In Experiment 1, salicylic acid was treated with an excess of acetic anhydride to synthesize aspirin. Once the reaction went to completion, water was added to the flask and the solution was heated. What was the primary reason for adding water

Answer:

a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :

There are no insoluble precipitate forms.

b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :                        

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.

                                          KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex]  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.

                                        [tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex]  ⇒ soluble.

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c)

Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.

                                        [tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex]  ⇒ insoluble.

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d)

Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.

                                        [tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex]  ⇒ soluble.

There are no insoluble precipitates forms.

Answer:

(B). it's metallic bonding

Answer:

77%

Explanation:

First we convert 3.0 g of cyclohexanol (C₆H₁₂O) to moles, using its molar mass:

Then we convert 1.9 g of cyclohexene (C₆H₁₀) to moles, using its molar mass:

Finally we calculate the theoretical yield:

Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

1.8×10⁻⁶ = x²

√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH

Answer:

*The molar volume of any gas at standard conditions of temperature and pressure is 22.4 L/mol

V= 101.3 L , n=? , Vm = 22.4 L/mol

101.3 = n × 22.4

n=101.3 / 22.4

n = 4.52 mol

I hope I helped you ^_^

Answer:

[tex]\boxed {\boxed {\sf 4.522 \ mol}}[/tex]

Explanation:

We are asked to find how many moles of gas occupy a volume of 101.3 liters.

1 mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 liters. We can use this information to make a proportion.

[tex]\frac {1 \ mol}{22.4 \ L}[/tex]

We are converting 101.3 liters to moles, so we multiply the proportion by that value.

[tex]101.3 \ L *\frac {1 \ mol}{22.4 \ L}[/tex]

The units of liters (L) cancel.

[tex]101.3 *\frac {1 \ mol}{22.4}[/tex]

[tex]\frac {101.3}{22.4} \ mol[/tex]

Divide.

[tex]4.52232143 \ mol[/tex]

The original value of liters (101.3 L) has 4 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 3 in the ten-thousandths place to the right tells us to leave the 2 in the thousandths place.

[tex]4.522 \ mol[/tex]

101.3 liters of gas is equal to approximately 4.522 moles of gas.

Answer:

1. Nitrate ions, NaNO3 - Sodium nitrate.

2. Sulphide ions, K2S - Potassium sulphide.

3. Sulphate ions, CaSO4 - Calcium sulphate.

4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.

5. Carbonate ions, CaCO3 - Calcium carbonate.

6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.

7. Phosphite ions, PH3 - Hydrogen phosphite.

8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).

9. Ethanoate ions, CH3COONa - Sodium ethanoate.

10. Methanoate ions, HCOONa - Sodium methanoate.

11. Fluoride ions, HF - Hydrogen fluoride.

12. Chloride ions, KCl - Potassium chloride.

13. Bromide ions, HBr - Hydrogen bromide.

14. Iodide ions, NaI - Sodium iodide.

15. Phosphate ions, K3PO3 - potassium phosphate.

Answer:

See explanation

Explanation:

Truly, the second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings.

When a solid is dissolved in a liquid, the solid dissociates into ions. These ions increases the number of particles and hence the entropy of the system thereby making the process spontaneous.

Hence, the dissolution of a substance via an endothermic process is spontaneous because of increase in the number of particles which in turn increases the entropy of the system.

Answer:

[tex]\boxed {\boxed {\sf 426 \ mL}}[/tex]

Explanation:

We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:

[tex]P_1V_1= P_2V_2[/tex]

The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.

[tex]450 \ mL * 720 \ mm \ Hg = P_2V_2[/tex]

The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.

[tex]450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2[/tex]

We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.

[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}[/tex]

[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2[/tex]

The units of millimeters of mercury (mm Hg) cancel.

[tex]\frac {450 \ mL * 720 }{760} = V_2[/tex]

[tex]\frac {324,000}{760} \ mL = V_2[/tex]

[tex]426.3157895 \ mL =V_2[/tex]

The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.

[tex]426 \ mL \approx V_2[/tex]

The volume at standard atmospheric pressure is approximately 426 milliliters.

Answer:

640 g

Explanation:

Step 1: Given and required data

Step 2: Calculate the moles of oxygen gas

We will use the ideal gas equation

P × V = n × R × T

n = P × V / R × T

n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol

Step 3: Calculate the mass corresponding to 20 moles of oxygen

The molar mass of oxygen is 32.00 g/mol.

