1.4mL
Explanation:
Given parameters:
Mass of object = 3.5g
density = 2.5g/mL
Unknown:
Volume of the object = ?
Solution:
Density is the mass per unit volume of a body. It is an intensive property that is the same for the same substances.
Density = 
since the volume is unknown; we make it the subject of the formula;
Volume = 
Volume = 
Volume of the object = 1.4mL
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Answer:

Explanation:
In its raw form, function notation essentially represents an equation with only one unknown variable, expressed in terms of another. Thus, f(x) = x² + 7x can be expressed as
g(x) = 2x + 3
f(g(x)) = (2x + 3)²
f(g(x)) = 4x² + 12x + 9
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The molecular formula : C₁₈H₁₈N₈
<h3>Further explanation</h3>
Given
62.41% C, 5.24% H, and 32.36% N
Required
The molecular formula
Solution
mol ratio
C : 62.41/12.0096 = 5.1967
H : 5.24/1.00784 = 5.1992
N : 32.36/14.0067 = 2.310
Divide by 2.310(smallest)
C : 5.1967/2.31=2.25
H : 5.1992/2.31 = 2.25
N : 2.31/2.31 = 1
Multiplied by 4
C : H : N = 9 : 9 : 4
The empirical formula : C₉H₉N₄
(C₉H₉N₄)n=346.40 g/mol
(12.0096 x 9 + 1.00784 x 9 + 14.0067 x 4)n=346.4
(108.0864+9.07056+56.0268)n=346.4
(173.184)n=346.4
n=2
<em>The molecular formula : C₁₈H₁₈N₈</em>
Answer : The atomic mass of chlorine is, 36.723 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:

As we are given that,
Mass of 35-Cl isotope = 35 amu
Percentage abundance of 35-Cl isotope = 13.85 %
Fractional abundance of 35-Cl isotope = 0.1385
Mass of 37-Cl isotope = 37 amu
Percentage abundance of 37-Cl isotope = 86.15 %
Fractional abundance of 37-Cl isotope = 0.8615
Now put all the given values in above formula, we get:
![\text{Average atomic mass }=[(35\times 0.1385)+(37\times 0.8615)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20%7D%3D%5B%2835%5Ctimes%200.1385%29%2B%2837%5Ctimes%200.8615%29%5D)

Therefore, the atomic mass of chlorine is, 36.723 amu