True/False: Chemical reactions can NOT proceed in the reverse direction

Answer:

1.Precipitation

2.Transpiration

3.Condensation

4.Evaporation

5.Evaporation

3.Condensation

Explanation:

Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

Answer is 9

pKb=−logK

b=−log10^-5=5

pOH=pKb+log[acid]*[salt]

Substitute values in the above expression.

pOH=5+log0.1*0.1=5

Hence, the pH of the solution is pH=14−pOH=14−5=9

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Answer:

molar heat of combustion = -5156 *10³ kJ/mol

Explanation:

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Step 1: Data given

Mass of naphthalene = 1.435 grams

Initial temperature of water = 20.28 °C

Final temperature of water = 25.95 °C

heat capacity of the bomb plus water was 10.17 kJ/°C

Molar mass naphtalene = 128.2 g/mol

Step 2:

Qcal = Ccal * ΔT

⇒with Qcal =the heat of combustion

⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C

⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C

Qcal = 10.17 kJ/°C * 5.67 °C

Qcal = 57.7 kJ

Step 3: Calculate moles

Moles naphthalene = 1.435 grams / 128.2 g/mol

Moles naphthalene = 0.01119 moles

Step 4: Calculate the molar heat of combustion

molar heat of combustion = Qcal/ moles

molar heat of combustion = -57.7 kJ/ 0.01119 moles

molar heat of combustion = -5156 *10³ kJ/mol

Answer:

Pharmaceutical laboratory helps in devloping and conducting research, vaccines. Various kinds of drugs and chemical substances used and are produced at a Pharmaceutical laboratory.

The pharmaceutical laboratories performs with various hazardous substances that results in exposure to various chemicals, biological substances and radiation. To avoid any injury or infection labs need to maintain all safety measures.

Spillage and relaseing chemical substances can be lethal during transportaions by safety measures for heling in for manufacturing of such therapeutic agents spillage and avoid wastage.

Maintaining good safety standards in the pharmaceuticals laboratory will help promote the health of technicians and workers which in turn will increase productivity and attain positive outcomes.

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

Answer:

Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.

[tex]2CO(g) + O2(g) <=> 2CO2[/tex]

Explanation:

When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.

So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.

Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.

Increase in concentration of CO2 favors the formation of CO and O2.

Decrease in product concentration also favors the formation of product.

Decrease in reactant concentration favors the formation of reactants only.

Answer:

Slower

Explanation:

The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.

This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.

Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;

Rate = k[alkyl halide] [iodide]

Answer:

2.07459

Explanation:

this is the correct answer.

ummm is that chemistry?

Answer:

is this chem

Explanation:

molarity = moles of solute / volume of solution

moles of solute = molarity × volume of solution

moles of solute = 0.245 mol/L × 0.1500 L

moles of solute = 0.03675 mol

moles of solute = 0.0368 mol

-----------------------------------------------------------

Step 1: Calculate the molar mass of solute.

molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)

molar mass of solute = 110.98 g/mol

Step 2: Calculate the mass of solute.

mass of solute = moles of solute × molar mass of solute

mass of solute = 0.03675 mol × 110.98 g/mol

mass of solute = 4.08 g

Note: The volume of solution must be expressed in liters (L).

Answer:

[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]

[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]

In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.

First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.

Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.

[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]

We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]

The units of liters cancel.

[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]

[tex]0.03675 \ mol \ CaCl_2[/tex]

The original measurements have 3 significant figures, so our answer must have the same.

We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.

[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]

We can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.

The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.

Notice that chlorine has a subscript of 2. We must multiply the molar mass by 2.

Add calcium's molar mass.

Use the molar mass as a ratio.

[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

Multiply the moles of calcium chloride we calculated above.

[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

The units of moles of calcium chloride cancel.

[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]

[tex]4.084064 \ g\ CaCl_2[/tex]

Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.

[tex]\bold {4.08 \ g \ CaCl_2}[/tex]

Answer:

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Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J  +  5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

Answer:

no reaction occurs .that is no product

Answer:

88.9%

Explanation:

Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:

Después calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:

Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:

Finalmente calculamos el rendimiento porcentual:

Answer:

All of the above.

