A 18.0 L gas cylinder is filled with 6.20 moles of gas. The tank is stored at 33 ∘C . What is the pressure in the tank?
Express your answer to three significant figures and include the appropriate units.

Answer:

See explanation

Explanation:

The IUPAC name of aspirin is 2-Acetoxybenzoic acid. It is composed of an acetoxy moiety and a benzoic acid moiety.

The compound can be hydrolysed under prolonged storage conditions to yield acetic acid which causes the vinegar like odour.

Also, one of the products of this hydrolysis bears a phenol group which reacts with FeCl3 to give a purple color.

Answer:

250ml

Explanation:

call it V

V*0.2=0.05 (moles)

so V=0.05/0.2 = 0.25l = 250ml

We know

[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]

Answer:

Once you have collected 40 mL of distillate, you should ________.

turn off the hot plate and carefully lower the lab jack, making sure that no cords or hoses get caught in it.

Explanation:

Distillate is the product obtained from the process of distillation.  Distillation is the separation of components of a liquid mixture based on different boiling points. Distillation can be used to purify alcohol, for desalination, refining of crude oil, and for obtaining liquefied gases.  A lab jack is an essential tool that supports and lifts hotplates, glassware, baths, and other small lab equipment requiring stable surfaces at a specific height.

Answer:

a. 6.15 mL b. 30.73 mL

Explanation:

a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?

Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.

So concentration of ketamine C = mass of ketamine, m/volume of water, V

m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL

So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL

Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg

Since mass, M = concentration ,C × volume, V

M = CV

V = M/C

The volume of ketamine required for the 0.400 mg/kg high is

V = 26 mg/4.23 mg/mL

V = 6.15 mL

b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg

Since mass, M' = concentration ,C × volume, V

M' = CV

V = M/C

The volume of ketamine required for the 2.00 mg/kg unconscious injection is

V = 130 mg/4.23 mg/mL

V = 30.73 mL

Answer:

d. amine.

It becomes an amine.

Explanation:

With general formular

[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]

R is the aryl group such as alkane

Answer:

a. True

Explanation:

The main information that gives an infrared absorption spectrum is the type of functional groups that are present in an organic compound. The infrared (IR) spectroscopy is based on the fact that functional groups absorb light in the IR region of the electromagnetic spectrum (approximately at 2,500-16,000 nm) and induces a vibrational excitation of the covalently bonded atoms in the group. The vibration of the atoms can be of different types, such as stretching, bending, etc. Each functional group (such as the carbonyl group) in an organic compound absorbs at a specific IR frequency so they can be distinguished from an IR spectrum.

Answer:

B) ethers

Explanation:

The functional group of an organic compound defines its specificity. The functional group is responsible for the chemical behavior of an organic compound. For example, alkenes are known to have a carbon-carbon double bond (C=C) functional group.

Likewise, organic compounds known as ETHERS are known to possess an ethoxy functional group i.e. oxygen atom bonded to two alkyl groups (R- OR; where R is an alkyl group). Members of ether functional group includes dimethyl ether (CH3-O-CH3), diethyl ether (C2H5-O-C2H5).

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

Answer:

11.9 pH

Explanation:

First, we need to find pOH

To find that, we use the formula -log[OH]

-log[7.7x10^-3] = 2.11351

To find the pH, we'll use this formula: 14 = pH + pOH

14 = pH + 2.11351

Subtract boths sides by 2.11351

14 = pH + 2.11351

-2.11351  -2.11351

pH = 11.88649

Answer:

cool or increase pressure

Explanation:

Explanation:

As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.

The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.

In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.

The compound in this reaction which is having the elemental ratio of 1:4:1 is NH₄Cl where nitrogen and chlorine are of one mole each with 4 hydrogens.

Elemental ratio of a compound is the ratio of number of atoms  of each  elements in that compound. The elemental ratio can be determined from the molecular formula of compounds.

