Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
<h3>Answer:</h3>
Phosphoric acid reacts with magnesium hydroxide to produce magnesium phosphate and water via the following reaction:
2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O
(solid) (solid) (solid) (liquid)
<h3>Explaination:</h3>
This is a typical neutralization reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat,
ΔH < 0 , and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.
Doing so we obtain :
2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O
(solid) (solid) (solid) (liquid)
<h3>hope it helps :)</h3>
Explanation:
SADMEP
-2(bx-5) = 16 distribute
-2bx +10 = 16 subtracte
-10 -10
-2bx = 6
divide by -2x (on both sides)
b = -3x
John Dalton's original atomic theory contained the following key ideas and the incorrect one is that elements are made of tiny indivisible particles called atoms and is denoted as option A.
<h3>What is Atom?</h3>
This is defined as the smallest unit of matter which forms a chemical element and Dalton proposed that it was indivisible which was later proved wrong.
It was later discovered that atom is made up of sub atomic particles such as proton, electron and neutron. This was therefore the reason why option A was chosen as the most appropriate choice.
Read more about Atom here brainly.com/question/6258301
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The options include:
A. elements are made of tiny indivisible particles called atoms
B. Atoms are unchanged in chemical reaction
C. Atoms can join together in whole number ratios to form compounds.
D. The atoms of each element are unique
A compound inequality<span> is an equation with two or more </span>inequalities<span> joined together with either</span>
