Answer: mg(OH)2 = +37
Rbl = 31
RbOH= -83
mgl2= -144
Explanation: just did it on edge2020
Answer : The resonance structure of
is shown below.
Explanation :
Resonance structure : It is defined as when more than one Lewis structure can be drawn, the molecule or ion is said to have resonance.
Resonance is the concept where electrons (bonds) are delocalized over three or more atoms which cannot be depicted with one simple Lewis structure.
First we have to draw Lewis-dot structure.
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that sulfur has '6' valence electrons, carbon has '6' valence electrons and nitrogen has '5' valence electron.
Therefore, the total number of valence electrons in
= 6 + 4 + 5 = 15
According to Lewis-dot structure, there are 7 number of bonding electrons and 8 number of non-bonding electrons.
In SCN, carbon atom is the central atom and sulfur and nitrogen are the neighboring atoms.
The resonance structure of
is shown below.
The answer would be the third option. (the angle at which the light hits the surface.)<span>
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Igneous rocks from cooling magma<span>. </span>Granite<span> is an igneous rock </span>formed<span> from </span>magma<span>that </span>cooled slowly<span> underground. As the </span>magma slowly cools<span>, large mineral crystals form.
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Answer:
The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml
Explanation:
Here we have the reaction of AgNO₃ and NaCl as follows;
AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)
Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,
Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃
Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;
0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL
Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.