In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=

%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
Answer:
<h2>15 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

But from the question
volume = final volume of water - initial volume of water
volume = 165 - 150 = 15 mL
We have

We have the final answer as
<h3>15 g/mL</h3>
Hope this helps you
Answer:
The Law of Conservation of Energy states that energy cannot be created or destroyed. In other words, the total energy of a system remains constant. This is an important concept to remember when dealing with energy problems. The two basic forms of energy that we will focus on are kinetic energy and potential energy.
Explanation:
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.
Im bad at these questions hope it helps and have a good day.
Answer:
The net energy is 2.196 eV
Explanation:
Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.
Using:
ΔE = h*c*(1/λ
- 1/λ
)
where:
ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*
J.
h = 4.135*
eVs
c = 3*
m/s
λ
= 300 nm = 300*
m
λ
= 640 nm = 640*
m
Thus:
ΔE = 4.135*
eVs*3*
m/s*(
)
ΔE = 4.135*
*3*
*1.77*
eV = 2.196 eV
Answer: Its average atomic mass is 114.9 amu
Explanation:
Mass of isotope 1 = 113 amu
% abundance of isotope 1 = 5% = 
Mass of isotope 2 = 115 amu
% abundance of isotope 2 = 95% = 
Formula used for average atomic mass of an element :

![A=\sum[(113\times 0.05)+(115\times 0.95)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%28113%5Ctimes%200.05%29%2B%28115%5Ctimes%200.95%29%5D)

Thus its average atomic mass is 114.9 amu