The given reaction is a double displacement reaction. Here, the Cl group is replaced to the ammonia and OH group is replaced to the water. Thus, two species is replaced in the reaction.

In NH₄Cl, there are one nitrogen, 4 hydrogens and one chlorine atom. Therefore, the elemental ratio of the compound is 1:4:1. The elemental ratio of water is 2:1 and HCl is 1:1 and that in NH₄OH is 1:5:1. Hence, option b is correct.

To find more on elemental ratio, refer here:

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Answer:

b) 0.75 mol, 0.9375 mol

Explanation:

According to this question, ammonia reacts with oxygen to produce nitrogen monoxide and water. The chemical equation is as follows:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Based on this balanced equation, 4 moles of ammonia (NH3) reacts with 5 moles of oxygen (O2).

A stoichiometric amount of the two reactants (NH3 and O2) must represent the ratio 4:5.

Given the provided options;

0.75 mol of ammonia (NH3) will react with (0.75 × 5/4) = 0.935 mol of O2 for them to be in stoichiometry.

N.B: 1 mol of NH3 will react with 1.25mol of O2 and not 1g, 1.25g.

Answer:

There are no options in this question, however, it can be answered based on general understanding

- The number of neutrons each isotope contain

Explanation:

Isotopes are atoms of an element with the same atomic number or number of protons but different mass number/atomic masses. Since isotopes have same proton numbers, they have similar chemical behavior or identity.

However, difference in atomic mass or mass number of the same atomic number indicates that the number of neutrons each isotope contain differs from one another. Hence, in two isotopes of chlorine given as follows: Cl-35 and Cl-37, the NUMBER OF NEUTRONS in each atom differentiates the two isotopes.

Cl-35 contains 18 neutrons while Cl-37 contains 20 neutrons.

Answer:

2 m/s

Explanation:

Applying the formulae of velocity,

V = d/t............. Equation 1

Where V = Velocity of the body, d = distance, t = time

From the question,

Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.

Substitute these values into equation 1

V = 600/300

V = 2 m/s.

Hence the velocity of the body when it travels is 2 m/s

Answer:

pH of buffer solution is 7.0

Initial pH of Weak acid is 3.27

Final pH of weak acid is 3.07

Amount of NaOH added is 1ml

Explanation:

Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.

Answer:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Explanation:

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.

Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Una sustancia homogénea es una sustancia que se compone de una sola fase.

Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.

El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.

Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.

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Solution :

Molecular      Molar Mass       Volume      Density       Mass      Moles      nmoles

formula            (g/mol)               (mL)          (g/mL)           (g)

[tex]$C_6H_8N_2$[/tex]            108.14                                                    0.108      0.001          1

HCOOH           46.02                0.064          1.22     0.07808     0.0017       1.7

mmoles of o-phenylenediamine = 1 mmoles

mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles

From the reaction of o-phenylenediamine and formic acid, we see,

1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.

But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.

The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.

molar mass of Benzimidazole = [tex]118.14[/tex] g/mol

mmoles of Benzimidazole formed = [tex]1[/tex] mmol

Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]

                                                    [tex]$=\frac{118.14 \times 1}{1000}$[/tex]

                                                     = 0.11814 g

So the theoretical yield of Benzimidazole is = 0.118 g = 118mg

Answer:

pH = 12.43

Explanation:

...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

Moles KOH:

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

Moles HN3:

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.

electron configuration is 2,8,6.

Belongs to group 6 and period group 3.

It forms an ion by accepting 2 electrons

Answer:

Boiling T° of solution  → 83.6°C

Explanation:

To solve this, we apply Elevation of boiling point, property

ΔT = Kb . m . i

As we talk about organic solute, i = 1. No ions are formed.

m = molality (moles of solute in 1kg of solvent)

We determine mass of solvent by density

D = m /V so D . V = m

950 mL . 0.877 g/mL = 833.15 g

We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg

Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol

m = 1.17mol / 0.833 kg = 1.41 mol/kg

We replace data:

Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1

Boiling T° of solution = 2.53°C/m . 1.41 m . 1  +  80.1°C → 83.6°C

Answer:

The answer is c or 17.1 g

Answer:

There is 13.48 grams of P2O5 formed

Explanation:

Step 1: Data given

A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.

