This is a true statement if it is density you are looking for... Density problem.....
Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits
The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.
Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.
The accepted value for the density of tin is 7.31 g/cm³
The answer is 14.22 mg / (mm^2)
Answer:
The uranium in the sample of the compound is radioactive
Explanation:
Some atoms can split on their own. Some split when bombarded by energetic particles. Such atoms are said to be radioactive.
Radioactivity is borne out of the drive of an atom to reach stabillity. Every atom have a specific neutron/proton ratio which ensures stability of the nucleus. A nucleus with a stability ratio different from that which makes it stable will become unstable and split into one or more other nuclei with emissons of energetic particles.
Note: neutrons and protons dictates the mass of an atom. They are located in the nucleus which is the site for nuclear radioactive reactions.
moles of CO gas : 1.545
<h3>Further explanation</h3>
Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions
There are 2 conditions that are usually used as a reference : STP and RTP
Assuming the STP state :
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
Then for 34.6 L of CO gas :
Answer:
<h2>1.89 atm</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>1.89 atm</h3>
Hope this helps you
