Answer:
a) AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Explanation:
a) AgNO3 + KI → Ag+ + NO3- + K+ + I-
Ag+ + NO3- + K+ + I- → AgI + KNO3
AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-
Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-
6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3
2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-
2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Answer:
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Answer:
Al₂(SO₄)₃ and Na
Explanation:
Al has a charge of +3, Na has a charge of +1 and SO₄ has a charge of -2. Since cations and anions will bond we know that Al will bond with SO₄ leaving Na by itself (since this is a single replacement reaction). When Al bonds with SO₄ it makes aluminum sulfate which is Al₂(SO₄)₃ and Na will be left by itself.
Answer:
No the substance is not water.
Explanation:
The balance chemical equation for the decomposition of water is as follow;
2 H₂O = 2 H₂ + O₂
Step 1: <u>Calculate moles of H₂O;</u>
Moles = Mass / M.Mass
Moles = 5.0 g / 18.01 g/mol
Moles = 0.277 moles of H₂O
Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>
According to equation,
2 moles of H₂O produced = 1 mole of O₂
So,
0.277 moles of H₂O will produce = X moles of O₂
Solving for X,
X = 0.277 mol × 1 mol / 2 mol
X = 0.138 moles of O₂
Also,
According to equation,
2 moles of H₂O produced = 2 mole of H₂
So,
0.277 moles of H₂O will produce = X moles of H₂
Solving for X,
X = 0.277 mol × 2 mol / 2 mol
X = 0.227 moles of H₂
Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>
For O₂:
Mass = Moles × M.Mass
Mass = 0.138 mol × 31.99 g/mol
Mass = 4.44 g of O₂
For H₂:
Mass = Moles × M.Mass
Mass = 0.227 mol × 2.01 g/mol
Mass = 0.559 g of H₂
Conclusion:
From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.