Answer:
Explanation:
As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)
ΔG = ΔG° + RTInQ
Q = 1
ΔG = ΔG°
ΔG = =nFE°
n=no of electrons transfered.
E° = 1.1v
ΔG° = -2 * 96500 * 1.10
= -212300J
ΔG° =-212.3kJ/mol
Therefore, the ΔG° = -212.3kJ/mol1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
a. with H_2 O
b. with H_2 SO_4
c. with 1 equiv of KOH
d. with Csl
e. with NaCN
f. with HCl
g. with (CH_3)_2 S
h. with 1 equiv of NH_3
j. with CI_2
k. with KF
Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
Aluminum metal reacts with sulfuric acid according to the equation:2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(s) + 3H2(g)If 46 g of aluminum reacts with excess sulfuric acid, and 108 g of Al2(SO4)3 are collected, what is the percent yield for the reaction?
Answer:
Percentage yield = 38.37%
Explanation:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃
2 moles of Al contains = (2 × 27) = 54g of Al
1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃
54g of Al produces 342.15g of Al₂(SO₄)₃,
46g of Al will produce x g of Al₂(SO₄)₃
Solve for x
X = (46 × 342.15) / 54
X = 15738.9 / 54
X = 281.46g
Theoretical yield of Al₂(SO₄)₃ = 281.46g
Percentage yield of a substance = (actual yield / theoretical yield) × 100
Actual yield = 108g
Theoretical yield = 281.46
% yield = (108 / 281.46) × 100
% yield = 0.3837 × 100
%yield = 38.37
The percentage yield of Al₂(SO₄)₃ is 38.37
Answer: Percentage yield = 38.37%
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂ from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃ .
2 moles of Al contains = (2 × 27) = 54g of Al
1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃
54g of Al produces 342.15g of Al₂(SO₄)₃,
46g of Al will produce x g of Al₂(SO₄)₃
Solve for x
X = (46 × 342.15) / 54
X = 15738.9 / 54
X = 281.46g
Theoretical yield of Al₂(SO₄)₃ = 281.46g
Percentage yield of a substance = (actual yield / theoretical yield) × 100
Actual yield = 108g
Theoretical yield = 281.46
% yield = (108 / 281.46) × 100
% yield = 0.3837 × 100
%yield = 38.37
The percentage yield of Al₂(SO₄)₃ is 38.37
The thyroid, parathyroid, and thymus are located in the
The adrenals, pancreas, testes, and ovaries are located in the
The pineal, hypothalamus, and pituitary are located in the
Answer:
The pineal, hypothalamus, and pituitary are located in the BRAIN
The thyroid, parathyroid, and thymus are located in the THROAT
The adrenals, pancreas, testes, and ovaries are located in the NEAR THE KIDNEY
Explanation:
The pineal, hypothalamus, and pituitary are located in the BRAIN
The thyroid, parathyroid, and thymus are located in the THROAT
The adrenals, pancreas, testes, and ovaries are located in the NEAR THE KIDNEY
Answer:
Throat
Abdomen
Brain
Explanation:
These are correct!
uppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.83 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 22.08 mL. What volume of base was required for the titration?
Answer:
20.25 mL.
Explanation:
The volume of base required for the titration can be derived by removing the base titrant volume from the volume at endpoint.
i.e Final volume - initial volume
= (22.08 - 1.83)mL
= 20.25 mL.
(Repeat and average volume results for accuracy.)
Can you please answer Question 9 please
As a quality control check, a sample of acetone is taken from a process to determine the concentration of suspended particulate matter. An 850 mL sample was placed in a beaker and evaporated. The remaining suspended solids were determined to have a mass of 0.001 g. The specific gravity of acetone is 0.79 g/cm.
(a) Determine the concentration of the sample as mg/L.
(b) Determine the concentration of the sample as ppm (problem from EPA Air Pollution Training Institute)
Answer:
The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,
Weight = volume × density = 850 × 0.79
= 671.5 grams
Weight of the suspended solids given is 0.001 gram
The concentration of the sample can be determined by using the formula,
Concentration = wt. of sample/volume
= [671.5 - 0.001) 10³ mg / 0.85 L
= 789998.82 mg/L or 789998.82 ppm
Now the concentration of suspended solids is.
Css = 0.001 × 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm
What is the Electronegativity difference of F and CI
The element fluorine is more electronegative than the element chlorine. Fluorine has an electronegativity value of 4.0 and Carbon has an electronegativity value of 2.5.
