Starting with 1.5 mL of isopentyl alcohol, 2.0 mL of concentrated acetic acid, and 0.4 mL of concentrated phosphoric acid, what is the theoretical yield (in g) of isopentyl acetate

The hypothesis should be; "Increasing the amount of ice increases the rate at which the ice will melt"

The term hypothesis refers to a tentative explanation that could be applied in order to explain an observation. When we give a hypothesis, it must be subjected to rigorous experimentation thus it can be confirmed or repelled by experiment.

Now, we can see that the question here is; "How does the amount of salt added to ice affect the rate at which the ice will melt?" This implies that the correct hypothesis here ought to have to do with the impact of the amount of ice in the affirmative.

Thus, the hypothesis should be; "Increasing the amount of ice increases the rate at which the ice will melt"

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Answer:

unbalanced

Explanation:

balanced: Al+HCl=AlCl+H

A 20. 0 l cylinder of oxygen gas is at a temperature of 27. 0°c and a pressure of 5. 00 atm. The density of the oxygen gas in the cylinder is 1.24 atm

The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa). Absolute, atmospheric, differential, and gauge pressures are different types of pressure.

By pressing a knife on some fruit, one can get a straightforward illustration of pressure. The surface won't be cut if you press the flat section of the knife against the fruit. The force is dispersed over a wide area (low pressure).

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Ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature.

The acetic acid and ethylene amine form salts and this result is expected same as form in the reaction of oxalic acid and ethylene diamine at ambient temperature.

The reaction of acetic acid with ammonium forms ammonium salt of carboxylic acid which on heating undergoes elimination to form amide as acetamine. The salt is form due to deprotonation of acid by the base. The amide is form on heating of the ammonium carboxylate salt. Such type of reaction is called aminooacide reaction.

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1.35 Litres of water need to be added to 750 ml of a 2.8 m hcl solution to make a 1.0 m solution.

Use the following relation:

M1V1=M2V2

Where M is molarity, V is volume and 1 is initial and 2 is the final conditions. Solving for V(2)

M1=2.8 M,V1=750 mL;M2=1.0 M

(2.8 M)×(750 mL)=(1.0 M)×V2

V2=(2.8 M)×(750 mL)(1.0 M) = 2100 mL = 2.1 L

Therefore, Volume of water to be added =2.1 L−0.75 L=1.35 L

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Ammonium chloride is a white solid that breaks down when heated and produces ammonia and hydrogen chloride.

A reversible reaction is a reaction in which the conversion of reactants to products and products to reactants occur at the same time. In the above example, the chemical shows a reversible reaction because it moves both forward and backward direction. In reversible reaction, equal amount of reactant is converted into product and product into reactant.

So we can conclude that Ammonium chloride is a chemical that represents a reversible reaction.

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A single slit with a width of 2019 * 109 m creates its initial minimal for 624 nm light at an angle of 18°.

How does diffraction work?

Waves spreading outward around obstructions are known as diffraction. Diffraction happens with sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as with incredibly minuscule moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities. Diffraction prevents the creation of sharp shadows as one of its effects. In order to spread out and illuminate regions at which a shadow is anticipated, light must be bent around corners, which is known as diffraction.

Calculation:

Provided for a single slit, m=1

λ = 624 *10⁻⁹

sinθ = sin 18⁰

Therefore,

asinθ=mλ

a = [tex]\frac{1 * 624 *10^{-9} }{sin 18}[/tex]

⇒a = 2019 *10⁻⁹ m

Therefore the width of a single slit is 2019 *10⁻⁹ m.

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The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:

Lead (ii) carbonate and  lead (ii) sulphate are both insoluble salts of lead.

In order to prepare lead (ii) sulphate, a two step process is performed.

In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.

In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.

In conclusion, lead (ii) sulphate is prepared in two steps.

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the first step in the mechanism is the acid-catalyzed generation of an enol and then electrophilic addition of bromine and cation is formed because of the destabilization effect of the electronegativity of oxygen

The ability of an atom or functional group to draw electrons to itself is known as an electronegativity in chemistry. An atom's electronegativity is influenced by both its atomic number and how far away from its charged nuclei its valence electrons are located.

