First calculate the electric charge used to deposit 1.0 g Pt
C = (1.0 g Pt) (1 mol Pt / 195.1 g Pt) ( 2 mol e / 1 mol Pt)
( 96485 C / 1 mol e)
C = 989.08 C
C = It
Where I is the current
T is the time
T = C / i
T = 989.08 C / 0.15 A
T = 6593.88 s
T = 1.83 hrs
<h2>Answer:</h2>
<u>Most porous is gravel</u>
<h2>Explanation:</h2>
Porosity tells us how much water rock or soil can retain. It tells us about the spaces among the grains. Gravel is more bigger than sand or silt. Its grains are arranged in such a manner that their integral spaces are very bigger so they are more porous. Clay is the most porous but since it is not mentioned in the option so after clay comes the gravel. Usually gravel and sand has equal porosity.
Answer:
B) One way of solving this is to use dimensional analysis and cancel out grams to find volume.
A) density over mass is volume. Mass is 84.7 g and volume is 49.6 cm cubed
Explanation:
Answer:
(i) 5-day BOD of the waste is 120 mg/l.
(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.
Explanation:
The dilution factor D is 0.05.
The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.
The BOD of the waste for an unseeded mixture is

The ultimate carbonaceous BOD (Lo) can be calculated as

If you’re asking if this is true or false, it is false . A HOMOGENEOUS mixture is one the where the substances are evenly distributed and have a uniform composition, while a HETEROGENEOUS mixture is uneven and the substances are part from each like oil and water.