Answer:
C) ball rollinflown a hill
Explanation:
The question asks to identify the endothermic process in the list of options. By way of elimination, we have;
A) condensation of water on a wind shield of a car
Condensation is an exothermic process. That is, heat is given out as the gases change into the liquid state of matter.
B) formation of copper
This is an exothermic process. Capture of electrons by a cation is always exothermic.
C) ball rollinflown a hill
This is the correct option. Energy is absorbed by the ball as it moves on the hill
D) formation of ice from liquid water
Freezing is an example of exothermic reaction. Heat is given off to the surroundings.
E) oxide from copper and oxygen
Formation of metal oxides and most reactions involving oxygen are exothermic reactions,
We can use the dilution formula to find the volume of the diluted solution to be prepared
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
Substituting the values in the equation
15 M x 25 mL = 3 M x v2
v2 = 125 mL
The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution
Lithium has the lowest. if fluorine is the highest then lithium is the lowest. i hope this helps you out!
d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