Mass of P = 5.89 grams

Molar mass of O2 = 32.0 g/mol

atomic mass of P = 30.97 g/mol

molar mass of P2O5 = 141.94 g/mol

Step 2: The balanced equation

4P(s)+5O2(g)⇔ 2P2O5(s)

Step 3: Calculate moles of P

Moles P = Mass P / atomic mass P

Moles P = 5.89 grams / 30.97 g/mol

Moles P = 0.190 moles

Step 4: Calculate moles of P2O5

For 4 moles P we need 5 moles O2 to produce 2 moles P2O5

For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5

Step 5: Calculate mass of P2O5

Mass P2O5 = moles P2O5 * molar mass P2O5

Mass P2O5 = 0.095 moles * 141.94 g/mol

Mass P2O5 = 13.48 grams

There is 13.48 grams of P2O5 formed

Answer:

0.886 J/g.°C

Explanation:

Step 1: Calculate the heat absorbed by the water

We will use the following expression

Q = c × m × ΔT

where,

Q(water) = c(water) × m(water) × ΔT(water)

Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J

According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.

Q(water) + Q(solid) = 0

Q(solid) = -Q(water) =  -1.89 × 10³ J

Step 2: Calculate the specific heat capacity of the solid

We will use the following expression.

Q(solid) = c(solid) × m(solid) × ΔT(solid)

c(solid) = Q(solid) / m(solid) × ΔT(solid)

c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C

Answer:

0.0922 g

Explanation:

Number of moles of acid present = 25/1000 × 0.0981

= 0.00245 moles

Number of moles of base = 5.83/1000 × 0.104

= 0.000606 moles

Since the reaction of HCl and NaOH is 1:1

Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles

= 0.001844 moles

From the reaction;

CaCO3 + 2HCl ----> CaCl2 + H2O + CO2

1 mole of CaCO3 reacts with 2 moles of HCl

x moles of CaCO3 reacts with 0.001844 moles ofHCl

x = 1 × 0.001844/2

= 0.000922 moles

Mass of CaCO3 = 0.000922 moles × 100 g/mol

= 0.0922 g

Answer:

velocity = distance/time

velocity = wavelength × frequency

Both of these are commonly known equations to calculate velocity with different variables.

Explanation:

The given balanced chemical equation is:

[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]

The value of Kc at 445oC is 0.020.

[HI]=1.5M

[H2]=2.50M

[I2]=0.05M

The value of Qc(reaction quotient ) is calculated as shown below:

Qc has the same expression as the equilibrium constant.

[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]

Qc>Kc,

Hence, the backward reaction is favored and the formation of Hi is favored.

Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.

Answer:

NaCl, Na⁺,Cl⁻.

MgCl₂, Mg²⁺, Cl⁻.

CaO, Ca²⁺, O²⁻.

Li₃P, Li⁺, P³⁻.

Al₂S₃, Al³⁺, S²⁻.

Ca₃N₂, Ca²⁺, N³⁻.

FeCl₃, Fe³⁺, Cl⁻.

FeO, Fe²⁺, O²⁻.

Cu₂S, Cu⁺, S²⁻.

Cu₃N₂, Cu²⁺, N³⁻.

ZnO, Zn²⁺, O²⁻.

Ag₂S, Ag⁺, S²⁻.

K₂CO₃, K⁺, CO₃²⁻.

NaNO₃, Na⁺, NO₃⁻.

Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.

Al(OH)₃, Al³⁺,OH⁻.

Li₃PO₄, Li⁺, PO₄³⁻.

K₂SO₄, K⁺, SO₄²⁻.

Explanation:

Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.

Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.

Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.

Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.

Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.

Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.

Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.

Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.

Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.

Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.

Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.

Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.

Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.

Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.

Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.

Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.

Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.

Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.