What is electronegativity ?The term electronegativity is defined as a chemical property that describes the capacity of an atom or a functional group to attract electrons toward itself.
Electronegativity is refer to the ability of an atom to attract shared electrons in a covalent bond. The grater the value of the electronegativity, the more strongly that element attracts the shared electrons.
In fluorine electronegativity value is more than carbon. Therefore, it strongly attracts the shared electrons.
Thus, The element fluorine is more electronegative than the element chlorine.
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The total enthalpy of the products in a reaction is 0 kJ, and the total enthalpy of the reactants is 100 kJ. What is the ΔH for the reaction? Choose the correct answer.
a.–393.5 kJ
b.–100 kJ
c.0 kJ
d.+100 kJ
Answer: b. -100 kJ
Explanation:
Heat of reaction is the energy released or absorbed during the course of the reaction. It is represented by the symbol [tex]\Delta H[/tex].
[tex]\Delta H=H_{products}-H_{reactants}[/tex]
[tex]\Delta H[/tex] = enthalpy of reaction = ?
[tex]H_{products}[/tex] = enthalpy of products = 0 kJ
[tex]H_{reactants}[/tex] = enthalpy of reactants = 100 kJ
[tex]\Delta H=0kJ-100kJ[/tex]
[tex]\Delta H=-100kJ[/tex]
Thus the ΔH for the reaction is -100kJ
what type of reaction is Cl2(g) + 2Kl(aq) arrow 2KCl(aq) + I2 (s)
Draw a Lewis structure for each covalent molecule. a. CH3F b. N2H4
Answer:
See attached picture.
Explanation:
Hello,
In this case, given the compounds we notice:
a. Since carbon has four valence electrons (group IVA), it has three hydrogen atoms around it as well as a fluorine atom which has seven valence electrons (group VIIA). Moreover, we notice both carbon and fluorine attaining the octet.
b. In this case, two nitrogen atoms which have five valence electrons each (groupVA) are bonded via a single bond and two hydrogen atoms are bonded to each nitrogen atom in order to attain the octet for nitrogen only.
Thus, structures are given on the attached picture.
Regards.
Lewis structure for each covalent molecule CH₃F and N₂H₄ are attached to the image below.
a. CH₃F:
Carbon (C) has 4 valence electrons, Hydrogen (H) has 1 valence electron, and Fluorine (F) has 7 valence electrons.
The total number of valence electrons in CH3F can be calculated as follows:
1 carbon atom (C) × 4 valence electrons = 4 valence electrons
3 hydrogen atoms (H) × 1 valence electron = 3 valence electrons
1 fluorine atom (F) × 7 valence electrons = 7 valence electrons
Total valence electrons = 4 + 3 + 7 = 14
b. N₂H₄:
Nitrogen (N) has 5 valence electrons, and Hydrogen (H) has 1 valence electron.
The total number of valence electrons in N2H4 can be calculated as follows:
2 nitrogen atoms (N) × 5 valence electrons = 10 valence electrons
4 hydrogen atoms (H) × 1 valence electron = 4 valence electrons
Total valence electrons = 10 + 4 = 14
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In an oven, 7000 kg of limestone are heated to over 900ºC to give lime and carbon dioxide. Assuming all calcium carbonate reacts. How much lime do you get? What volume of CO2 is emitted into the atmosphere? Atomic weights: Ca: 40; C: 12; O: 16.
Answer:
3920kg of lime(CaO) 1568m^3 of CO2
Explanation:
I attached the explanation!
also for calculating the volume I assumed Standard Temperature and Pressure(STP).
b) How many kilograms of carbon dioxide are formed when 24.42 g of iron is
produced?
Answer:
0.0289 kg of CO2 will be formed
Explanation:
Step 1: Data given
Mass of iron produced = 24.42 grams
Atomic mass iron = 55.845 g/mol
Molar mass CO2 = 44.01 g/mol
Step 2: The balanced equation
Fe2O3 + 3CO → 2Fe + 3CO2
Step 3: Calculate moles iron
Moles iron = mass iron / molar mass iron
Moles iron = 24.42 grams / 55.845 g/mol
Moles iron = 0.437 moles
Step 4: Calculate moles CO2
For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2
For 0.437 moles Fe we'll have 3/2 * 0.437 = 0.6555 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.6555 moles * 44.01 g/mol
Mass CO2 = 28.85 grams = 0.0289 kg
0.0289 kg of CO2 will be formed
HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)
If the value of Kc for this process is 5.2 x 10-5, there is no H3O+ or BrO- initially present, and the initial [HBrO] = 0.749 M, what will be the concentration of BrO- at equilibrium?
