<span>he density of a material is defined as its mass per
unit volume. In this example, each volume of water is different and
therefore has a specific and unique mass. The mass of water is expressed
in grams (g) or kilograms (kg), and the volume is measured in liters (L), cubic centimeters (cm 3), or milliliters (mL).</span>
Covalent bond two non metals
Molecular weight it stands for molecular weight
984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.
Explanation:
From the question data given is :
volume of strontium in sea water= 9.84x10^8 cubic meter
(1 cubic metre = 1000000 ml)
so 9 .84x10^8 cubic meter
= 984 ml.
density of sea water = 1 gram/ml
from the formula mass of strontium can be calculated.
density = 
mass = density x volume
mass = 1 x 984
= 984 grams of strontium will be recovered.
98400 centigram of strontium will be recovered.
Strontium is an alkaline earth metal and is highly reactive.
Answer:
21.5 g.
Explanation:
Hello!
In this case, since the reaction between the given compounds is:

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

Now, the moles of Li3P consumed by 15 g of Al2O3:

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

Therefore, the total mass of products is:

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.
Best Regards!