Answer:
the range should be 2.2 to 4.3
Explanation:
I think so because the numbers at the left side of the scale from 1 are more acidic so as it increases it's still acidic but lesser so 1 is more acidic than 2 so I used 2.2 as the beginning of the range because it's less acidic than A even though its a greater number and 4.3 is lesser than 4.4 but its still greater on the scale. frankly speaking I don't feel so correct because it's in decimal so try and compare facts thank you
The full sentences are given below:
1. During the process of erosion and deposition, sediments that are the SMALLEST in size will be carried the greatest distance before being deposited.
Erosion and deposition are the methods by which sand and rock particles are moved from one place to another. The erosion can be caused by water or wind. Water and wind have the capacity to transport particle from one location and deposit them in another location. How far the erosion is able to move the particles depend on the weight of the particles. It is easier for erosion to carry small particles over a long distance than for it to carry large particles over the same distance.
2. Most METAMORPHIC rocks form under conditions found a few kilometer under the earth surface.
Metamorphic rocks generally are formed from existing rocks. The existing rocks are usually subjected to heat and pressure, which cause radical changes in the chemical and physical properties of the rock. Metamorphic rocks can be formed underneath the earth surface if they are subjected to high temperature and pressure by the rock layers above them.
Answer:
Explanation: The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color.
Hope this helps
Answer:
13.8072 kj
Explanation:
Given data:
Mass of water = 100.0 g
Initial temperature = 4.0 °C
Final temperature = 37.0°C
Specific heat capacity = 4.184 j/g.°C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 37.0°C - 4.0 °C
ΔT = 33.0°C
Q = 100.0 g ×4.184 j/g.°C × 33.0°C
Q = 13807.2 j
Joule to KJ:
13807.2 j × 1kj /1000 j
13.8072 kj
