<span>To find the volume of the plate without accounting for the hole firstly
V = (15.0 cm)(12.5 cm)(0.250 cm) = 46.875 cm^3
and the volume of the hole is
(pi)(1.25 cm)^2(0.250 cm) = 1.2272 cm^3
we will subtract the volume of the hole from the rest 45.648 cm^3
the multiply this by the density of the alloy to find the mass
(8.80 g/cm^3)(45.648 cm^3) = 401.701 g.
0.044% of this is Si, so (0.00044)(401.701 g) = 0.17675 g is silicon.
by the number of atoms and using average atomic mass of silicon and Avogadro's number to find the number of silicon atoms:
(0.17675 g)(1 mol/28.0855 g)(6.022E23 atoms/1 mol) =3.794E21atoms of Si
3.10% of these are Si-30:(0.0310)(3.794E18 atoms)=1.176E20 atoms of Si-30 and with two significant figures, 1.2E20 atoms.
hope this helps
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<h2>
Answer:</h2>
The magnesium ribbon, <u>D. It forms a material to cast the tool mark</u>.
<h2>
Explanation:</h2>
When a magnesium ribbon is burnt in the presence of oxygen it gives out strong light and heat is produced. Apart from it, it leads to the production of substance called as magnesium oxide which is formed as the product due to the reaction of magnesium with the oxygen present in the air.
Tool marks are the mark which is created by tools while using them. In order to identify or locate them castes made up of magnesium oxide is utilized. When this is pasted on the suspected area, the tool mark of the suspected tool gets pasted on it.
2x 6.022x10^23= 1.204x10^24
Answer:
molar mass of carbon tetrafluoride (CF4) is
(12.01 × 1 ) + ( 4 × 19.00)
= 12.01 + 76
= 88.01u
= 88u
Hope this helps
I would assume so.
Given
, we can simplify the fraction to
Both would obtain the same proportions, so I don't see why putting a half cup of sugar would make things any different.
Hope this is the answer you are looking for.
