Answer:
29260J
Explanation:
Given parameters:
Mass of water sample = 100g
Initial temperature = 30°C
Final temperature = 100°C
Unknown:
Energy required for the temperature change = ?
Solution:
The amount of heat required for this temperature change can be derived from the expression below;
H = m c (ΔT)
H is the amount of heat energy
m is the mass
c is the specific heat capacity of water = 4.18J/g°C
ΔT is the change in temperature
Now insert the parameters and solve;
H = 100 x 4.18 x (100 - 30)
H = 100 x 4.18 x 70 = 29260J
A). The number of solute particles increases
This was answered before on Brainly!
I used it for one of my chemistry courses before as well.
Hope this helps!
Transport of Na+ from a place of low concentration to a place of higher concentration. <u>This is the right answer.</u>
<u />
The sodium-potassium pump is the most common and well-known example of active transport. At the cell membrane, the sodium-potassium pump moves 3 sodium ions out of the cell and two potassium ions into the cell per ATP. Examples of active transport include the uptake of glucose in the human intestine and the uptake of minerals and ions into the root hair cells of plants.
One of the greatest examples of active transport is the movement of calcium ions out of cardiomyocytes. Cells secrete proteins such as enzymes, antibodies, and various other peptide hormones. Amino acids are transported across the intestinal mucosa of the human intestine. The movement of ions or molecules across cell membranes to regions of a higher concentration is assisted by enzymes and requires energy.
Learn more about Active transport here:-brainly.com/question/25802833
#SPJ1
Answer:
4FeS + 7O₂ ----> 2Fe₂O₃ + 4SO₂
Explanation:
You have to drag the elements shown on the right under the formula. For example, you would have to drag 4 of the FeS molecules under the 4FeS text in the formula. Then place 7 O₂ molecules under the 7O₂ text, etc.
When we have Ksp CaSO4 = 9.1x10^-6 so by substitution in Ksp formula:
Ksp CaSO4 = [Ca+2][SO4]
1.9x10^-6 = [Ca+2][1.5x10^-2]
∴ [Ca+2] M in the solution =1.27x10^-4
when percent remaining = [Ca in solution]/[C original value] * 100
= [(1.27x10^-4*(50ml+50ml)]/[1.28x10^-2*50ml]*100
≈ 2%