All categories

2 years ago

Use the periodic table to classify each of the elements below.

Chemistry

2 answers:

Alika [10]2 years ago

5 0

Answer:

Explanation:

Cadmium:(Cd)

Cadmium is transition metal present in group twelve. It is soft metal and properties are similar to the other group members like zinc and mercury. Its atomic number is forty eight and have two valance electrons.

Electronic configuration:

Cd₄₈ = [Kr] 4d¹⁰ 5s²

Vanadium: (V)

It is present in group five. It is malleable and ductile transition metal. Its atomic number is twenty three. Vanadium have five valance electrons.

Electronic configuration:

V₂₃ =[Ar] 3d³ 4s²

Xenon: Xe

Xenon is present in group eighteen. It is noble gas. Its outer most valance shell is complete that's why it is inert. its atomic number is fifty four. Xenon have eight valance electrons.

Electronic configuration:

Xe₅₄ = [Kr] 4d¹⁰ 5s² 5p⁶

Iodine: (I)

Iodine is present in group seventeen. Its outer most valance shell have seven electrons. Iodine is the member of halogen family. It gain one electron to complete the octet. its atomic number is fifty three.

Electronic configuration:

I₅₃ = [Kr] 4d¹⁰ 5s² 5p⁵

Potassium: (K)

Potassium is present in group one. it is alkali metal. Its atomic number is nineteen. Its valance shell has one electron. Potassium loses its one valance electron and gets stable electronic configuration.

Electronic configuration:

K₁₉ = [Ar] 4s¹

Strontium: Sr

Strontium is present in group two. it is alkaline earth metal. its atomic number is thirty eight and have two valance electrons.

Electronic configuration:

Sr₃₈ = [Kr] 5s²

dalvyx [7]2 years ago

5 0

Answer:

cadmium (Cd): transition metal

vanadium (V): transition metal

xenon (Xe): noble gas

iodine (I): halogen

potassium (K): alkali metal

strontium (Sr): alkaline earth metal

Explanation:

You might be interested in

On the basis of molecular structure and bond polarity, which of the following compounds is most likely to have the greatest solu

Drupady [299]

Answer:

$\boxed{\text{c) NH$_{3}$; hydrogen bonding}}$

Explanation:

For each of these molecules, you must determine their VSEPR structure and then identify the strongest intermolecular forces.

Remember that water is a highly polar molecule.

a) CH₄

Electron geometry: tetrahedral

Molecular geometry: tetrahedral

Bond polarity: C-H bond nonpolar

Molecular polarity: nonpolar

Strongest IMF: London dispersion forces

Solubility in water: low

A nonpolar molecule is insoluble in a polar solvent.

b) CCl₄

Electron geometry: tetrahedral

Molecular geometry: tetrahedral

Bond polarity: C-Cl bond nonpolar

Molecular polarity: nonpolar (symmetrical molecule. All bond dipoles cancel)

Strongest IMF: London dispersion forces

Solubility in water: low

A nonpolar molecule is insoluble in a polar solvent.

d) PH₃

Electron geometry: tetrahedral

Molecular geometry: trigonal pyramidal

Bond polarity: P-H bonds are polar

Molecular polarity: polar (all P-H bond dipoles point towards P)

Strongest IMF: dipole-dipole

Solubility in water: soluble

A polar molecule is soluble in a polar solvent.

c) NH₃

Electron geometry: tetrahedral

Molecular geometry: trigonal pyramidal

Bond polarity: N-H bonds are highly polar

Molecular polarity: highly polar (all N-H bond dipoles point towards N)

Strongest IMF: hydrogen bonding

Solubility in water: highly soluble

NH₃ is so polar that it can form hydrogen bonds with water.

$\boxed{\textbf{The compound with the greatest solubility in water is NH$_{3}$}}$

8 0

2 years ago

For each reaction, identify the precipitate, or lack thereof. 2nacl(aq)+ba(oh)2(aq)⟶bacl2+2naoh naoh bacl2 no precipitate agclo3

Paha777 [63]

1) Answer is: <span>no precipitate.
</span>Chemical reaction: 2NaCl + Ba(OH)₂ → BaCl₂ + 2NaOH.
Barium chloride is salt that dissolves in water, sodium hydroxide is strong base that dissolves in water. This is example of double replacement reactions(double displacement or metathesis reactions)<span>, two ionic compounds are exchanged, making two new compounds.</span>

2) Answer is: AgI.
Chemical reaction: 2AgClO₃ + MgI₂ → 2AgI + Mg(ClO₃)₂.
Silver chloride form yellow precipipate, magnesium chlorate is salt that dissolves in water. This is also example of double replacement, Mafnesium has oxidation number +2 and siilver has oxidation number +1, iodine has -1 and chlorate also -1.

7 0

2 years ago

Read 2 more answers

Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react w

iogann1982 [59]

Answer:

$C_{3}H_{12}N_{2}Cl_{2}$

Explanation:

Lets say the formula of the compound is:

$C_{a}H_{b}N_{c}Cl_{d}$

Experiment 1:

With 1.00 g of the compound you get 1.95 g of AgCl

The molar mass of AgCl is 143.32 g/mol.

The moles of AgCl in 1.95g are:

$moles = \frac{1.95}{143.32} = 0.0136$

Also the moles of Cl are 0.0136 because 1 mol of AgCl has 1 mol of Cl

Experiment 2:

With 1.00 g of the compound you get 0.900 of CO2 and 0.735 g of H2O

The molar mass of CO2 is 44 g/mol

The molar mass of H2O is 18 g/mol

$moles\ CO_{2} = \frac{0.900}{44} = 0.0205 \\moles\ H_{2}O = \frac{0.735}{18} = 0.0408$

moles of C are 0.0205

moles of H are 0.0816 (2 times the moles of H2O)

With the given information so far you get that in 1.00 g of the compound there are 0.0136 moles of Cl, 0.0205 moles of C and 0.0816 moles of H.

In mass you have:

Mass Cl = 0.0136 * 35.5 = 0.4828 g

Mass C = 0.0205 * 12 = 0.246 g

Mass H = 0.0816 * 1 = 0.0816 g

Total mass = 0.4828 + 0.246 + 0.0816 + mass N

1.00 = 0.8104 + Mass N

Mass N = 0.1896

$moles\ N =\frac{0.1896}{14} =0.0135$

Now you divide all moles for the least

Cl: 0.0136 / 0.0135 = 1.0007

C: 0.0205 / 0.0135 = 1.52

H: 0.0816 / 0.0135 = 6.04

N: 0.0135 / 0.0135 = 1

multiplying by 2 to get an integer for C

Cl = 2

C = 3

H = 12

N = 2

$C_{3}H_{12}N_{2}Cl_{2}$

7 0

2 years ago

A sample of H2SO4 contains 2.02 g of hydrogen, 32.07 g of sulfur, and 64.00 g of oxygen. How many grams of sulfur and grams of o

Gemiola [76]

From the chemical formula of sulfuric acid, we can see the molar ratio:

H : S : O
2 : 1 : 4

Now, we convert the mass of hydrogen given into the moles of hydrogen. This is done using

Moles = mass / Mr
Moles = 7.27 / 1
Moles = 7.27

Therefore, the moles will be:

S = 7.27 / 2 = 3.64 moles
O = 7.27 * 2 = 14.54 moles

Now, the respective masses are:
S = 32 * 3.64 = 116.48 grams
O = 16 * 14.54 = 232.64 grams

7 0

1 year ago

A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound

Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5

C:H:O = 2 : 6 : 1

The empirical formula of compound will be C₂H₆O

5 0

2 years ago