Answer:
N = N1V1 +N2V2 +N3V3
(resultant ) ________________
V1+V2+V3
Here are 5 examples of suspensions mixtures:
<span>1-mercury shaken in oil</span>
<span>2-oil shaken in water</span>
<span>3-powdered chalk in water </span>
<span>4-dust in air</span>
<span><span>5-soot in air</span></span>
<span><span>hope this helps
</span></span>
Answer:
C. SOIL THAT IS A MIXTURE OF STILT AND CLAY
PLS MAKE ME BRAINLEST!
Answer:
see explanation below
Explanation:
In this case, we have the following reaction:
Mg + 1/2O₂ -------> MgO
Now, according to this reaction we want to know the percent composition of MgO. The problem is not providing the mass of the initial reactants and the product, so we can use the atomic weights of the components, to do this
The molecular weight of Mg is 24.305 g/mol, and O is 15.999 g/mol, so, let's calculate the molar mass of MgO:
MM MgO = 24.305 + 15.999 = 40.304 g/mol
Now with this weight, let's see the percent composition of this compound:
%Mg = 24.305 / 40.304 * 100 = 60.304 %
%O = 15.999 / 40.304 * 100 = 39.696 %
And this would be the percent composition of MgO
