbl2 are HCl , HBr and HI which are strong , so these are slightly soluble in acidic  solution .

* But  the conjugate acid of BaCO3 is H2CO3 wich is weak acid, more soluble in acidic solution than in neutral solution.

Hope this helps!

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.

Best regards.

Answer: 204.2 g/mol

Explanation:

Elevation in boiling point is given by:

[tex]\Delta T_b=k_b\times m[/tex]

[tex]\Delta T_b[/tex] =  Elevation in boiling point =  [tex]3.68^0C/m[/tex]

[tex]k_b[/tex] = boiling point constant  = [tex]3.63^0C/m[/tex]

m = molality

[tex]Molality=\frac{n\times 1000}{W_s}[/tex]

where,

n = moles of solute = [tex]\text {given mass}{\text {molar mass}}=\frac{0.207}{M}[/tex]

 [tex]W_s[/tex] = weight of solvent in g  = 1g

Molality =[tex]\frac{0.207\times 1000}{M\times 1g}[/tex]

Putting in the values we get,

[tex]3.68=3.63\times \frac{0.207\times 1000}{M\times 1g}[/tex]

[tex]M=204.2g/mol[/tex]

The molar mass of caryophyllene is 204.2 g/mol.

Answer:

36.37% is the percent yield of the reaction.

Explanation:

[tex]4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)[/tex]

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

[tex]PV=nRT[/tex]

[tex]n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol[/tex]

2) Moles of ammonia gas=[tex]\frac{2.53 g}{17 g/mol}=0.1488 mol[/tex]

Moles of oxygen gas =[tex]\frac{3.53 g}{32 g/mol}=0.1101 mol[/tex]

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

[tex]\frac{4}{3}\times 0.1101 mol=0.1468 mol[/tex] of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

[tex]\frac{2}{3}\times 0.1101 mol=0.0734 mol[/tex] of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

[tex]\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100[/tex]

Percentage yield of the reaction:

[tex]\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%[/tex]

36.37% is the percent yield of the reaction.

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:

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 : 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 %