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Answer:
12.8 g of
must be withdrawn from tank
Explanation:
Let's assume
gas inside tank behaves ideally.
According to ideal gas equation- 
where P is pressure of
, V is volume of
, n is number of moles of
, R is gas constant and T is temperature in kelvin scale.
We can also write, 
Here V, T and R are constants.
So,
ratio will also be constant before and after removal of
from tank
Hence, 
Here,
and 
So, 
So, moles of
must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of
= 32 g/mol
So, mass of
must be withdrawn = 
The conclusion best supported by the data can either be high temperature, low temperature, and normal temperature. Since there are diverse substances included in the chart, It is expected to also have diverse temperatures.
Answer:
Explanation:
sp² hybridization is found in those compounds having double bond .
Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .
Rest have sp³ hybridization because they are saturated compounds .