The bottom of group 1. Francium (or Fr) is the element with the greatest metallic properties.
Francium is not a naturally-occurring element, however. It is man-made. There is an isotope of francium that exists naturally, but it's half life is so short that it decays almost instantly into a different element.
The naturally-occurring element with the highest metallic properties is cesium (or Cs), located right above francium.
Metallic characteristics decrease as you move from left to right on the periodic table.
Explanation:
It is known that equation for ideal gas is as follows.
PV = nRT
The given data is as follows.
Pressure, P = 1500 psia, Temperature, T = 
= 104 + 460 = 564 R
Volume, V = 2.4 cubic ft, R = 10.73 
Also, we know that number of moles is equal to mass divided by molar mass of the gas.
n = 
m = 
= 
= 9.54 lb
Hence, molecular weight of the gas is 9.54 lb.
- We will calculate the density as follows.
d = 
= 
= 3.975 
- Now, calculate the specific gravity of the gas as follows.
Specific gravity relative to air = 
= 
= 51.96
Answer:
BRAINLIEST?
Explanation:
Ammonia is a typical weak base. Ammonia itself obviously doesn't contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. My findings said that ammonia is a weak base, potassium hydroxide is a strong base, vinegar is a weak acid and ethyl alcohol is a weak acid.
Vinegar and ethyl alcohol are eliminated as they are acids. The question is on bases.... Potassium hydroxide is a strong base. So we are left with ammonia, being a weak base.
A is your answer
Answer:
Cyclopropane has a planar carbon back bone while propane does not
Explanation:
We have to recognize that in straight chain saturated organic compounds, carbon atoms have a tetrahedral geometry. Each carbon atom is bonded to four other atoms.
However, carbon atoms in cyclic compounds are also sp3 hybridized with each carbon bonded to only four other atoms but the ring system is highly strained.
Cyclopropane is a necessarily planar molecule with a bond angle that is far less than the expected tetrahedral bond angle due to strain in the molecule. Hence, the carbon atoms may have have a "planar backbone".
