Explanation:
Pure subsance is a substance that is made up of only one type of particle - each piece is the same throughout.
Being present before the reaction but not after means it's no the same (it couldve evaporated)
Answer:
They all have the same number of protons but different numbers of neutrons.
Explanation:
Elements will always have the same number of protons no matter the isotopes. Isotopes only change the number of neutrons. Silicon will always have 14 protons. So silicon-28 has 14 protons and 14 neutrons. Silicon-29 has 14 protons and 15 neutrons. Silicon-30 has 14 protons and 16 neutrons.
the formation of cations by using electron dot structures are :
a) Al
.
Al . losing the three valence electrons makes the Al³⁺
.
b) Sr :
Sr : losing the two valence electrons makes Sr²⁺
c) Ba
: Ba , losing the two valence electrons makes it Ba²⁺
A Lewis electron dot diagram is a representation of the valence electrons of an atom that employments specks around the image of the element. The number of dots equals the number of valence electrons within the molecule. These dots are arranged to the right and left and over and underneath the symbol, with no more than two dots on a side. Cations are the positive ions shaped by the loss of one or more electrons. The foremost commonly shaped cations of the representative elements are those that include the loss of all of the valence electrons.
To know more about the lewis electron dot diagram refer to the link brainly.com/question/14191114?referrer=searchResults.
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The moon orbiting around the earth.
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
