Answer:
A. it contains a lot of matter
Answer: Chloroplasts
Reasoning: I just had my 830th lesson in school about these
Answer:
D. ionic sodium phosphate (Na3PO4)
Explanation:
Molecule for molecule, the solute that raises the boiling point of water the most is the one that makes the most particles in the solution. Lithium chloride breaks up into two ions (Li+ and Cl-). So does sodium chloride (Na+ and Cl-). Molecular molecules don't break up at all, so sucrose has only 1 particle per molecule. Sodium phosphate makes 4 total particles (3 Na+ ions and 1 PO4^3-). And magnesium bromide would make 3 particles (1 Mg2+ and 2 Br-). So the most is 4.
Answer: a. +2, cation and magnesium ion .
b. -1, anion, chloride
c. -2, anion, oxide
d. +1. cation , potassium ion
Explanation:
When an atom accepts an electron negative charge is created on atom and is called as anion.
When atom loses an electron positive charge is created on atom and is called as cation.
Magnesium (Mg) with atomic number of 12 has electronic configuration of 2,8,2 and thus it can lose 2 electrons to form
cation and becomes magnesium ion.
Chlorine (Cl) with atomic number of 17 has electronic configuration of 2,8,7 and thus it can gain 1 electron to form
anion and becomes chloride.
Oxygen (O) with atomic number of 8 has electronic configuration of 2,6 and thus it can gain 2 electrons to form
anion and becomes oxide.
Potassium (K) with atomic number of 19 has electronic configuration of 2,8,8,1 and thus it can lose 1 electron to form
cation and becomes potassium ion.
Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
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:
- Al: 4 moles
- O₂: 3 moles
- Al₂O₃: 2 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- O₂: 32 g/mole
- Al₂O₃: 102 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 4 moles ×27 g/mole= 108 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- Al₂O₃: 2 moles ×102 g/mole= 204 grams
<h3>Mass of Al₂O₃ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?

<u><em>mass of Al₂O₃= 102 grams</em></u>
Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
Learn more about the reaction stoichiometry:
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