Dimension analysis is to be used to solve this problem. First convert 1L to milliliters. That is equivalent to 1000 ml. Then by dimension analysis, multiply the volume ( 1000ml) to the density of oil (0.92 g/ml) resulting to the answer: 920 grams.
Complete question is;
A drop of water has a volume of approximately 7 × 10⁻² ml. How many water molecules does it contain? The density of water is 1.0 g/cm³.
This question will require us to first find the number of moles and then use avogadro's number to get the number of water molecules.
<em><u>Number of water molecules = 2.34 × 10²¹ molecules</u></em>
We are given;
Volume of water; V = 7 × 10⁻² ml
Density of water; ρ = 1 g/cm³ = 1 g/ml
Formula for mass is; m = ρV
m = 1 × 7 × 10⁻²
m = 7 × 10⁻² g
from online calculation, molar mass of water = 18.01 g/mol
Number of moles(n) = mass/molar mass
Thus;
n = (7 × 10⁻²)/18.01
n = 3.887 × 10⁻³ mol
from avogadro's number, we know that;
1 mol = 6.022 × 10²³ molecules
Thus,3.887 × 10⁻³ mol will give; 6.022 × 10²³ × 3.887 × 10⁻³ = 2.34 × 10²¹ molecules
Read more at; brainly.in/question/17990661
Hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9
