Asealed tank containing seawater to a height of 10.7 m also contains air above the water at a gauge pressure of 3.20 atm. water flows out from the bottom through a small hole. how fast is this water moving?

28.97 m/s

Explanation:

height of water (H1) = 10.7 m

Gauge pressure (Pg) = 3.2 x 1.01325 x 10^{5} = 324240 N/ m^[2}

pressure at the top (P1) = gauge pressure (Pg) + atmospheric pressure(Pa)  (Pg + Pa)

acceleration due to gravity (g) = 9.8 m/s^{2}

height at the bottom (H2) = 0

pressure at the bottom (P2) = atmospheric pressure = (Pa)

density of sea water (ρ) = 1030 kg/m^{3}

speed of water at the top (V1) = 0 (assuming the tank is big, the speed of water at any point on the surface is almost 0 and also the hole is very small compares to the size of the tank)

speed of water at the top (V2) = ?

applying Bernoulli's equation we can find the value of V2.

(take note that represents ρ whcih is the density of sea water)

recall that P1 = Pg + Pa and P2 = Pa

)

)

V2 =

V2 =

V2 = 28.97 m/s

ammonium chloride & tallow

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