The ability of an atom to draw shared electrons in a covalent connection is referred to as electronegativity. The stronger an element attracts the shared electrons, the higher its degree of electronegativity.

The propensity of an atom to attract other atoms when it is combined is known as an element's electronegativity. Additionally, a pair of bound electrons are shared. In contrast, an element's electropositivity refers to an atom's propensity to contribute electrons while also withdrawing from covalent connections.

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The solubility product, ksp of PbBr₂ is 2.102 × 10⁻⁶

The solubility of a solute is defined as the maximum amount of that solute that can be dissolved in a known quantity of solvent at a given temperature.

Some salts are sparingly soluble in a solvent. For them, we calculate the solubility product.

It is an equilibrium constant that defines the relationship between a solid and its respective ions in an aqueous solution in equilibrium.

The greater the solubility product, the greater the solubility and vice versa.

Here, the solubility of PbBr₂ = 2.96 g/l

Molar solubility of  PbBr₂ = [tex]\frac{2.96}{molar mass of PbBr_2}[/tex] = 2.96/367 = 8.07 × 10⁻³

At equilibrium,

[tex]PbBr_2\rightarrow Pb^{2+} + 2Br^-[/tex]

1 mole of PbBr2 dissociates into 1 mole of Pb²⁺ ions and 2 moles of Br⁻

Let the molar concentration of Pb²⁺ be x, then the molar concentration of Br⁻ is 2x

Ksp = x.(2x)²

       = 4x³

Substitute, x = 8.07 × 10⁻³

Ksp = 4 (8.07 × 10⁻³)³

      = 2.102 × 10⁻⁶

Thus, The ksp of PbBr₂ is 2.102 × 10⁻⁶

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Answer:

Below I have listed a write up that helped me!

Explanation:

Here are some general rules regarding significant figures (or sigfigs!)

1) any number that is not a 0 is always a sigfig. In your example, the significant figures would be 4, 6, 7,3,5 and *0. (*there are special rules regarding the number 0)

2) If a 0 is between two non-zero digits (sigfigs), it will be a significant figure. For example, in the number 305, the 0 between the 3 and 5 would be considered a significant figure.

3) 0's that "lead", or are to the left of significant figures are NOT significant figures. For example, in the number 0.002, there would be only one significant figure, which is the 2.

4) 0's that "trail", or are to the right of a significant figure behind a decimal place, are significant figures. For example, in the number 3.1250, the 0 at the end would be considered significant. There are five sigfigs total in this number. The trailing 0 in your example would be significant!

5) However, regarding the rule above, if the number DOES NOT have a decimal place and has trailing 0's, those 0's will not be considered significant figures. For example, in the number 5,320, the 0 at the end would not be significant and there would be three sigfigs in this number.

Answer:

[tex]\frac{g}{mL}[/tex]

Explanation:

You can determine the units in the final answer by identifying which of the units cancel out. Units are eliminated (cancelled out) when they are located both in the numerator and denominator of proportions being multiplied.

In this case, these units are cancelled out.....

-----> milligrams (mg) = (1st and 2nd proportions)

-----> decaliters (dL) = (1st and 3rd proportions)

-----> liters (L) = (3rd and 4th proportions)

As these units are not cancelled out, you are left with grams (g) in the numerator and milliliters (mL) in the denominator.

The final units should be represented by:

[tex]? \frac{g}{mL}[/tex]

The value of r is 0.0868 and the error percentage is 5.78%.

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There are 372 grams of calcium phosphate are theoretically produced if we start with 3. 40 moles of calcium nitrate and 2. 40 moles of lithium phosphate .

Calculation,

The reactant or reagent that produces the least moles of the products is called limiting reagents. When  limiting reagents used up , the reaction stops.

