Answer:
8.47 x 10^3
Explanation:
So since we only want 3 sig figs we have to round our last digit up and then it is 10^3 because we moved the decimal place to the left 3 times to get 8.47
Comment if you have any other sig fig questions, happy to answer them!
For a polar molecule<span>, your bonds will not cancel out. This means that in a </span>polar<span> bond, the electronegativity of the atoms will be different. For </span>nonpolar<span> bonds the electronegativity of the atoms will be equal. In a </span>polar<span> bond you will have an unequal sharing of electron pairs which causes a molecular dipole.</span>
Answer: The angle between the carbon-chlorine bonds in phosgene will be 120°
Explanation: Formula used to calculate the hybridization:
where,
V = number of valence electrons present in central atom i.e. carbon = 4
N = number of monovalent atoms bonded to central atom = 2
C = charge of cation = 0
A = charge of anion = 0
For
The number of electron pairs are 3 which means that the hybridization will be and the electronic geometry of the given molecule will be trigonal planar.
The bond angle in this electronic geometry is 120°
The structure of phosgene is attached below.
You want us to answer all that and only get 5 points that’s not fair :(
Answer:
P₂ = 0.6406 atm
Explanation:
Given data:
Initial volume = 24.0 L
Initial pressure = 1.00 atm
Initial temperature = 203.0 K
Final temperature = 298.0 K
Final volume = 55.0 L
Final pressure = ?
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 1 atm × 24.0 L × 298.0 K / 203.0 K × 55.0 L
P₂ = 7152 atm .L. K / 11165 K.L
P₂ = 0.6406 atm
When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value.
Thus we will include four significant figures in answer because 7152 have four significant figure.