Answer:
1. BF3 This is a trigonal planar molecule; the electron density is drawn into a cloud that circles the Boron, this is made nonpolar by the geometrically equivalent structure of the surrounding electronegative Fluorines.
2. H2O The 2 lone pairs of e- of Oxygen makes the O partially negative, the H’s, partially positive. Polar.
3. NF3 Lone pair on Nitrogen overwhelmed by the 3 incredibly electronegative Fluorines. Polar
4. CH3Br The “Soft Ion” of Bromine is negative; it is electronegative. Polar.
5. SO2 the lone pairs of Oxygen, at approximately 119°-120° angles to one another will form a reasonance structure; there will be more lone pairs about the Oxygen than the Sulfur; the Sulfur will be partially positive compared to the oxygens. Polar.
What is the question that needs an answer?
Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2
(p1)(V1)/(T1) = (p2)(V2)/(T2)
(1.00 atm)(V) / (273 + 25K) = (40.0 atm)(V/10) / (273 + T)
273 + T = (40.0)(1/10)(273 + 25K) / (1.00)
T = 919°C
Answer:
Explanation:
use the equation
moles = mass/mr
=19.9/79.5
=0.250moles of CuO
then do the same for
H = 2.02/1
=2.02
so CuO is the limiting reagent because there is less amount of it.
Hope this helps :)
