Answer:
Models help scientists understand processes in the natural world by providing simpler representations of those processes.
Answer:


Explanation:
Given that;
diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m
length (l) = 10 cm = 0.1 m
porosity = 50%
height (h) = 30 cm = 0.3 m
time (t) = 5 s
volume (v) = 60 cm³ = 60 × 10⁻⁶ m³
Q (flow rate) = 
Q = 
Q = 
From constant head method, we use the relation K =
to determine the hydraulic conductivity ; we have:


Seepage velocity 
where; velocity = 
=
= 


<span>Na2CO3 (aq) + CaCl2H4O2 (aq) = CaCO3 (s) + 2 NaCl (aq) + 2 H2O (l)</span>
Answer:
459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution
Explanation:
Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by the following expression:

Molarity is expressed in units 
In this case:
- Molarity: 1.56 M= 1.56

- Number of moles of calcium chlorine= ?
- Volume= 2.657 liters
Replacing:

Solving:
Number of moles of calcium chlorine= 1.56 M* 2.657 liters
Number of moles of calcium chlorine= 4.14 moles
In other side, you know:
- Ca: 40 g/mole
- Cl: 35.45 g/mole
Then the molar mass of the calcium chloride CaCl₂ is:
CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole
Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

mass= 459.126 g
<u><em>459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution</em></u>
Answer: assume pathogens are present and treat the samples accordingly
Explanation:
When investigators are unable to conclusively ascertain the source of a biological sample found at a crime scene, the correct thing to do is to treat it as if pathogens are present in it and handle it according to set rules on how to handle pathogens.
This is done to ensure that if a pathogen is indeed present, it would not cause a health emergency by infecting those who come in contact with the samples at the scene.