The lithium ion[Li⁺]=1.642 M
<h3>Further explanation</h3>
Ionic compounds will dissociate into ions when dissolved in water. The strength of the dissociation depends on the strength of the electrolyte properties. The stronger it is, the more ions will dissociate
Reaction
Li₂S⇒2Li⁺+S²⁻
mol ratio Li₂S : Li⁺: S²⁻ = 1 : 2 : 1
so the lithium-ion (Li⁺) concentration :
Answer:
6.94 x 10²⁴molecules
Explanation:
Given parameters:
Mass of water = 207.45g
Unknown:
Number of molecules = ?
Solution:
To solve this problem, we must understand:
1 mole of a substance contains 6.02 x 10²³ molecules
Now, first find the number of moles of water;
Number of moles = 
Molar mass of water = 2(1) + 16 = 18g/mol
Number of moles = 
= 11.53moles
So;
1 mole of a substance contains 6.02 x 10²³ molecules
11.53moles of carbon dioxide will contain 11.53 x 6.02 x 10²³
= 6.94 x 10²⁴molecules
164. You have excess hydrogen- you need double the number of moles hydrogen as oxygen because water has two hydrogens and only one oxygen while both hydrogen gas and oxygen gas are diatomic.
<span>Because oxygen is diatomic, you spit it in half and basically double the number of moles, tack on your excess hydrogen, and you have 164 moles of water. </span>
<span>Edit: Lets look at it in one more way, taking into account what a mole is: </span>
<span>A mole is Avogadro's Constant number of particles- or about 6.02x10^23. So, we can multiply the number of moles you have to figure out how many particles of each we have. You can skip this step, because we will convert back to moles, but its illustrative of the concept. </span>
<span>6.02x10^23 x 174 moles H2 = 1.05 x 10^26 particles H2 </span>
<span>6.02x10^23 x 82 moles O2 = 4.94 x 10^25 particles O2 </span>
<span>I'm going to take one more step here- take them out of diatomic form. This will double the number of particles of each. </span>
<span>2.10 x 10^26 particles H+ </span>
<span>9.88 x 10^25 particles O(2-) </span>
<span>Now, we know we need a 2:1 oxygen atom to hydrogen atom ration, so we devide and see oxygen is our limiting reagent. So, we have 9.88 x 10^25 single oxygen atoms, all of which will become water the hydr</span>
Percent error in team 1 value:
(0.392 - 0.385)/0.385 x 100
= 1.81%
Percent error in team 2 value:
(0.395 - 0.385)/0.385 x 100
= 2.60%
B. Team 2 has an unresolved issue in on their hands, resulting in greater than acceptable error.
Explanation:
When iodine heptafluoride reacts rapidly with water to give a mixture of periodic acid and hydrofluoric acid. Reaction equation for the same is as follows.
As it is given that there are 
moles of iodine heptafluoride are present. Molar mass of 
is 259.9 g/mol. Molar mass of 
is 227.94 g/mol and molar mass of HF is 20.01 g/mol.
Now, according to the reaction equation M gives M 
.
Also, the volume is given as 795 ml or 0.795 L (as 1 mL = 0.001 L).
Hence, calculate the concentration of into the solution as follows.
Concentration = 
= 
= 
mol/L
or, = 0.047 M
Now, as 1 mol of produces 7 mol HF. So, 
M HF.
Therefore, concentration of HF will be calculated as follows.
Concentration of HF = 
= 0.33 M
Thus, we can conclude that concentration of periodic acid is 0.047 M and concentration of hydrofluoric acid is 0.33 M.
