Answer:
<h2>Density = 1.67 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 50 g
volume = 30 mL
Substitute the values into the above formula and solve for the density
That's

Wr have the final answer as
<h3>Density = 1.67 g/mL</h3>
Hope this helps you
Question 3 is uniformitarianism
Question 4 is sedimentary rock
Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
![pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28acetic%20acid%29%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7Bmmol%20of%20CH_%7B3%7DCOOH%20%7D%5D)
Here pH is 5.31,
(acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
![5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]](https://tex.z-dn.net/?f=5.31%3D4.74%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7B10%7D%5D)
or, mmol of
= 37
So 37 mmol of acetate need to add to this solution.
Conditions for rusting
• moisture
•oxygen
Prevention
•Painting
• Galvanisation
• Making Alloys
• greasing
Hope this helps you ;)(:
Answer: B) calcium nitrate