Following the key in the diagram (see the attached image), the only particle diagram that represents a mixture of three substances is diagram 2.
To simplify it, let us replace the key in the diagram as follows;
- atom of one element = A
- atom of different element = B
Diagram 1 consists of only AA and AB
Diagram 2 consists of AA, BB, and AB.
Diagram 3 consists of AA and ABA
Diagram 4 consists of AA and BAB
Thus, only diagram 2 has a mixture of 3 substances.
More on mixtures can be found here: brainly.com/question/6594631
Answer:
Al2(SO4)3 and Mg(OH)2
Explanation:
1. Al has a charge of 3-, and SO4 of 2-
when you cross multiply the charges you get
Al2 and (SO4)3
*the reason theres a bracket around the sulfate ion is that the charge 3 is not for oxygen only, but the entire sulphate ion*
Hence, Al2(SO4)3
2. Mg has a charge of 2- and OH of 1-
again cross multiply
Mg (you dont need to add the 1) and (OH)2
again, the bracket around OH means the charge appiles to Oxygen AND hydrogen
hence, Mg(OH)2
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer:
Hi there I think the awnser is B. or A. lol but im 90% its A.
Explanation:
xoxo hope this helps
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