Answer:
full octets live in noble gas world so you need to be whatever your charge is away from there in steps on the periodic table so s-1 is the one that falls short as it is 2 moves away from Ar which is its closest Noble gas
Explanation:
Answer:
Pressure
Gases are the only state of matter that are affected by changes in pressure when it comes to rate of reaction. Solids and liquids are not affected by changes in pressure.
Answer:If each side of the equation has the same number of atoms of a given element, that element is balanced. If all elements are balanced, the equation is balanced. - online resource
Explanation: if not im sorry
Answer:
85.5%
Explanation:
To get the experimental value, you need to convert 15.0 grams of Na2SO4 to grams of Na3PO4. You do this with stoichiometry.
Convert grams of Na2SO4 to moles with molar mass. Then convert to moles of Na3PO4 with the mole-to-mole ratio according to the balanced chemical equation. Then convert moles of Na3PO4 to grams with the molar mass.
15.0 g Na2SO4 x (1 mol/142.04 g) x (2 Na3PO4/3Na2SO4) x (163.94 g/1 mol) = 11.7 g Na3PO4
Percent Yield = (actual value/experimental value) x 100
Actual Value = 10.0 g
Experimental Value = 11.7 g
10.0g/11.7 g = 85.5%
Answer:
355.5 g.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (87.7 g) of Al:
no. of moles of Al = mass/atomic mass = (87.7 g)/(26.98 g/mol) = 3.251 mol.
<em><u>Using cross multiplication:</u></em>
2.0 mol of Al need to react → 6.0 mol of HCl, from stichiometry.
3.251 mol of Al need to react → ??? mol of HCl.
∴ the no. of moles of HCl needed to react with (87.7 g) 3.251 mol of Al = (3.251 mol)(6.0 mol)/(2.0 mol) = 9.752 mol.
- Now, we can get the mass of HCl needed to react with (87.7 g) 3.251 mol of Al:
<em>mass of HCl = (no. of moles)(molar mass) </em>= (9.752 mol)(36.46 g/mol) = <em>355.5 g.</em>