20 mol × 32.00 g/mol = 640 g

 9 is the element Florine

Florine has 9 electrons as well as the 9 protons that determine its atomic number.

The ground state configuration is the lowest energy configuration.

Explanation:

1. boiling and melting point

2. electrical conductivity

3. Bond strength

4. bond length

A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.

Answer:

Increase

Decrease

Explanation:While in solution, ionic substances produce ions. The ions in solution determine the conductivity of the solution.

The ionic strength of a solution shows the concentration of ions in a given solution. The more the number of ions in the solution, the greater the ionic strength of the solution and vice versa.

When HNO3 is added to a solution of KOH, the number of ions in the solution increases and so does the ionic strength of the solution.

When KOH is added to a solution of CaCl2 then Ca(OH)2 is formed. The formation of a solid precipitate decreases the concentration of ions in solution as well as the ionic strength of the solution.

Answer:

A) Forms a racemic mixture of the two possible enantiomers

When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and  new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

There are three types of chemical changes:

1) inorganic changes

2)organic changes

3) biochemical changes

During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.

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Answer:

Mole = molecular weight / molecular mass

Mole = 16/16

Mole= 1

The volume of 1.5 moles of gas at STP  is 33.6 L.

The volume of the gas at STP is calculated as follows;

I mole of gas at STP = 22.4 L

1.5 moles of the gas at STP = ?

= 1.5 moles  x 22.4 L/mole

= 33.6 L

Thus, the volume of 1.5 moles of gas at STP  is 33.6 L.

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Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

Answer:

B

Explanation:

Answer:

[tex]K=1.7x10^{-3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:

[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]

Next, we plug in the given concentrations on the data table to obtain:

[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]

Regards!

Answer:

MoO2

Explanation:

The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.

The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239.9g/mol-

12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo

Mass Mo -95.95g/mol-:

0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo

Mass oxygen in the oxide:

13.197 - 9.895g = 3.302g Oxygen

Moles oxygen -Molar mass: 16g/mol-:

3.302g Oxygen * (1mol / 16g) = 0.206 moles O

Now, the ratio of moles O / moles Mo is:

0.206 moles O / 0.1031 moles Mo = 2

That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

this is the difference between organic and inorganic if this doesn't help you can research more on it

Explanation:

here's the answer to your question

Answer:

There is no difference between the two.

Explanation:

They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments

Answer:

4.858 g

Explanation:

Start with the formula

density = [tex]\frac{mass}{volume}[/tex]

density = 1.98 g/mL

volume = 2.45 mL

mass = ??

rearrange the formula to solve for mass

(density) x (volume) = mass

Add in the substitutes and solve for mass

1.98 g/mL x 2.45 mL = 4.858 g

Answer:

Explanation:

maleic acid ⇒ fumaric acid

ΔHreaction = ΔHproduct - ΔHreactant

ΔHproduct = -1336.0 kJ mol⁻¹

ΔHreactant = - 1359.2 kJ mol⁻¹.

ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)

=   1359.2 kJ mol⁻¹   -1336.0 kJ mol⁻¹

= 23.2 kJ mol⁻¹ .

Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.

The properties of solids, liquids and gases regarding their shapes and volumes are:

                       Shape             Volume

Solids              Fixed                Fixed

Liquids           Variable            Fixed

Gases             Variable           Variable

Solids have strong attraction forces between their molecules. Thus, the molecules are closely packed with little movement. As a consequence, both shape and volume are fixed.

In liquids, attraction and repulsion forces are similar. They have a little more movement than the solid state. Then, they do have a fixed volume but they adopt the shape of the container.

Gases have very weak attraction forces between their molecules. They move very freely and expand trying to occupy as much volume as possible. So, they have a variable volume and shape (adopt the shape of the container).

You can learn more about states of matter here:

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Answer:

[tex]C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:

[tex]C_9H_8O_4+NaC_2H_3O_2\rightarrow C_2H_4O_2+NaC_9H_7O_4[/tex]

Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:

[tex]C_9H_8O_4+Na^++C_2H_3O_2^-\rightarrow C_2H_4O_2+Na^++C_9H_7O_4^-[/tex]

Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:

[tex]C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-[/tex]

Regards!

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