Option 4. ratio of electrons to protons
Isotope that has atomic number of 82 is stable. An element that has an atomic number ∠82 more stable except Tc and Pm. Also there is the concept that isotopes consisting a combination of even-even,even-odd,odd-even, and odd-odd are all stable. Many isotopes with no magic numbers of nucleons are stable
Answer:
order = SrS > SrCl2 > RbCl > CsBr
Explanation:
Comparison of the melting points of compounds is basically dependent on the charge on their cation and anion, the more the charges on the cation and anion, the stronger and greater the force of attraction and as such the melting point will be relatively higher as well.
The ionic radii is also another factor to be considered, the more the distance between ions, the lesser the bond strength and the lesser the melting point.
from the options, in terms of ionic radii SrS > SrCl2 and RbCl > CsBr
also both SrS and SrCl2 have more charges on their ions compared to RbCl and CsBr and as such the arrangement of the highest melting point will be in the order SrS > SrCl2 > RbCl > CsBr.
At divergent boundaries, plates separate, forming a narrow rift valley. Here, geysers spurt super-heated water, and magma, or molten rock, rises from the mantle and solidifies into basalt, forming new crust. Thus, at divergent boundaries, oceanic crust is created.
Answer:
Kc = 3.90
Explanation:
CO reacts with 
to form 
and 
. balanced reaction is:
No. of moles of CO = 0.800 mol
No. of moles of = 2.40 mol
Volume = 8.00 L
Concentration = 
Concentration of CO = 
Concentration of = 
Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium 
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium = 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium = 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)
Within the options, we have four organic compounds. Let's see what the skeletal structure of the compounds is in order to identify them better:
The first compound CH3OCH3 has two methyl groups linked by a carbon atom, this type of compound is called an Ether
The second compound has a double bond, it is badly written but it seems that is an alkene.
The third compound has two methyl groups linked by nitrogen atoms, therefore will be an amine.
The last compound has a hydroxyl group, therefore it is an alcohol
Answer:
CH3OCH3 Ether
CH2CH2CHCH2CH3 Alkene
CH3NHCH3 Amine
CH3CH(OH)CH₂CH3 Alcohol
