How much of nacl is in 1.67 l of 0.400 m nacl? answer in units of mol.

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

What is the mass of 0.513 mol Al2O3? Give your answer to the correct number of significant figures. (Molar mass of Al2O3 = 102.0

prisoha [69]

.513mol x (102g/1mol)

Essentially, this is .513 x 102
Which equals: 52.326
But because you can only have 3 significant figures, your answer is:
52.3 grams

I hope this Helps!

7 0

3 years ago

Read 2 more answers

Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h

zubka84 [21]

1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3

5 0

3 years ago

What is the pH for a 0.10 M HCI solution at 25 degreesCelsius?

11Alexandr11 [23.1K]

Answer:

1.0

Explanation:

Hydrochloric acid is a strong acid, that is, an acid that dissociates completely, according to the following reaction.

HCl(aq) → H⁺(aq) + Cl⁻(aq)

Then, the concentration of H⁺ will be equal to the initial concentration of the acid, i.e., 0.10 M.

We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 0.10 = 1.0

3 0

3 years ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

3 years ago