Answer:
d. 60.8 L
Explanation:
Step 1: Given data
Heat absorbed (Q): 53.1 JExternal pressure (P): 0.677 atmFinal volume (V2): 63.2 LChange in the internal energy (ΔU): -108.3 JStep 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge.
b. A beta particle contains neutrons.
c. A beta particle is less massive than a gamma ray.
d. A beta particle is a high-energy electron.
Answer:
a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.
Explanation:
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)
b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.
c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.
d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.
If 7 mol of copper reacts with 4 mol of oxygen, what amount of copper (II) oxide is produced? What amount of the excess reactant remains?
Answer:
7 mol CuO
0.5 mol O₂
Explanation:
Step 1: Write the balanced equation
2 Cu + O₂ ⇒ 2 CuO
Step 2: Identify the limiting reactant
The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.
The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.
Since TMR > EMR, Cu is the limiting reactant
Step 3: Calculate the amount of CuO produced
7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO
Step 4: Calculate the excess of O₂ that remains
The amount of O₂ that reacts is:
7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂
The excess of O₂ that remains is:
4 mol - 3.5 mol = 0.5 mol
All of the following statements concerning real cases is correct EXCEPT Group of answer choices molecules of real gases are attracted to each other. molecules of real gases occupy no volume. nonideal gas behavior is described by the Van der Waals Equation. the pressure of a real gas is due to collisions with the container. the pressure of a real gas at low temperatures is lower than for ideal gases.
Answer:
molecules of real gases occupy no volume.
Explanation:
As all the real gases are composed of particles that occupy the non-zero volume that is the excluded volume. If the gas is behaving in an ideal manner. The correction becomes negatable and is relative to the total volume. The extended volume is volume that is taken by the non ideal gas particles.HELP ASAP PLS
Reactions, products and leftovers
Answer:
See the answer below
Explanation:
From the original equation in the image, the mole ratio of C:CO2:CO is 1:1:2. This means that for every 1 mole of C and CO2, 2 moles of CO would be produced.
Now, looking at the simulation below the equation of the reaction, 3 moles of C and 8 moles of CO2 were supplied as input. Applying this to the original equation of reaction, C seems to be a limiting reagent for the reaction because the ratio of C to CO2 should 1:1.
Hence, taking all the 3 moles of C available means that only 3 moles out of the available 8 for CO2 would be needed. 3 moles c and 3 moles CO2 means that 6 moles CO would be produced (remember that the ratio remains 1:1:3 for C, CO2, and CO). This means that 5 moles CO2 would be leftover.
In other words, all the 3 moles C would be consumed, 3 out of 8 moles CO2 would be consumed, and 6 moles CO would be produced while 5 moles CO2 would be leftover.
Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is started with 2.6 grams of NH3 (g) added into a reaction vessel with CO2 (g).
Write the balanced equation for this reaction, being sure to include physical states. Based on the balanced equation above, calculate the following:
a. the theoretical yield of urea in grams that can be made from the NH3
b. the actual amount of urea made if the percent yield for this reaction is 34%.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
) The C O bond dissociation energy in CO2 is 799 kJ/mol. The maximum wavelength of electromagnetic radiation required to rupture this bond is ________.
Answer:
λ = 150 nm
Explanation:
For C-O bond rupture:
The required energy to rupture C-O bond = bond energy of C-O bond
= 799 kJ/mol
[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]
[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]
[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]
Recall that the wavelength associated with energy and frequency is expressed as:
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]\lambda = \dfrac{hc}{E}[/tex]
[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]
[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]
λ = 150 nm
Select the number of valence electrons for hydrogen.
Answer:
Vanlency of hydrogen - 11
Electrons of hydrogen - 1
Answer:
The answer is: 1
Hope this helps :) <3
Explanation:
Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in % wt/v concentration.
Known; Mass of CaCl2 present in original solution, based on actual yield= 1.77g moles
CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles
Total Volume of solution =V, which is 80ml
Answer:
2.21% wt/v
Explanation:
The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL
The %wt/v of this solution is:
%wt /v = 1.77g / 80mL * 100
%wt/v = 2.21% wt/v
Standard hydrogen electrode acts as both anode and cathode.Explain.
Answer:
A Standard Hydrogen Electrode is an electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. It is important to have this common reference electrode, just as it is important for the International Bureau of Weights and Measures to keep a sealed piece of metal that is used to reference the S.I. Kilogram.
Answer:
The role of an electrode as cathode or anode depends on the nature and electrode potential of the other electrode with which it forms the complete electrochemical cell.
When a cell is to be made with zinc electrode and hydrogen electrode, the hydrogen electrode will behave as a cathode and the zinc electrode will behave as anode because zinc is present above hydrogen in the activity series. That is zinc is more electropositive than hydrogen.
