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.
Answer:
The answer to your question is: 234.7 cans
Explanation:
data
caffeine concentration = 3.55 mg/oz
10.0 g of caffeine is lethal
there are 12 oz of caffeine in a can
Then
3.55 mg ----------------- 1 oz
x mg -----------------12 oz (in a can)
x = 42.6 mg of caffeine in a can
Convert it to grams 42,6 mg = 0.0426 g of caffeine in a can
Finally
0.0426 g of caffeine ------------------ 1 can
10 g of caffeine ----------------- x
x = 10 x 1/0.0436 = 234.7 cans
Answer:
n = 2.58 mol
Explanation:
Given data:
Number of moles of argon = ?
Volume occupy = 58 L
Temperature = 273.15 K
Pressure = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K
58 atm.L = n × 22.43 atm.L/ mol.
n = 58 atm.L / 22.43 atm.L/ mol
n = 2.58 mol
Answer:
I think the answer would be A.