The balanced equation is given as,

[tex]3Ca(NO_{3} )_{2} + 2Li_{3} ( PO_{4} )[/tex] → [tex]3LiNO_{3} + Ca_{3}( PO_{4} )_{2}[/tex]

Multiply the moles of each reactant by the mole ratio between it and calcium phosphate in the balanced equation . so that the moles of the reactant cancel , leaving moles of calcium phosphate.

3.4 mol of calcium nitrate × 1 mol calcium phosphate / 3 mol calcium nitrate = 1.13 mol calcium phosphate

2.4 mol of  lithium phosphate× 1 mol calcium phosphate / 2 mol  lithium phosphate = 1.02 mol calcium phosphate

So, calcium nitrate  is limiting reactant .

Calculation of mass of 1.02 mol calcium phosphate.

Multiply the moles of  calcium phosphate by its molar mass.

molar mass of  calcium phosphate = 3×40.078 g/mol calcium ion+2×30.9 g/mol phosphorus + 8×15.99 g/mol calcium phosphate = 310.178 g/mol calcium phosphate

1.20 mol calcium phosphate × 310.178 g/mol = 372 gram

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The equilibrium constant, Keq for the reaction, given the data from the question is 0.084

The equilibrium constant, Keq for a reaction is the ratio of the concentration of the products raised to their coefficients to the ratio of the concentration of the reactants raised to their coefficients. For example, the equilibrium constant for the reaction nA + mB <=> eC can be obtained as follow

Keq = [C]^e / [A]^n × [B]^m

The equilibrium constant, Keq for the above reaction can be obtained as illustrated below:

Keq = [SO₂]^2 × [O₂] / [SO₃]^2

Keq = (0.17)^2 × (0.701) / (0.491)^2

Keq = 0.084

Thus, the equilibrium constant, Keq for the reaction is 0.084

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The  [H3O+] in step 1 is 0.0034 M while the  [H3O+] in step 2 is 0.00039 M

Let us set up the ICE table in each case, for K1;

         H2A(aq) + H2O(l)-------->   H3O^+(aq) + HA^-(aq)

I        0.12                                     0                    0

C       -x                                        +x                   +x

E      0.12 - x                                x                     x

Ka1= [H3O^+] [HA^-]/[ H2A]

Ka1= x^2/  0.12 - x  

1.0×10^−4 = x^2/  0.12 - x  

1.0×10^−4(0.12 - x ) = x^2

1.2 * 10^-5 - 1.0×10^−4x =  x^2

x^2 +  1.0×10^−4x - 1.2 * 10^-5  = 0

x =0.0034 M

[H3O+] = 0.0034 M

Again;  [H3O+] = [HA^-] = 0.0034 M

          HA^-(aq) + H20(l)    -------> A^-(aq)   + H3O^+

I       0.0034                                  0               0

C       -x                                          + x            +x

E    0.0034 - x                               x                x

Ka2= [A^-] [H3O^+]/[HA^-]

5.0×10^−5 = x^2/ 0.0034 - x  

5.0×10^−5 (0.0034 - x ) = x^2

1.7 * 10^-7 - 5.0×10^−5x =  x^2

x^2 + 5.0×10^−5x - 1.7 * 10^-7 = 0

x=0.00039 M

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The chemoreceptors located on the tongue for the detection of taste are found in structures called taste buds.

What are chemoreceptors?

What are taste buds or taste receptors?

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The descriptive term that is applied to the type of diene represented by 1,5-octadiene is isolated diene. The correct option is C.

Diene is a compound that contains two or more double bonds, usually  carbon bonds, which are separated by a single bond. They are covalent compounds. Alkene units are surely present in these compounds, whose quantity is two.

1,5-octadiene is a polymer of diene, which are generally elastomers, and they made of vulcanized rubber. They are isolated diene.

Thus, the correct option is C) Isolated diene.

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The question is incomplete. The complete question is given below:

A) Conjugated diene.

B) Cumulated diene.

C) Isolated diene.

D) Alkynyl diene.

E) None of the above.

The molecular weight of this gas will be 45 g/mol .

