Answer:
17.5 g/cm³
Explanation:
We can solve this particular problem by keeping in mind the <em>definition of density</em>:
As the problem gives us both <em>the mass and the volume</em> of the box, we can now proceed to <u>calculate the density</u>:
The density of the box is 17.5 g/cm³.
Answer:
<h3>The International Union of Pure and Applied Chemistry (IUPAC) defines alkanes as "acyclic branched or unbranched hydrocarbons having the general formula CnH2n+2, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms". ... The number of carbon atoms may be considered as the size of the alkane.</h3>
It helps to map out how you will navigate through your unit analysis problem before setting it up.
You are given moles and need grams. What can be used as a conversion factor from moles to grams? Molar mass. We are working with aluminum, so we will need the molar mass of aluminum. My Periodic Table tells me the molar mass of aluminum is approximately 27 g/mol. Now we are ready to set up the unit analysis.
Moles must go on the bottom so that they cancel. Notice how our number of significant figures is 2, so the answer must round to 16 g Al.
<h3>
Answer:</h3>
16 grams
Answer:
Four (4) hydrogen atoms
Explanation:
C=C-C(triple bond)C
only four hydrogen atoms can bond with the vacant space
Answer:
E = 12.92 × 10^(-16) J
Explanation:
Formula for energy is;
E = hc/λ
Where;
h is Planck's constant = 6.63 x 10^(-34) J.s
c is speed of light = 3 × 10^(8) m/s
λ is wavelength = 0.154 nm = 0.154 × 10^(-9) m
E = (6.63 x 10^(-34) × 3 × 10^(8))/(0.154 × 10^(-9))
E = 12.92 × 10^(-16) J