HINT: you should not use the quadratic formula for this problem.
7.67 x 10-3 M
9.01 x 10-3 M
5.17 x 10-3 M
4.29 x 10-3 M
6.24 x 10-3 M
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
[tex]Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}[/tex]
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change [tex]x[/tex], we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:
[tex]5.2x10^{-5}=\frac{x*x}{0.749-x}[/tex]
Thus, we obtain a quadratic equation whose solution is:
[tex]x_1=-0.00627M\\x_2=0.00624M[/tex]
Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.
Answer:
6.24×10¯³ M
Explanation:
Step 1:
The equation for the reaction is given below:
HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)
Step 2:
The following data were obtained from the question:
Equilibrium constant, Kc = 5.2x10^-5
Concentration of HBrO, [HBrO] = 0.749 M
Concentration of BrO-, [BrO-] =?
Step 3:
Determination of the concentration of
BrO-.
We can obtain the concentration of BrO- as follow:
Initial concentration:
[HBrO] = 0.749 M
[H3O+] = 0
[BrO-] = 0
During reaction:
[HBrO] = - y
[H3O+] = +y
[BrO-] = +y
After reaction:
[HBrO] = 0.749 - y
[H3O+] = y
[BrO-] = y
Applying the equation constant equation:
Kc = [H3O+] [BrO-] / [HBrO]
5.2x10^-5 = y × y / 0.749
5.2x10^-5 = y² / 0.749
Cross multiply
y² = 5.2x10^-5 x 0.749
Take the square root of both side
y = √(5.2x10^-5 x 0.749)
y = 6.24×10¯³ M
[BrO-] = y = 6.24×10¯³ M
Therefore, the the concentration of BrO- at equilibrium is 6.24×10¯³ M
Explain why airborne and water-borne chemicals are both local and global issues.
How many moles of iron(III) sulfide,
Fe2S3, would be produced from the
complete reaction of 449 g iron(III)
bromide, FeBr3?
Answer:
0.76 mole of Fe2S3.
Explanation:
Step 1:
Determination of the number of mole in 449g iron(III)bromide, FeBr3. This is illustrated below:
Mass of FeBr3 = 449g
Molar mass of FeBr3 = 56 + (80x3) = 296g/mol
Mole of FeBr3 =..?
Mole = Mass /Molar Mass
Mole of FeBr3 = 449/296
Mole of FeBr3 = 1.52 moles
Step 2:
The balanced equation for the reaction. This is given below:
2FeBr3 + 3Na2S —> 6NaBr + Fe2S3
Step 3:
Determination of the number of mole of Fe2S3 produced from the reaction of 449g ( i.e 1.52 moles) of FeBr3. This is illustrated below:
From the balanced equation above,
2 moles of FeBr3 reacted to produce 1 mole of Fe2S3.
Therefore, 1.52 moles of FeBr3 will react to produce = (1.52 x 1)/2 = 0.76 mole of Fe2S3.
Therefore, 0.76 mole of Fe2S3 is produced from the reaction.
Answer: 0.76
Explanation:
A chemist performed a calibration method analysis to determine the concentration of vitamin c in a fruit sample. The standard vitamin c samples were prepared by mixing a complexing agent with the standard solution. The absorbance of the resulting solution was obtained. Once the absorbance of each standard solution was measured, he prepared the following calibration curve.y = 1.6421x -0.1113The unknown sample was prepared by taking 2.0 g of crushed fruit and extracting the concentrated juice. The extracted juice was added to 100.0 mL volumetric flask and the final volume was adjusted 100.0 mL with DI water. 10.0 mL of diluted fruit extra was pipetted into 50.0 mL volumetric flask and after adding 5.0 mL of complexing agent, the DI water was added until the final volume of the solution reaches 50.0 mL. This solution gave an absorbance of 0.461. I. Find the concentration of vitamin C in the original fruit sample. II. What is the total mass of vitamin C in the sample? III. Calculate the wt% of vitamin C in the fruit sample.
