The rate law for the following reaction: H2SiF6(aq)+2NaF(aq)+3H2O(aq)⟶Na2SiO3(s)+8HF(aq) is second order in H2SiF6, zero order in NaF and first order in H2O. By what factor will the reaction rate change if the concentrations of all reactants are tripled?

Answer:

At higher temperatures, particles move faster and collide more, increasing solubility rates.

Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute

The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.

Explanation:

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:

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:

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:

C₃H₆O₃

Explanation:

To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).

With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:

Moles C -Molar mass:12.0g/mol-

0.24g * (1mol/12.0g) = 0.020 moles C

Moles H = Mass H because molar mass = 1g/mol:

0.040 moles H

Moles O -Molar mass: 16g/mol-

Mass O: 0.60g - 0.24g - 0.040g = 0.32g O

0.32g O * (1mol/16g) = 0.020 moles O

Ratio of atoms (Dividing in moles of C: Lower number of moles):

C = 0.020 moles C / 0.020 moles C = 1

H = 0.040 moles H / 0.020 moles C = 2

O = 0.020 moles O / 0.020 moles C = 1

Empirical formula:

CH₂O.

Molar mass CH2O:

12g/mol + 2*1g/mol + 16g/mol = 30g/mol

As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:

The molecular formula of the organic acid would be C3H6O3

Molecular formula = [empirical formula]n

Where n = molar mass/mass of empirical formula

Empirical formula

C = 0.24/12 = 0.02

H = 0.040/1 = 0.04

O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02

Divide by the smallest

C = 1

H = 2

O = 1

Empirical formula = CH2O

Empirical formula mass = 12 + 2 + 16 = 30

n = 90/30 = 3

Molecular formula = [CH2O]3

                               = C3H6O3

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Answer:

The Density Of Ar (g) At -11°C And 675 MmHg (R =0.08206 L·atm/mol·K, 1 Atm = 760mmHg).

Answer:

i have no answer for part A

part B

the one that has a 4 can be divided by 2 because reducing

part c

you can determine if an equation is written in the correct way by balancing the equation as if it had not been done already.

Answer:

0.74 M

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 5.90 M

Volume of stock solution (V₁) = 0.250 L

Volume of diluted solution (V₂) = 2 L

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.90 × 0.250 = M₂ × 2

1.475 = M₂ × 2

Divide both side by 2

M₂ = 1.475 / 2

M₂ = 0.74 M

Thus, the molarity of the diluted solution is 0.74 M

Answer:

Object: To demonstrate phototropism

equipments: A black box, window with light source, a well watered potted plant.

Experiment: A wellwatered potted plant is placed inside a darkened black  box that is having a small window on one side. This is called as phototropic chamber. Window is remain closed for a day the plant shows normal growth.

Whereas if window is opened atwo days it will be found that shoot tip bends and grows towards light proving that it is positively phototropic.

Answer:

6.52g = Actual yield (g)

Explanation:

The yield of a reaction is:

Percent yield = Actual yield (g) / Theoretical Yield (g) * 100

As 1 mol of isopentyl alcohol produce 1 mol of isopentyl acetate (Theoretical Yield), the theroretical yield of isopentyl acetate is 65.0mmol = 0.0650mol. To solve this question we need to convert the moles of isopentyl acetate to mass using its molar mass (130.19g/mol).

With the equation of percent yield we can find the mass obtained as follows:

Theoretical yield:

0.0650mol * (130.19g/mol) = 8.46g of isopentyl alcohol

Mass produced:

77 = Actual yield (g) / 8.462g * 100

6.52g = Actual yield (g)

The mass of isopentyl acetate that can be produced is 6.52 g

See attached photo

From the balanced equation,

1 mole of isopentyl alcohol reacted to produce 1 mole of isopentyl acetate.

Therefore,

65 mmole (i.e 0.065 mole) of isopentyl alcohol will also react to produce 0.065 mole of isopentyl acetate.

Actual yield = percent × theoretical

Actual yield = 77% × 0.065

Actual yield = 0.05005 mole

Mass = mole × molar mass

Mass of isopentyl acetate = 0.05005 × 130.19

Mass of isopentyl acetate = 6.52 g

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Answer:

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Explanation:

The total volume of the solution is:

4mL + 5.00mL + 12.0mL = 21mL

As the volume of the SbCl3 is 5.00mL, the dilution factor is:

21mL / 5.00mL = 4.2 times

The concentration of SbCl3 is:

0.10M SbCl3 / 4.2 times = 0.0238M SbCl3

The concentration of H+ = [HCl]:

4.5M / 4.2 times = 1.07M H+

The initial concentration of Cl- is:

3 times SbCl3 + HCl = 0.10M*3 + 4.5M =

3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-

4.8M Cl- / 4.2 times = 1.14M Cl-

Answer:

Butanoic acid

Explanation:

The IUPAC name of CH3CH2CH2COOH is:

The IUPAC name for a given compound is Butanoic acid.

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:

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.

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:

(A) 1.962x10^-10 M solubility in pure water

(B) 4.0 x 10^-33 M solubility

(C) 4.0 x 10^-27 M solubility

Explanation:

(A) Fe(OH)3 would give (Fe3+) and (3OH-)

Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

Let y = [Fe^3+]

Let 3y = [OH-]

4x10^-38 = (y)(3y)^3

4x10^-38 = 27y^4

y^4 = 4x10^-38 ÷ 27

y^4 = 1.481 x 10^-39

y = 1.962x10^-10 M solubility in pure water

(B) pH = 5.0

5.0 = - log [OH-]

-5.0 = log [OH-]

[OH-] = 10^-5.0 =  1.0 x 10^-5 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5

= 4.0 x 10^-33 M solubility

(C) pH = 11.0

11.0 = - log [OH-]

-11.0 = log [OH-]

[OH-] = 10^-11.0 =  1.0 x 10^-11 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11

= 4.0 x 10^-27 M solubility

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

khanacademy

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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#SPJ2

Answer:

The appropriate answer is "1.96 mol".

Explanation:

The given values are:

Volume,

V = 26.5 L

Pressure,

P = 1.7 atm

Temperature,

T = 280 K

Now,

The number of moles (n) will be:

= [tex]\frac{PV}{RT}[/tex]

By substituting the values, we get

= [tex]\frac{1.7\times 26.5}{0.0821\times 280}[/tex]

= [tex]\frac{45.05}{22.988}[/tex]

= [tex]1.96 \ mol[/tex]

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:

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

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:

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:

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.

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:

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.