Hey there,
Answer:
4 valence electrons.
Hope this helps :D
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Answer:
Empirical formula of C₈H₈ = CH
Explanation:
Data Given:
Molecular Formula = C₈H₈
Empirical Formula = ?
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule
C₈H₈ Consist of Carbon (C), and Hydrogen (H)
Now
Look at the ratio of these two atoms in the compound
C : H
8 : 8
Divide the ratio by two to get simplest ratio
C : H
8/8 : 8/8
1 : 1
So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1
So the empirical formula will be
Empirical formula of C₈H₈ = CH
Answer:
2HgS + 3O2 → 2HgO + 2SO2
The coefficients are: 2, 3, 2, 2
Explanation:
HgS + O2 → HgO + SO2
The equation can be balance as follow:
Put 3 in front of O2 as shown below:
HgS + 3O2 → HgO + SO2
Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:
HgS + 3O2 → 2HgO + 2SO2
Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:
2HgS + 3O2 → 2HgO + 2SO2
Now the equation is balanced.
The coefficients are: 2, 3, 2, 2
The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
The given compound is Aluminum sulfate, Al2(SO4)3:
Molar masses:
Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol
The total molar mass is 342 g/mol
The ratio by mass of the elements:
Aluminum = 27*2/342
= 0.16
Sulfur = (32*3)/342
= 0.28
Oxygen = (16*12)/342
= 0.56
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