The state equilibrium equation for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it represents a decent approximation of the activity of many gases under various conditions.

Ideal gas law can be expressed as:

PV =nRT

Calculation of molecular weight by using ideal as law.

Given data:

P = 3 atm

T = 127 °c

Density =  4. 20 grams per liter

PV =nRT

where p is pressure , T is temperature and R is gas constant.

PV = gram / molecular weight RT

Molecular weight = (g/v)( 1/P) RT

Putting the given data in above equation.

Molecular weight =4.20 × 1/ 3× 400 × 0.0831

Molecular weight = 45 g/mol.

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Standard enthalpy change of neutralization is the enthalpy change when 1 mole of water is produced by the reaction of an acid and a base under standard conditions.

Enthalpy change of neutralization:

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Soil particles, soot, lead (pb), asbestos, sea salt, and sulfuric acid droplets are all types of particulate matter.

The atmosphere is dusty .The solid and liquid particles suspended in the atmosphere .

Examples : Soil particles, soot, lead (pb), asbestos, sea salt, and sulfuric acid droplets are all types of particulate matter

There are different categories of particulate matter. Such as mold spores , bacteria , dust , smoke and airborne viral paricles. The major sources of particulates matter are wildfires , dust storms , sea spray , volcanic eruption , and industrial process.

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The pH of the solution is used to estimate the acidic and the alkaline condition. The pH paper can be used to determine the conditions. The compound with pH 13.3 is basic.

The concentration of the hydrogen or the hydroxide ion in the water gives the estimate of the pH. The potential and the amount of hydrogen decide the acidic and the basic compound.

The pH scale ranges from 0-14 where the scale of 0-6 is acidic, 7 is neutral and 8-14 is basic. If the substance shows a pH of 13.3 then it will lie in the basic range.

Therefore, the compound with a pH of 13.3 is basic.

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1. The mass of RbCN required to prepare the solution is 4.97 g

2. The mass of RbCN required to prepare the solution is 3.31 g

3. 1 molar solution of CaCl₂ contains 1 mole of CaCl₂ in 1 L of the solution

4. The empirical formula of the compound is Li₂SO₄

5. The empirical formula of the compound is Sc₂O₃

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

We'll begin by calculating the mole of RbCN in the solution

Mole = molarity × volume

Mole = 0.12 × 0.37

Mole = 0.0444 mole

Finally, we shall determine the mass of RbCN required as illustrated below

Mass = mole × molar mass

Mass of RbCN = 0.0444 × 112

Mass of RbCN = 4.97 g

We'll begin by calculating the mole of RbCN in the solution

Mole = molarity × volume

Mole = 0.12 × 0.246

Mole = 0.02952 mole

Finally, we shall determine the mass of RbCN required as illustrated below

Mass = mole × molar mass

Mass of RbCN = 0.02952 × 112

Mass of RbCN = 3.31 g

This is a solution that contains 1 mole of CaCl₂ in 1 L of the solution

Divide by their molar mass

Li = 12.6 / 7 = 1.8

S = 29.2 / 32 = 0.9125

O = 58.2 / 16 = 3.6375

Divide by the smallest

Li = 1.8 / 0.9125 = 2

S = 0.9125 / 0.9125 = 1

O = 3.6375 / 0.9125 = 4

Thus, the empirical formula of the compound is Li₂SO₄

Divide by their molar mass

SC = 65.2 / 45 = 1.449

O = 34.8 / 16 = 2.175

Divide by the smallest

SC = 1.449 / 1.449 = 1

O = 2.175 / 1.449 = 3/2

Multiply by 2 to express in whole number

SC = 1 × 2 = 2

O = 3/2 × 2 = 3

Thus, the empirical formula of the compound is Sc₂O₃

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The volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.

If 400 ml of CO2 is produced at 30°C at 740 torr, then number of moles can be calculated as:

By using ideal gas equation:

P1V1 = N1R1T1

P1 = pressure = 740torr

V1 = 400 ml = volume of CO2

R = Gas constant = 8.314

T = 273+30 = 303 k

740×400 = N1×8.314×303

N1 = (740×400) /(8.314×303) =117.5.