Answer:
I. 1.743mg/mL
II. 0.174g
III. 8.7 wt%
Explanation:
As absorbance is directly proportional to concentration. The calibration curve was:
Y = 1.6421X - 0.1113
Where Y is absorbance of sample and X its concentration in mg/mL
The diluted sample obtains an absorbance of 0.461, that means its concentration is:
0.461 = 1.6421X - 0.1113
0.3485mg/mL = X
I. The sample was diluted to 50.0mL from an aliquot of 10.0mL, thus, concentration in the original fruit sample was:
0.3485mg/mL × (50.0mL / 10.0mL) = 1.743mg/mL
II. The total mass of vitamin C in the sample (100.0mL) is:
100.0mL × (1.743mg/mL) = 174mg = 0.174g
III. As the fruit sample was of 2.0g, the wt% is:
0.174g / 2.0g = 8.7 wt%
Identify which subatomic particles match each of these descriptions. One of the descriptions describes two particles. Make sure to include both particles it describes.
Description Particle(s)
have a relative charge of +1
have a relative charge of −1
have no charge
located in the nucleus of an atom
have a much lower mass than the other two types of particles
Explanation:
First off, it's important to know that the three subatomic particles we have are the proton, neutron and electron.
have a relative charge of +1 : This is the proton. It is the postive subatomic particle present in an atom.
have a relative charge of -1 : This is the electron. It is the negative subatomic particle present in an atom.
have no charge : This is the neutron. It is the neutral subatomic particle present in an atom.
located in the nucleus of an atom : Twoo subatomic particles are located in the nucleus, the neutron and proton.
have a much lower mass than the other two types of particles: This is the electron. The proton and neutron are almost similar in mass. The electron is abut 1 / 1840 times the mass of the proton.
how many significant figures are in 11.005 g
Answer:
Number of Significant Figures: 5
The Significant Figures are 1 1 0 0 5
Explanation:
How many mL of a 0.4 M nitric acid solution are required to neutralize 200 mL of a 0.16 M potassium hydroxide
Answer:
80 ml.
Explanation:
The balanced equation is:
KOH + HNO3 ----> KNO3 + H2O
So 1 mole of KOH reacts with 1 mole of HNO3
200ml of 0.16M KOH contains 0.16 * 200/1000
= 0.16 * 0.2
= 0.032 M KOH.
So we need the volume (V) of 0.4 M HNO3 which contains 0.032 moles.
0.4 * V = 0.032
V = 0.032 / 0.4
V = 0.08 liters
= 80 mls.
Calculate the empirical formula of a compound with the following
percentage composition: 31,8% K; 29,0% C1; 39,2% O
The empirical formula of ethane is CH. Its molar mass is 30 g.mol-
Answer:
The empirical formula for this compound is thus CH2.
Explanation:
A balloon is cooled by removing 0.784 kJ of heat. It shrinks on cooling, and the atmosphere does 655 J of work on the balloon. Calculate the change in internal energy, ∆E, of the system.
Answer:
The correct answer is -0.129 kJ
Explanation:
In the given case, the cooling of the balloon is done by withdrawing 0.784 J of heat, and the work done by the atmosphere on the balloon is 655 J. First, there is a need to transform kJ into J, 1 kJ = 1000 J. So, 0.784 kJ would be 784 J.
The ΔE or the change in the internal energy can be calculated by using the formula, ΔE = q + w ----- (1).
In the given case, q refers to the heat moved out of the system, that is, the value of q would be less than 0 or will be a negative quantity. Therefore, the heat moved out of the system will be -784 J. On the other hand, as the work or w is done on the system, therefore, the value of w would be more than 0 or will be a positive quantity. Thus, the value of w will be +655 J.
Now putting the values in the equation (1) we get,
ΔE = -784 J + 655 J
ΔE = -129 J or -0.129 kJ
As the change in internal energy comes out to be a negative value, therefore, the process is considered exothermic.
A certain compound is made up of one carbon (C) atom, two fluorine (F) atoms, and one oxygen (O) atom. What is the chemical formula of this compound?
Answer:
CF₂O
Explanation:
To determine how the molecule will be made think about the valence number of each element in molecule:
F= -1 , O= -2 , C= +4
because the C is positive atom it will be the central atom with a double bond to the O and a single bond to each F.
this way all atoms completed the octet and stability
three requirements to start a fire
1.09 grams of H2 to moles
Explain why the benzene molecule usually reacts with electrophiles
Answer:
Explanation:Because of the delocalised electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules.
Answer:
Benzene has a pi electron system that is electron rich.
Electrophiles are molecules that are attracted to electron rich molecules as they are electron deficient in order to form a more stable complex.
Hence, as benzene is electron rich, electrophiles will be attracted to it.
What is the volume in a 2. 00M solution containing 8.00 moles of solute?
Answer:
4000cm³
Explanation:
2.0moles-1000cm³
8.0moles-?
