Balanced there are 2 Al, 6 oxygens on both sides.
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
150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
Answer:
D
Explanation:
If the pressure remains constant then the temperature and Volume are all that you have to consider.
Givens
T1 = 19oC = 19 + 273 = 292o K
T2 = 60oC = 60 + 273 = 333oK
V1 = 250 mL
V2 = x
Formula
V1/T1 = V2/T2
250/292 = x/333
Solution.
The solves rather neatly. Multiplly both sides by 333
250*333 / 292 = 333 *x / 333
Do the multiplication
250 * 333 / 292 = x
83250 / 292 = x
Divide by 292
x = 285.1 mL
The answer is D
Answer:
theory is diffrent from law
Explanation:
a Theory can never be proven to be true nd a law can usually be expressed