If the cell is made with a copper electrode and hydrogen electrode, the hydrogen electrode will behave as anode and the copper electrode as a cathode. This is due to the fact that Cooper is present below hydrogen in the activity series. Copper is less electropositive than hydrogen.
Explanation:
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
https://brainly.com/question/24310467
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
Complete the sentences by identifying the correct missing words. Alph and beta particles originate from the Choose... . Protection from radiation is necessary because if radiation passes through the body it can damage Choose... . Exposure to radiation can be limited by increasing the Choose... from the radioactive source.
Answer:
Alpha and beta particles originate from the nucleus, protection from radiation is important because if the radiation passes through the body it can damage cells. Exposure to radiation is often limited by increasing the distance from the radioactive source.
Explanation:
Alpha and beta particles come from unstable atoms during their decay. This radiation is extremely harmful which may damage DNA, causing a high rate of mutation. If we increase the distance of the source of radioactive exposure we will prevent damage.Which statement best describes what happens during a chemical reaction?
A. Reactants change into products.
B. Reactants change into new reactants.
C. Products change into reactants.
D. Products change into new products.
Answer:
A. Reactants change into products
Question 14
2 pts
A chemist wants to make 100 mL of a 0.500 M solution of NaCl. They have a
stock solution of 1.2 M NaCl. How much of the original stock solution do they
need to make their new dilute solution?
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 1.2 M
Molarity of diluted solution (M₂) = 0.5 M
Volume of diluted solution (V₂) = 100 mL
Volume of stock solution needed (V₁) =?The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂1.2 × V₁ = 0.5 × 100
1.2 × V₁ = 50
Divide both side by 1.2
V₁ = 50 / 1.2
V₁ ≈ 42 mLThus, 42 mL of the stock solution is needed.
Learn more: https://brainly.com/question/24219233
Answer:
They need 41.7 mL of the original stock solution.
Explanation:
We can use the following equation for dilutions:
Cc x Vc = Cd x Vd
Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.
The concentrated solution is the original stock solution, and it has:
Cc = 1.2 M
The diluted solution must be:
Cd = 0.500 M
Vd = 100 mL
So, we have to calculate Vc. For this, we replace the data in the equation:
[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]
Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100 mL of a diluted solution with a concentration of 0.500 M.
A saturated solution of potassium iodide contains, in each 100 mL, 100 g of potassium iodide. The solubility of potassium iodide is 1 g in 0.7 mL of water. Calculate the specific gravity of the saturated solution
Answer:
Specific gravity of the saturated solution is 2
Explanation:
The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:
Specific gravity = Density
The density is the ratio between the mass of the solution and its volume.
In 100mL of water, the mass of KI that can be dissolved is:
100mL * (1g KI / 0.7mL) = 143g of KI
That means all the 100g of KI are dissolved (Mass solute)
As the volume of water is 100mL, the mass is 100g (Mass solvent)
The mass of the solution is 100g + 100g = 200g
In a volume of 100mL, the density of the solution is:
200g / 100mL = 2g/mL.
The specific gravity has no units, that means specific gravity of the saturated solution is 2
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
a) potassium tetracyanonickelate(II)
b) sodium diamminedicarbonatoruthenate(III)
c) diamminedichloroplatinum(II)
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
Match each land resource to its use.
clay - used to make steel
iron ore - used to make batteries
salt - used to make pottery and tiles
aggregate - used in construction
graphite - used as a flavoring in food
i will give 10 points and brainliest!!!
Answer:
Explanation:
We are to match each land resource to what they are being used for.
Clay →→→ used to make pottery and tiles
iron ore →→→ used to make steel
Salt →→→ used as a flavoring in food
aggregate →→→ used in construction
graphite →→→ used to make batteries
Clay is a kind of soil particle that forms as a result of weathering processes. Examples include; pottery clays, glacial clays, and deep-sea clays e.t.c. The presence of one or more clay minerals, as well as variable quantities of organic and detrital components, characterizes all of them. Clay is usually sticky and moist when wet, but hard when dry. They are used in the making of tiles and potteries.
Iron ore: The iron ore deposits are found in the Earth's crust's sedimentary rocks. They're made up of iron and oxygen that mix during the chemical process in marine and freshwater. iron ores are used to produce almost every iron and steel product that we use today.
Aggregate: are utilized in construction activities. It is a material used to mix cement, gypsum, bitumen, or lime to produce concrete in the construction industry.
Graphite: Graphite is a mineral that occurs in both igneous and metamorphic rocks. It is generally generated on the earth's surface when carbon is exposed to high temperatures and pressures. It is mainly used in the production of batteries and electrodes,
What is the oxidation number of the metal ion in the coordinate complex [Fe(CN)6]3–?
A. NCO-
B. -OH
C. -CN
D. -SCN
Answer:
The options are incorrect.................
Explanation:
The Oxidation no. is +3
What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
A sample of nitrogen gas occupies 117 mL at 100°C. At what
temperature would it occupy 234 mL if the pressure does not
change? (express answer in K and °C)
47
Page
8 I 8
- Q +
Answer:
The new temperature of the gas is 746 K.
