Good evening, Laneyc!
Granite is not known as a solution because it has little pieces inside of it, so it cannot form all the way to an actual "solution".
I hope this answer has satisfied your needs, and if you have anymore questions, feel free to ask. Have a good night!
BROO DEE YOU ALREADY KNOW WHAT IS STOP DA CAP, o.81
Answer:
The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At) are the elements found along the step like line between metals and non-metals of the periodic table.
Elements: Germanium; Boron; Arsenic
Explanation:
Answer:
<u>Explanation</u>:
<u>Number of molecules for
</u>

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol



<u>Number of molecules for for
</u>

= Atomic mass of 3(Na) + P + 4(O)
= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

