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:
B) THE DEPTH OF THE LAKE IS 0.060 m
Explanation:
b) Determine the depth of the lake in metres
1. Using the general gas law, we will calculate the initial pressure of the air bubbles.
P1V1 /T1 = P2V2/T2
P1 = Unknown
T1 = 5.24 °C
T2 = 18.73 °C
P2 = 0.973 atm
V1 = V1
V2 = 6V1
P1 = P2 V2 T1 / V1 T2
P1 = 0.973 * 6V1 * 5.24 / V1 * 18.73
P1 = 5.09852 * 6 / 18.73
P1 = 30.59112 / 18.73
P1 = 1.633 atm.
2. Calculate the depth of the lake:
Pressure = length * density * acceleration
length = Pressure / density * acceleration
Pressure = 1.633 atm = 1.633 * 101, 325 Nm^2 = 165, 463.725 Nm^2
Density = 1.02 g/cm3 = 1.02 * 10^3 kg/m^3
Acceleration = 9.8 m/s^2
So therefore, the length in metres is:
Length = density * acceleration / pressure
Length = 1.02 *10^3 * 9.8 / 165, 463.725
Length = 9.996 * 10^3 / 165 463.725
Length = 0.06 m
Hence, the depth of the lake is 0.06 m
I think the answer might be 51.14% since the formula had to equal a 100% just add 41.86 and 6.98 and subtract the sum to 100.
Ca (Clo3)2(s) ⇒ Ca Cl2 (s) + 3 O2 (g)
Answer:
We actually output other gases like methane lol
Explanation:
Well since most of the air we breathe is made up of nitrogen we also breathe out nitrogen
Source: da brain
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