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Amiraneli [1.4K]
3 years ago
8

25. Compare the compositions of binary ionic and binary molecular compounds. 26. Write the general chemical formula for an acid,

and discuss what each component of the formula represents.
Chemistry
2 answers:
QveST [7]3 years ago
7 0
Ionic binary compounds -- hold a metal cation and an anion made of a nonmetal anion. Example: calcium phosphate: Ca = 2+ (Group IIA) PO4 = 3- (phosphate) the lowest shared multiple of 2+ and 3- is 6. You must have 3 Ca2+ to make +6; you must have 2 PO43- to make -6. The sum of the positive and negative charges must be equivalent in a compound for the reason that they are electrically neutral. Ca3(PO4)2 -- calcium phosphate 

Binary molecular compounds -- has two or more nonmetals fused covalently. There are no ions current for the reason that the atoms share electrons. Prefixes are used to designate the number of atoms of each kind present. Example: SO2 -- sulfur dioxide  

The overall formula for an acid is HX, where X is an anion. Binary acids cover H and a nonmetal and are called with the prefix "hydro" and the suffix "ic".
Murljashka [212]3 years ago
5 0

Explanation:

25. Binary ionic compounds: They are known as the compounds which contains a cation formed from a metal and an anion formed from a non-metal or a polyatomic anion. For example : Barium sulfate

Ba : +2 ( metal of group II-A)

SO_4 : -2 ( sulfate)

These two ions combine to form a binary ionic compound having formula BaSO_4

Binary Molecular compounds: They are known as the compounds which contain two or more non-metals that are bonded covalently. In these compounds, no ions are present because the electrons are shared by the atoms. Examples are: PCl_3 : phosphorous trichloride

SO_2 : sulfur dioxide

26. The general chemical formula for an acid is HX where H is the hydrogen and X is the non-metal or polyatomic ions. Examples of acids are:

HF : Hydrofluoric acid

H_2SO_4 : Sulfuric acid

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Answer:

The molarity of the NaOH solution is 0.076 M

Explanation:

Step 1: Data given

Mass of oxalic acid = 192 mg = 0.192 grams

volume = 250 mL = 0.250 L

Molar mass oxalic acid = 90.03 g/mol

Step 2: The balanced equation

H2C2O4 + 2NaOH → Na2C2O4 + H2O

Step 3: Calculate moles of oxalic acid

Moles oxalic acid = 0.192 grams / 90.03 g/mol

Moles oxalic acid = 0.00213 moles

Step 4: Calculate molarity of oxalic acid

Molarity = Moles / volume

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Molarity = 0.00852 M

Step 5: Calculate Molarity of NaOH

2 Ca*Va = Cb*Vb

with Ca = Molarity of oxalic acid = 0.00852 M

with Va = volume of oxalic acid = 0.250 L

with Cb = Molarity of NaOH = TO BE DETERMINED

with Vb = volume of 0.0558 L

Cb = (2*0.00852 * 0.25) / 0.0558

Cb = 0.076 M

The molarity of the NaOH solution is 0.076 M

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