For the answer to the question above, <span>ater weights 1000kg per meter cubed. the volume of the pool is part A is 5*4*3 = 60 meters cubed </span>
<span>60*1000 = 60 000kg. the force from this is m*g = 60 000 * 9.81 = 588kN </span>
<span>part B: </span>
<span>volume: 4*4*3 = 48 meters cubed </span>
<span>48 * 1000 = 48000kg </span>
<span>F = 9.81*48000 = 470kN
I hope this helps.</span>
A) Limiting reactant
You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)
a) Molar ratios:
6 mol HF : 1 mol SiO2 : 1 mol H2SiF6
2) Molar masses:
Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol
=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol
3) convert data in grams to moles
21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2
70.5 g HF / 20 g/mol = 3.525 mol HF
4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.
6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10
=> There is more HF than the needed to react with 0.35mol of SiO2 =>
SiO2 is the limiting reactant (HF is in excess)
b) Mass of excess reactant.
1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted
0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF
2.1 mol HF * 20 g/mol = 42 gram of HF
2) Subtract the quantity that reacted from the original quantity:
70.5 g - 42 g = 28.5 g of HF in excess
c) Theoretical yield of H2SiF6
1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6
Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams
d) % yield
% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
Answer:
is the change in enthalpy associated with the combustion of 530 g of methane.
Explanation:
Mass of methane burnt = 530 g
Moles of methane burnt = 
Energy released on combustion of 1 mole of methane = -890.8 kJ/mol
Energy released on combustion of 33.125 moles of methane :
is the change in enthalpy associated with the combustion of 530 g of methane.
Answer:
endoskeleton
Explanation:
most organisms that we know of have an endoskeleton
(but trust Ariellijo46 more, he probably knows more than that of my knowledge)
This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.
ΔQ/Δt = (1000 W/m²)∈Acosθ
A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²
ΔQ is the heat of melting ice.
ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
ΔQ = (916.7 kg/m³)(1 m²)(1.8 cm)(1m/100 cm)(<span>333,550 J/kg)
</span>ΔQ = 5,503,780 J
5,503,780 J/Δt = (1000 W/m²)(0.05)(1.072 m²)(cos 33°)
<em>Δt = 122,434.691 s or 34 hours</em>
