First, we need the balanced equation: H₂ + Cl₂ ---> 2HCl
since not much information is given, I am assuming we are at STP and that 22.4 Liters= 1 mol
1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.
8 cm³ (1 ml/ 1 cm³)(1 L/ 1000 mL) (1 mol/ 22.4 Liters)= 3.6x10⁻⁴ moles of H₂
2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl
3.6x10⁻⁴ mol H₂ (2 mol HCl/ 1 mol H₂)= 7.1x10⁻⁴ mol HCl
3) lastly, we convert the moles of HCl to grams using the molar mass.
molar mass of HCl= 1.01 + 35.5= 36.51 g/mol
7.1x10⁻⁴ mol HCl (36.51 g/mol)=<span> 0.026 grams HCl</span>
Elements Y and elements Z would have similar properties due to the fact that they both posses the same number of valence electrons. They both have a single valence electron that determines the corresponding elements bonding properties and the fact that it can either donate 1 valence electron to produce an ion that would be attracted to another atom, that is also an ion. Assuming that these elements are group 1 elements, they do not undergo in covalent bonding.
The Chemical fomulla for Zinc Phosphate
Zn3(PO4)2
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
