<span>Which
of the following best describes the electron cloud model?
THE ELECTRON CLOUD MODEL WAS DEVELOPED BY SCHRODINGER. IT STATES THE THE ELECTRONS ARE NOT PARTICLES MOVING AROUND THE NUCLEUS IN FIXED ORBITY BUT THEIR LOCATIONS CAN ONLY BE STATED BY A PROBABILITY DENSITY IN FORM OF CLOUD AROUND THE NUCLEUS.
THEN THE MAIN POINT OF THE CLOUD MODEL IS THAT THE ELECTRONS ARE NOT IN FIXED ORBITS AROUND THE NUCLEUS BUT THEIR LOCATION IS STATED BY A PROBABILITY FUNCTION THAT IS LIKE A CLOUD REGION.
A. It shows
that electrons usually carry a negative charge.
FALSE: ELECTRONS ALWAYS CARRY NEGATIVE CHARGE
B. It shows that
electrons remain in high-energy subshells.
FALSE: ELECTRONS OCCUPY THE LOWEST-ENERGY SUBSHELLS AVAILABLE UNLESS THEY ARE EXCITED (ABSORB ENERGY)
C. It shows that electrons
move quickly in circular orbits.
FALSE: ELECTRONS DO NOT MOVE IN CIRCULAR ORBITS.
D. It shows that the electrons within
an atom do not have sharp boundaries.
TRUE. THE IDEA OF A CLOUD IS A DIFFUSSE REGION WHERE IS A 90% OF PROBABILITIES TO FIND THE ELECTRON, AND THEY DO NOT HAVE SHARP BOUNDARIES.
</span>
Answer:
14 ft 7 inches
step-by-step explanation:
since each model inch corresponds to 25 real inches, 7 model inches will correspond to
7×25" = 175"
at 12" per foot, that is 14 feet, 7 inches.
Explanation:
The solubility of a given solute in a given solvent typically depends on temperature. Many salts show a large increase in solubility with temperature. ... A few, such as cerium(III) sulfate, become less soluble in water as temperature increases.
Answer:
The new volume after the temperature reduced to -100 °C is 0.894 L
Explanation:
Step 1: Data given
Volume of nitrogen gas = 1.55 L
Temperature = 27.0 °C = 300 K
The temperature reduces to -100 °C = 173 K
The pressure stays constant
Step 2: Calculate the new volume
V1/T1 = V2/T2
⇒with V1 = the initial volume of the gas = 1.55 L
⇒with T1 = the initial temperature = 300 K
⇒with V2 = the new volume = TO BE DETERMINED
⇒with T2 = the reduced temperature = 173 K
1.55 L / 300 K = V2 / 173 K
V2 = (1.55L /300K) * 173 K
V2 = 0.894 L
The new volume after the temperature reduced to -100 °C is 0.894 L
