answer= 0.912 L or 912 mL
M(KClO3) = 122.55 g/mol
3.00 g KClO3 * 1 mol/122.55 g = 3.00/122.55 mol =0.02449 mol
2KCIO3(s)=2KCI(s) + 3O2(g)
from reaction 2 mol 3 mol
given 0.02449 mol x
x = 0.02449*3/2 =0.03673 mol O2
T = 24 + 273.15 = 297.15 K
PV = nRT
V= nRT/P = (0.03673 mol*0.082057 L*atm/K*mol*297.15 K)/0.982 atm =
= 0.912 L or 912 mL
Answer:
The answer to your question is Ionic bonding. Magnesium loses 2 electrons and Chlorine gains 1 electron
Explanation:
An Ionic bond is formed between a metal and a nonmetal and the difference of electronegativity is higher than 1.7. In this type of bond, metals lose electrons and nonmetals gain them.
Covalent bonding is formed between two nonmetals and the difference of electronegativity is between 0 and 1.7. Elements share electrons.
Process
Electronegativity of Mg = 1.31
Electronegativity of Cl = 3.16
Difference = 3.16 - 1.31
= 1.85
1.85 > 1.7 then the bond formed is Ionic
Magnesium loses 2 electrons and Chlorine gains 1 electron
Answer:
9.47 mL
Explanation:
The reaction that takes place is:
- 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
First we <u>calculate how many KOH moles reacted</u>, using <em>the given concentration and volume of KOH solution</em>:
- 0.061 mol/L = 0.061 mmol/mL
- 0.061 mmol/mL * 26.7 mL = 1.6287 mmol KOH
Then we <u>convert KOH moles into H₂SO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 1.6287 mmol KOH *
= 0.8144 mmol H₂SO₄
Finally we <u>calculate the required volume of the H₂SO₄ solution</u>, using<em> the number of moles and given concentration</em>:
- 0.8144 mmol ÷ 0.086 mmol/mL = 9.47 mL
