Answer:
A strong acid completely ionizes in water.
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


Answer:
<h2>15 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

But from the question
volume = final volume of water - initial volume of water
volume = 165 - 150 = 15 mL
We have

We have the final answer as
<h3>15 g/mL</h3>
Hope this helps you
U won’t get hyperthermia or a heat stroke when ur hot u use up more energy but when ur cold u use up less
Proteins make up many many important structures in our cells ( ex : receptors , Enzymes , integral proteins and so on )