(i) We start by calculating the mass of sugar in the solution:
mass of sugar = concentration × solution mass
mass of sugar = 2.5/100 × 500 = 12.5 g
Then now we can calculate the amount of water:
solution mass = mass of sugar + mass of water
mass of water = solution mass - mass of sugar
mass of water = 500 - 12.5 = 487.5 g
(ii) We use the following reasoning:
If 500 g solution contains 12.5 g sugar
Then X g solution contains 75 g sugar
X=(500×75)/12.5 = 3000 g solution
Now to get the amount of solution in liters we use density (we assume that is equal to 1):
Density = mass / volume
Volume = mass / density
Volume = 3000 / 1 = 3000 liters of sugar solution
Answer:
ethier a dessert or a plains
though plains can get rain in the summer
from me living in both
it seems more like a dessert
Answer:
Percent error = 1.5%
Explanation:
Given data:
Measured value of density of graphite = 2.3 g/cm³
Percent error = ?
Solution:
Formula:
Percent error = [Measured value - Actual value / actual value] × 100
Actual/accepted value of density of graphite = 2.266 g/cm³
Now we will put the values:
Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100
Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100
Percent error = 0.015 × 100
Percent error = 1.5%
Answer:
18 O, 17 O, and 16 O
Explanation:
three naturally stable isotopes
Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
- For a weak acid like HF, the dissociation of HF will be:
<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>
[H₃O⁺] = [F⁻].
<em>∵ [H₃O⁺] = √Ka.C,</em>
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>
<em></em>
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>