Explanation:

For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.

Answer: Molar mass of the unknown gas is 73.153 g/mol.

Explanation:

Given: Mass of each gas = 8.7 g

Volume = 17.28 L

Let us assume that the molar mass of gas is m g/mol.

Molar mass of Ar is 40 g/mol and Ne is 20 g/mol.

Hence, total moles of each gas are as follows.

[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol[/tex]

At STP, the total volume of these gases is as follows.

[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol \times 22.4 L = 17.28 L\\(\frac{8.7}{40} + \frac{8.7}{20})22.4 L + \frac{8.7}{m} \times 22.4 L = 17.28 L\\14.616 + \frac{8.7}{m} \times 22.4 L = 17.28 L\\\frac{8.7}{m} \times 22.4 L = (17.28 L - 14.616)\\\frac{8.7}{m} \times 22.4 L = 2.664 \\m = 73.153 g/mol[/tex]

Thus, we can conclude that molar mass of the unknown gas is 73.153 g/mol.

Answer: Hello the options related to your question are attached below

The slope is related to the rate constant so all four trials should have the same slope since the reactions are all the same ( Option C )

Explanation:

It is important when choosing the order of the reaction because the concentration of the bleaches used in the four trials are in excess hence their slopes have to be roughly the same and also because the reactions are similar and they where done at the same temperature, hence the slope of the first and second-order graphs will be the same.

Answer:

e. HCOOH and NaCHOO

Explanation:

For a buffer solution, both an acid and its conjugate base are required.

With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.

Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:

The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO

Answer:

Percent yield = 48.3%

Explanation:

The reaction is:

CCl₄  +  2HF → CF₂Cl₂ + 2HCl

1 mol of CCl₄ reacts with 2 moles of hydrofluoric acid in order to produce 1 mol of CF₂Cl₂ and 2 moles of hydrogen chloride.

HF is in excess, so the limiting reagent is the CCl₄.

We convert mass to moles:

32.9 g . 1mol / 153.8g = 0.214 moles

Ratio is 1:1. In conclussion: 0.0813 moles of CCl₄ can produce 0.0813 moles of CF₂Cl₂. We convert moles to mass, to determine the theoretical yield:

0.214 mol . 120.91g /mol = 25.8 g

Percent yield = (Yield produced /Theoretical yield) . 100

Percent yield = (12.5 g/ 25.8g) . 100 = 48.3%

Answer:

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

Answer:

9.36 g

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)

Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles

1 mole of octane yields 9 moles of water

0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water

Number of moles of oxygen = 23g/32g/mol = 0.72 moles

12.5 moles of oxygen yields 9 moles of water

0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water

Hence oxygen is the limiting reactant;

Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g

Explanation:

Esterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.

Answer:

The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.

The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.

The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.The methyl acetate that was formed is an ester. In this image, the green circle represents what was the carboxylic acid (in this case acetic acid), and the red circle represents what was the alcohol (in this case methanol):

This reaction lost an -OH from the carboxylic acid and a hydrogen from the alcohol. These two also combine to form water. So any esterification reaction will also form water as a side product.

Answer:

True

Explanation:

The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.

The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.

Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.

Answer:

Explanation:

/ means divided by

* means multiply

1. formula is

partial pressure = no of moles(gas 1)/ no of moles(total)

0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->

.3/(.6+.3+.1) =

.3/1 =

.3 =

partial pressure of CO

2.

.3 * .8 atm = .24

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quizlet

The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.

Dalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.

Dalton's Law of partial pressure using mole fraction of gas

Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure

Now, we have to find the first mole fraction of CO

Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]

                                                                  = [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]

                                                                  = [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]

                                                                  = [tex]\frac{0.30}{1}[/tex]

                                                                  = 0.3

Now, put the value in above equation, we get that

Partial pressure of carbon monoxide (CO)

= Mole fraction of carbon monoxide (CO) × Total pressure

= 0.3 × 0.8

= 0.24 atm

Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.

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Answer:

1, 1, 2, 3

Explanation:

The numbers 1 and 8 both have 1 sig. fig.

The number 13 has 2 sig. figs.

The number 104 has 3 sig. figs.