Answer:
Manganese trinitrate or manganese(III) nitrate
Explanation:
Answer:
If mass increases, force increases.
Explanation:
hope this helps, pls mark brainliest :D
Answer:
The answer is
<h2>11.73 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula

From the question
mass of object = 30.5 g
Density = 2.6 g/cm³
The volume is

We have the final answer as
<h3>11.73 mL</h3>
Hope this helps you
Answer: A volume of 500 mL water is required to prepare 0.1 M
from 100 ml of 0.5 M solution.
Explanation:
Given:
= 0.1 M,
= ?
= 0.5 M,
= 100 mL
Formula used to calculate the volume of water is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M
from 100 ml of 0.5 M solution.