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.
An ambient pressure of 1 atm
The answer is D, reactant.
Answer:
Detail is given below
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.
In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm
In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N
In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.
Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:
Moles of nitric acid are:
As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>