(8.0×1000)÷2
=4000cm³
moles of Fe in 3.13×1022 atoms of Fe
Answer: 0.05 moles
Explanation:
To convert from atoms to moles, you will need Avogadro's number.
Avogadro's number: 6.022×10²³atoms/moles
(3.13×10²²atoms)×(moles/6.022×10²³ atoms)=0.05 moles
The atoms would cancel out and we will be left with our moles.
How many grams of silver sulfide are formed when 0.280 g of hydrogen sulfide reacts with excess silver and oxygen?
Answer:
[tex]2.03~g~Ag_2S[/tex]
Explanation:
For this question we have to start with the reaction:
[tex]Ag~+~H_2S~+~O_2~->~Ag_2S~+~H_2O[/tex]
Now, we can balance the reaction, so:
[tex]4Ag~+~2H_2S~+~O_2~->~2Ag_2S~+~2H_2O[/tex]
With this in mind, we have to start with the amount of [tex]H_2S[/tex]. The first step is to convert from grams to moles. For this, we need to find the molar mass of [tex]H_2S[/tex]. If we check the periodic table we will find the atomic masses for Ag and H; H: 1 g/mol and A: 32 g/mol, so:
(1*2)+ (32*1) = 34 g/mol.
Now we can calculate the moles of [tex]H_2S[/tex]:
[tex]0.280~g~H_2S\frac{34~g~H_2S}{1~mol~H_2S}=0.0082~mol~H_2S[/tex]
With the moles of [tex]H_2S[/tex] we can calculate the moles of [tex]Ag_2S[/tex] if we check the molar ratio in the balanced equation, [tex]2~mol~H_2S=2~mol_Ag_2S[/tex], so:
[tex]0.0082~mol~H_2S\frac{2~mol_Ag_2S}{2~mol~H_2S}=0.0082~mol~H_2S[/tex]
With the molar mass of [tex]Ag_2S[/tex] we can convert from moles to grams (Ag: 107.86 g/mol, S: 32 g/mol), so:
(107.86*1)+(32*2)=247.80 g/mol
[tex]0.0082~mol~H_2S\frac{247.8~g~Ag_2S}{1~mol~H_2S}=2.03~g~Ag_2S[/tex]
I hope it helps!
A certain atom has 22 protons and 19 electrons. This atom loses an electron. The net charge on the atom is now ___.If this same
atom with 22 protons and 19 electrons were to gain 3 electrons, the net charge on the atom would be ____
Answer:
I think first one has net charge 4+
second one has earlier 3+ charge after gaining net charge becomes zero
I think this is correct
Explanation:
if wrong means pls correct me bro
An atom has 22 protons and 19 electrons. The net charge on the atom is now elementary, 1e+. If the same atom with 22 electrons and 19 electrons were to gain 3 electrons, the net charge on the atom would be, negative, -3.
What are electrons?Electrons are the positively charged particles that revolve around the negatively charged nucleus. Electrons revolve around the nucleus, and protons and neutrons are also present.
When the electrons gain electron they acquire negative charge and when they donate electrons it acquires positive charge.
It gains 3 electrons that why -3 charge.
Thus, the correct elementary charge, negative, -3.
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(a) Give systematic names or the chemical formulas for the following. Remember, names must
be spelled correctly!
1.) [Ni(CO)2C2O4]
2.) K3[Fe(NCO)3(OCN)3]
3.) diammineaquahydroxoplatinum(II) ion
4.) Al[Co(CO3)2(CN)2]
5.) ethylenediaminedinitritozinc(II)
Answer:
[Ni(CO)2C2O4]- Dicarbonyloxalatonickel(0)
K3[Fe(NCO)3(OCN)3-potassiumtricyanatotriisocyanatoferrate II
diammineaquahydroxoplatinum(II) ion- [Pt(NH3)2(H2O)(OH)]^+
Al[Co(CO3)2(CN)2- Aluminiumdicarbonatodicyanocobalt III
ethylenediaminedinitritozinc(II)- Zn(en)(NO2)2
Explanation:
The international Union of Pure and applied chemistry published a set rules for the nomenclature of inorganic compounds in 1985. This set of rules have been consistently revised ever since.
The rules stipulate the order for naming positive ions, negative ions and neutral ligands and the allowed order of preference for naming. It also stipulates the proper endings for ligands which are positive ions, negative ions or neutral ligands as the case may be.
The compounds named above were named in accordance with the revised IUPAC nomenclature for inorganic compounds.
