I need help with this question

Name the following ionic compounds, keeping in mind that a transition metal cation must include its charge. ____________________

Lilit [14]

Answer :

(a) The name for $TiO_2$ is titanium(IV) oxide.

(b) The name for $BaCl_2$ is barium chloride.

(c) The name for $CuCl_3$ is copper(III) chloride.

(d) The name for $KI$ is potassium iodide.

(e) The name for $SrCl_2$ is strontium chloride.

(f) The name for $CuBr_2$ is copper(II) bromide.

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive ion is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(a) $TiO_2$

$TiO_2$ is an ionic compound because titanium element is a metal and oxygen element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on tin is (+4) and the the charge on oxygen is (-2). Thus, the name for $TiO_2$ is titanium(IV) oxide.

(b) $BaCl_2$

$BaCl_2$ is an ionic compound because barium element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on barium is (+2) and the the charge on chlorine is (-1). Thus, the name for $BaCl_2$ is barium chloride.

(c) $CuCl_3$

$CuCl_3$ is an ionic compound because copper element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on copper is (+3) and the the charge on chlorine is (-1). Thus, the name for $CuCl_3$ is copper(III) chloride.

(d) $KI$

$KI$ is an ionic compound because potassium element is a metal and iodine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on potassium is (+1) and the the charge on iodine is (-1). Thus, the name for $KI$ is potassium iodide.

(e) $SrCl_2$

$SrCl_2$ is an ionic compound because strontium element is a metal and chlorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on strontium is (+2) and the the charge on chlorine is (-1). Thus, the name for $SrCl_2$ is strontium chloride.

(f) $CuBr_2$

$CuBr_2$ is an ionic compound because copper element is a metal and bromine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on copper is (+2) and the the charge on bromine is (-1). Thus, the name for $CuBr_2$ is copper(II) bromide.

3 0

3 years ago

A discovery only brings extensive change when

dsp73

Answer is: it comes at the right time.
Some of the chemistry discoveries that come in right time (with enough informations, knowledge and right technology to use that discovery) are: Periodic Table of the Elements, discovery of electron, proton, neutron, radioactivity, pasteurization, the Haber-Bosch process.

7 0

2 years ago

Read 2 more answers

A chemist needs 0.550 mol selenium for a reaction. What mass of selenium should a chemist use?

-BARSIC- [3]

Answer:

The chemist would require to use 43.43 grams.

Explanation:

In order to solve this problem we need to know how much do 0.550 moles of selenium weigh. To do that we use selenium's molar mass and multiply it by the given number of moles:

0.550 mol * 78.96 g/mol = 43.43 g

The chemist would require to use 43.43 grams.

4 0

2 years ago

When combing one's hair, the comb becomes positively charged. What happens to the hair being combed?

TEA [102]

Since the comb is positively charged, the hair will stick to the comb.

I think

6 0

3 years ago

Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume

kvv77 [185]

*** 2 ***
if we assume volume NaCl + volume H2O = volume H2O.. i.e.. NaCl does not effect volume 

therefore.. the units of.. 
.. M = moles NaCl / L solution ≈ moles NaCl / L H2O 
.. density = grams NaCl / L solution ≈ grams NaCl / L H2O 
again.. that is our assumption 

so we can readily see that 
.. M = (1 mol NaCl / ___g NaCl) x (__g NaCl / L H2O) + 0 
ie.. 
.. M = (1 mol NaCl / 58.5g NaCl) x density solution + 0 

so.. we would expect.. 
.. m = 0.01709 mol / g 
.. b = 0

3 0

3 years ago