Identify the type of bonding within each substance. Co ( s ) ionic covalent metallic CoCl 2 ( s ) covalent ionic metallic CCl 4
1 answer:
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Taking into account the reaction stoichiometry, P₄ will be the limiting reagent.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
8 P₄ + 3 S₈ → 8 P₄S₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- P₄: 8 moles
- S₈: 3 moles
- P₄S₃: 8 moles
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 3 moles of S₈ reacts with 8 moles of P₄, 4 moles of S₈ reacts with how many moles of P₄?
<u><em>moles of P₄= 10.667 moles</em></u>
But 10.667 moles of P₄ are not available, 4 moles are available. Since you have less amount of moles than you need to react with 4 moles of S₈, P₄ will be the limiting reagent.
Learn more about the reaction stoichiometry:
#SPJ1
The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ +
.
45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains ×1000 = 14.411 gm-mole. Thus the solubility product
of AgBr = [
]×
.
Or, 5.0× =
(the given value of solubility product of AgBr is 5.0×
and the charge of the both ions are same).
Thus S = (5.00×)
= 7.071×
g/mL.
Thus the concentration of Br or HBr is 7.071×
g/mL.
Answer:
1) 0.3g Mg
2)0.5g MgO
3)0.2g O
4)0.01mol Mg & 0.01mol O
5)0.01mol MgO
6) Empirical formula MgO
Explanation:
The mass og Mg is obtained by substracting 24.36g from 24.66g:
24.66 - 24.36 = 0.3g Mg
The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.
We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:
*
= 0.2g O
Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO
We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:
*
= 0.01mol O
*
= 0.01mol Mg
The moles of MgO can be obtained from:
*
= 0.01mol MgO
To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.
The result for both number of Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the formula unit of the compound.
The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.