Please answer this it's due today!!!!!!

How many sulfate ions are there in 321 grams of iron lll sulfate

nikitadnepr [17]

Here, we use the mole as we would use any other collective number: a dozen eggs; a Bakers' dozen; a Botany Bay dozen.
Of course, the mole specifies a much larger quantity, and if I have a mole of stuff then I have
6.022
×
10
23
individual items of that stuff. We can also specify an equivalent mass, because we also know the mass of a mole of iron, and a mole of oxygen etc........The mole is thus the link between the macro world of grams and kilograms and litres, that which we can measure out in the lab, to the micro world of atoms, and molecules, that which we can perceive only indirectly.
Here we have the formula unit
F
e
2
(
S
O
4
)
3
. If there is a mole of formula units, there are necessarily 2 moles of iron atoms, 3 sulfate ions,.......etc.

7 0

2 years ago

What is the Net Ionic equation for this chemical reaction: FeBr2+Na2S=FeS+2NaBr

Artyom0805 [142]

Answer: Fe(aq)+S(aq)=FeS(s)

Explanation: The Sodium and Bromine are spectator ions because they don't react with anything, you can see this by writing the ionic equation like so:

1.) Molecular formula (given): FeBr2 (aq)+Na2S (aq)= FeS(s)+2NaBr(aq)

Each dissolved FeBr2 breaks up into one Fe with a charge of 2+ and two Br with a negative charge. This gives you:

Fe(aq)+ 2Br(aq)+Na2S(aq)=FeS(s)+2NaBr

2.) Now repeat what was shown with the other compounds in the given molecular formula, and pay attention to the states that each ion is in (solid, liquid, aqueous, gas) because this will give you the ionic equation, which from there you can get rid of any ions that don't change amount or state.

3.) Ionic formula: Fe(aq)+ 2Br(aq)+2 Na(aq)+S (aq)=FeS(s)+2 Na(aq)+2Br(aq)

4.)When you've derived a total ionic equation (above), you'll find that some ions appear on both sides of the equation in equal numbers. For example, in this case two Na cations and two Br anions appear on both sides of the total ionic equation. What does this mean? It means these ions don't participate in the chemical reaction. They're present before and after the reaction. Nothing happens to them. So those are removed and you're left with the net ionic: Fe(aq)+S(aq)=FeS(s)

Hope this helps :)

7 0

3 years ago

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

You and your friends decide to travel, on foot, to your favorite beach at the Jersey Shore. Your speed for the first part of the

Juli2301 [7.4K]

Answer:

Sorry dont know

Explanation:

8 0

3 years ago

Read 2 more answers

How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?

ozzi

The number of Ml of a 0.40 %w/v solution of ,nalorphine that must be injected to obtain a dose of 1.5 mg is calculated as below

since M/v% is mass of solute in grams per 100 ml

convert Mg to g
1 g = 1000 mg what about 1.5 mg =? grams
= 1.5 /1000 = 0.0015 grams

volume is therefore = 100 ( mass/ M/v%)

= 100 x( 0.0015/ 0.4) = 0.375 ML

6 0

3 years ago