Explanation:
Given that,
The volume of the gas, V₁ = 117 mL
Temperature, T₁ = 100°C = 373
Final volume of the gas, V₂ = 234 mL
We need to find the final temperature. The relation between temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{373\times 234}{117}\\\\T_2=746\ K[/tex]
So, the new temperature of the gas is 746 K.
which primitive organic molecule was essential to form lipid bilayer?
a)protenoid
b)phospholipid
c)autocatalytic RNA
d)aminoacids
Answer:
c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.
A eudiometer is used to collect hydrogen gas in a chemical reaction, as in your Exp 7. The volume of the gas in the tube (when pressure is held
constant) is 479.10 mL. The pressure of the atmosphere during the experiment is 758.3 mmgHg, and the temperature of the water and gas is
19.0*C. The water vapor pressure at this temperature is 16.5 torr.
Calculate the mass of hydrogen, in mg, collected.
Answer:
39.29 mg
Explanation:
Step 1: Calculate the partial pressure of hydrogen
The pressure of the atmosphere is equal to the sum of the partial pressures of the water and the hydrogen. (1 Torr = 1 mmHg)
P = pH₂O + pH₂
pH₂ = P - pH₂O = 758.3 mmHg - 16.5 mmHg = 741.8 mmHg
We will convert it using the conversion factor 1 atm = 760.0 mmHg.
741.8 mmHg × 1 atm/760.0 mmHg = 0.9761 atm
Step 2: Convert 19.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 19.0 + 273.15 = 292.2 K
Step 3: Calculate the mass of hydrogen
First, we will calculate the moles of hydrogen using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 0.9761 atm × 0.47910 L / (0.08206 atm.L/mol.K) × 292.2 K = 0.01950 mol
The molar mass of hydrogen is 2.015 g/mol. The mass of hydrogen is:
0.01950 mol × 2.015 g/mol = 0.03929 g = 39.29 mg
Help me in this question!!!
Answer:
d. End product is that product with a ketone and carboxylic acid.
Explanation:
[tex]{ \sf{NaBH_{4} : }}[/tex]
Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.
[tex]{ \sf{H _{2} O \: and \: H {}^{ + } }}[/tex]
Then acidified water is an oxidising mixture which reverses the reduction reaction.
Explanation:
Option D is your answer
Hope it helps
81.5 g of metal was heated from 11 degrees Celsius to 69 degrees Celsius. If 6739 joules of heat energy were used, what is the specific heat capacity of the metal?
Answer:
the metal become red hot
How does the number of molecules in one mole of carbon dioxide compare with the number of molecules in one mole of water?
ОА.
There are four times as many molecules in one mole of carbon dioxide as there are in one mole of water.
ОВ.
There are twice as many molecules in one mole of carbon dioxide as there are in one mole of water.
OC
There are three times as many molecules in one mole of carbon dioxide as there are in one mole of water.
OD
There are the same number of molecules in one mole of carbon dioxide as there are in one mole of water.
Answer:
d
Explanation:
A given solution has 42.5 g NaNO3 in 1500 mL of water. What is the molarity of this solution? Show your work, rounding the atomic mass of each element to the nearest tenth and your final answer to the nearest hundredth
Explanation:
here's the answer to your question
Because the double bond in an alkene is rigid, alkenes can exist as geometric isomers. To clarify geometric isomers, IUPAC uses cis- and trans- as part of a compound name. If the substituents around the double bond are on the same side of the double bond, this is called
cis, cis.
cis.
cis, trans.
trans.
Answer:
cis
Explanation:
Cis isomers are formed when the substituents on the carbons of the double bond are on the same side of the double bond, forming a U. Trans isomers have substituents on opposite sides of the double bond, forming a sideways Z.
A student was asked to determine the percent mass of sodium nitrate in a mixture of sodium nitrate (NaNO3) and calcium carbonate (CaCO3). The mass of the mixture used was 3.2 g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the solution by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighted 0.45 g. The student dried the insoluble residue of CaCO3 and found that it weighted 2.23 g. Calculate the percent mass of NaNO3 in the mixture and round your final answer to the correct number of significant figures.
Answer:
自分の仕事をする translate to english
Explanation:
FORMULAS OF IONIC COMPOUNDS
FIND: POSITIVE ION, NEGATIVE ION AND FORMULA IN:
NAME:
Sodium chloride
Magnesium chloride
Calcium oxide
Lithium phosphide
Aluminum sulfide
Calcium nitride
Iron(III)chloride
Iron(II)oxide
Copper(I)sulfide
Copper(II)nitride
Zinc oxide
Silver sulfide
Potassium carbonate
Sodium nitrate
Calcium bicarbonate
Aluminum hydroxide
Lithium phosphate
Potassium sulfate
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⁻.