Chemical equation

C2H6 ---- 2CO2 + 3H2O.

As we noticed from the equation that

2 moles of CO2 = 3 moles of H2O

1 moles of CO2 × 1 moles of H2O

Then N2 = 117.5 moles of CO2 = 3/2 × 117.5 moles of H2O

By using ideal gas equation:

P2V2 = N2RT2

V2 = 3/2 × 117.5 × 8.314 × 292/ 780

= 548.5ml.

Thus, we found that the volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.

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Answer:

It's too short. Write at least 20 characters to explain it well.

Explanation:

It's too short. Write at least 20 characters to explain it well.

The same amount of energy is released when they recombine.

Like energy components, electrolyzers comprise of an anode and a cathode isolated by an electrolyte. Different electrolyzers capability in various ways, fundamentally because of the different kind of electrolyte material included and the ionic species it conducts.

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The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

= 4.75

Therefore, 4.75 is the pH of the buffer.

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Question:

If butane had a volume of 500 ml. at a pressure of 1.5 atm and a temperature of 20 °C, what would the new volume of the gas be at a temperature of 30 °C and a pressure of 500 Torr?

Solution Given:

Let P be the pressure V be volume and and T be temperature.

Volume of Butane [[tex]V_1[/tex]] = 500 ml

Pressure of Butane [[tex]P_1[/tex]] = 1.5 atm

Temperature [[tex]T_1[/tex]] =20°C=20+273=293K

New Volume of Butane [[tex]V_2[/tex]] = ?

New Pressure of Butane [[tex]V_2[/tex]] =500 Torr=500*0.00131579=0.657895 atm

Note: 1 Torr= 0.00131579 atm

New Temperature of Butane [[tex]V_2[/tex]] =30°C=30+273=303K

Now

By using combined gas law equation:

[tex]V_2=V_2*\frac{P_1*T_2}{P_2*T_1}[/tex]

[tex]V_2=500*\frac{1.5*303}{0.657895*293}=1178.9 ml[/tex]

The new volume of Butane is 1178.9 ml

There are  0.94 g mass of [tex]CaCO_{3}[/tex] is required to react completely with 25. 0 ml of 0. 750 m HCl .

Calculation ,

Mass of [tex]CaCO_{3}[/tex] = ?

The 1000ml of HCl  = 27.375 g

then the 1 ml of solution contains HCl = 27.375 g/1000×1

25 ml  of solution contains HCl = 27.375 g/1000 × 25 = 0.684 g

The chemical equation can be given as :

[tex]CaCO_{3} +HCl[/tex]  → [tex]CaCl_{2} +CO_{2} +H_{2} O[/tex]

2 mol of HCl reacts with 1 mol of [tex]CaCO_{3}[/tex]

The amount of [tex]CaCO_{3}[/tex] reacted is given by ,

100/71 × 0.684 g = 0.9639 g = 0.94 g

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Answer:

  3.0×10⁻¹³ M

Explanation:

The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.

The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.

  Ksp = [Al₃⁺³]×[OH⁻]³

The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.

This means the solubility product equation is ...

  Ksp = x(10⁻⁷ +3x)³

For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...

  3.0×10⁻³⁴ = x(10⁻⁷)³

Then x = [Al₃⁺³] will be ...

  [tex][\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}[/tex]

We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.

__

Additional comment

The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.

Equilibrium constant for the reaction is

[tex]\sf K_{sp}=[Al^{3+}][OH^-]^3[/tex]

So

[tex]\\ \rm\Rrightarrow s=\dfrac{3.0\times 10^{-34}}{(10^{-7})^3}[/tex[

[tex]\\ \rm\Rrightarrow s=\dfrac{3.0\times 10^{-34}}{10^{-21}}[/tex]

[tex]\\ \rm\Rrightarrow s=3.0\times 10^{-13}M[/tex]