Homogeneous mixture that does not settle out upon standing but which will reflect light is called COLLOIDS.
There are three types of homogeneous mixtures, these are: solutions, colloids and suspension. Colloids are usually distinguished by Tyndall effects. Light passing through a colloidal dispersion will be reflected.<span />
D. A set of steps that help scientists gather knowledge
Answer:
Investigating a data breach
Explanation:
A data breach usually involves data exfiltration over a computer network. the other options involve data being stored on a device locally which isn't volatile data like text messages, photos or rearranging data in defragmentation all of which does not require a network.
Explanation:
Fusion vs Fission
In fission, energy is gained by splitting apart heavy atoms, for example uranium, into smaller atoms such as iodine, caesium, strontium, xenon and barium, to name just a few. However, fusion is combining light atoms, for example two hydrogen isotopes, deuterium and tritium, to form the heavier helium. Both reactions release energy which, in a power plant, would be used to boil water to drive a steam generator, thus producing electricity.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
