BeCl2 = Beryllium Chloride
2H2O = 2 molecules of H2O (water)
LiNO2 = Lithium Nitrate
H2O = water
Answer:
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Explanation:
Energy absorbed by pork,E =
(assuming)
Total energy produced by barbecue = Q
Percentage of energy absorbed by pork = 10%


Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.
Q = 
Moles of propane burnt to produce Q energy =n


According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:
carbon dioxide gas.
Mass of 23.49 moles of carbon dioxide gas:
23.49 mol × 44 g/mol =1,033.56 g
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Answer:
Δx ≥ 1.22 *10^-10m
Explanation:
<u>Step 1:</u> Data given
The E. coli bacterial cell has a mass of 1.80 fg ( = 1.80 * 10^-15 grams = 1.80 * 10^-18 kg)
Velocity of v = 8.00 μm/s (= 8.00 * 10^-6 m/s)
Uncertainty in the velocity = 3.00 %
E. coli bacterial cells are around 1 μm = 10^−6 m in length
<u>Step 2:</u> Calculate uncertainty in velocity
Δv = 0.03 * 8*10^-6 m/s =2.4 * 10^-7 m/s
<u>Step 3:</u> Calculate the uncertainty of the position of the bacterium
According to Heisenberg uncertainty principle,
Δx *Δp ≥ h/4π
Δx *mΔv ≥ h/4π
with Δx = TO BE DETERMINED
with m = 1.8 *10^-18 kg
with Δv = 2.4*10^-7
with h = constant of planck = 6.626 *10^-34
Δx ≥ 6.626*10^-34 / (4π*(1.8*10^-18)(2.4*10^-7))
Δx ≥ 1.22 *10^-10m