Answer:
2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20
Explanation:
Since there's no "?" shown in the equation, let's balance it and solve it entirely.
Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20
1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.
To do this, let's start by looking at the elements that are only presnet once on each side:
On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H → to balance it, you need 4 water molecules on the products side.
Answer: It will take 11.775 seconds.
Explanation: As a sphere with a diameter of 0.1 mm, the area of an alveolus is
A = 4.π.r²
r for an alveolus would be: r = 0.00005m or r = 5.
m
Finding the area:
A = 4.3.14.(5.)²
A = 3.14.
m²
The concentration change is to be 90% of the final, so
c = 0.9.3.14.
c = 28.26.
The oxygen diffusivity is 2.4.m²/s, that means in 1 second 2.4. of oxygen spread in one alveolus area. So:
1 second = 2.4.m²
t seconds = 28.26.m²
t = 
t = 11.775s
For a concentration change at the center to be 90%, it will take 11.775s.
Answer:
36
Explanation:
products and reactants have to be equal
Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
