Answer:
PubChem CID 7991
Structure Find Similar Structures
Chemical Safety Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula C5H10O2 or CH3(CH2)3COOH
Synonyms Valeric acid PENTANOIC ACID n-Valeric acid 109-52-4 n-Pentanoic acid More.
Answer:
Stone A
Explanation:
Measuring density is an easiest way to determine if two similar looking substances are same or not. Here also we need to perform the density test for each stone that is suspected to be fake diamond. We will calculate the density of each stone and compare it with the density of original diamond.
Density is calculated using the formula

It has been given in the question that both the substances have same volume of 0.15
.
Density of stone A =
= 3.47
or after rounding off we get 3.5 
Density of stone B =
= 2.8 
It is clear from the above calculation that the stone A has same density as the diamond but stone B lacks behind in density.
So, stone A could be the real diamond.
Only 0.7% of natural uranium is 'fissile', or capable of undergoing fission, the process by which energy is produced in a nuclear reactor. The form, or isotope, of uranium which is fissile is the uranium-235 (U-235) isotope. ... Most is then converted into uranium hexafluoride, ready for the enrichment plant.
<span>This question cannot be answered in actual centimeters unless at least one edge has a known length however it can be written as (length^3) or (length cubed). As an example if the sample of gold cube had one length of 3cm then all lengths would be 3cm on the cube.</span>
Answer:

Explanation:
To convert from representative particles to moles, Avogadro's Number: 6.02*10²³, which tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance.
We can use it in a ratio.

Multiply by the given number of molecules.

Flip the ratio so the molecules of water cancel out.



Divide.

The original number of molecules has 2 significant figures: 3 and 1, so our answer must have the same. For the number we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

There are about 5.1 moles of water in 3.1*10²⁴ molecules of water.