Answers
Answer:
See detailed explanation.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:
[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]
Then, we divide all of them by 1.55 to realize the empirical formula is:
[tex]CHOCl[/tex]
Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.
The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.
Regards!
Related Questions
A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 
Answers
Explanation:
here's the answer to your question
Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Answers
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.
which of the following is indicated by the ph value of a solution?
a- it's hydrogen ion concentration
b- its ammonium ion concentration
c- ability to undergo chemical reaction
d- its ratio of solute amount to solvent volume
Answers
Answer:
c- ability to undergo chemical reaction
The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.
a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl
Answers
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.
If the starting material has no stereogenic centers, when carbonyl compounds are reduced with a reagent such as LiAlH4 or NaBH4 and a new stereogenic center is formed, what will the composition of the product mixture be?
A) Forms a racemic mixture of the two possible enantiomers.
B) Forms more of one enantiomer than another because of steric reasons around the carbonyl.
C) Forms more of one enantiomer than another depending on the temperature of the reaction.
D) Forms different products depending on the solvent used.
Answers
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.
What is a chemical change?
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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name a factor tht affects the value of electron affinity
Answers
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationwhat is the difference between 25ml and 25.00ml
Answers
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
Which phenomenon explained below is an example of deposition?
Select the correct answer below:
A) Hail is formed from water droplets lifted by air currents to an altitude where they turn into pellets of ice.
B) Frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
C) In the winter, the top few inches of a pond turn to ice.
D) The visible cloud arising from a boiling tea kettle is not actually steam, but droplets of liquid water that form as the
steam cools in the air.
Answers
Answer:
b
Explanation:
deposition is when water turns from gas to solid. b is the only one that fits
Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
What is deposition?Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.
Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
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For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb
Answers
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
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
Answers
calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume
Answers
Answer :
volume of a gas = weight * 22.4 l / gram molecular weight
volume of o2 = ?
weight given = 20.5 g
gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )
volume of oxygen = 20.5 * 22.4 / 32
volume of oxygen = 14.35 liters
Explanation:
hope this helps you
if wrong just correct me
Which redox reaction would most likely occur if silver and copper metal were added to a solution that contained silver and copper ions?
A. Cu + Agt Cu2+ + 2Ag
B. Cu2+ + 2Ag* → Cu + 2Ag
C. Cu2+ + 2Ag → Cu + 2Ag+
D. Cu + 2Ag Cu²+ + 2Ag+
give the wrong answer and I'm reporting
Answers
Answer:
B
Explanation:
b/c copper is readuction agent
The most likely redox reaction that would occur if silver and copper metal were added to a solution that contained silver and copper ions is [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]. The correct answer is option C.
Redox reaction is a reaction in which reduction and oxidation takes place simultaneously.
In this reaction:
[tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]
Copper metal has a higher reduction potential than silver metal, which means that it will be oxidized to [tex]\rm Cu^{2+}[/tex] ions before silver metal is oxidized to [tex]\rm Ag^+[/tex] ions.
The [tex]\rm Cu^{2+}[/tex] ions in the solution will then react with the silver metal to form [tex]\rm Ag^+[/tex] ions and Copper metal. This reaction is an example of a displacement reaction, where a more reactive metal removes a less reactive metal from its compound.
Therefore, option C. [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex] is the correct answer.
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How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
Answers
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. What is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3
Answers
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:
c. C3H9N2O2
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answers
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
cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?
Answers
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Percentage remaining after 60 days =?Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Number of half-lives (n) =?
n = t / t½
n = 60 / 9.7Finally, we shall determine the percentage remaining. This can be obtained as follow:
Let the original amount be N₀
Let the amount remaining be N
Number of half-lives (n) = 60 / 9.7
N = N₀ / 2ⁿ
Divide both side by N₀
N/N₀ = 1/2ⁿ
N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾
N/N₀ = 0.0137
Multiply by 100 to express in percentage
N/N₀ = 0.0137 × 100
N/N₀ = 1.37%Therefore, the percentage remaining after 60 days is 1.37%
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A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide
Answers
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:
MoO2What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answers
Answer:
(B). it's metallic bonding
what are the properety of covalent bond
Answers
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.
how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?
Answers
Answer:
how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?
Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?
Answers
Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5
B)5,5
C)6,5
D)7,5
E)9,5
Answers
The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:
maleic acid ----> fumaric acid
Answers
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.
Starting from (R)-3-methylhex-1-yne as the substrate at the center of your page, draw a reaction map showing the regiochemical and stereochemical outcome or outcomes for each of the following series of reagents. Name each of your products, including stereochemical designations for any chirality centers that are generated.
a. HgSO4, H2SO4, H2O
b. 1. 9-BBN; 2. H2O2, NaOH
c. Br2, CCl4
d. HBr
Answers
Solution :
A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.
Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.
In the context, the products including the stereochemical designations for any chirality centers starting from the (R)-3-methylhex-1-yne as the substrate are attached below.
11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Answers
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answers
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.
Answers
The structures are shown in the image attached.
A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.
Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.
I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2) and tetrahydrofuran (image 3).
All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.
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How do I do this? What are the answers to the 5 questions shown?
Answers
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
Which subshells are found in each of the following shells
electron subshell - M shell
Answers
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
here is the question
Answers
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.
