Single bond: 1 pair of electrons
Double bond: 2 pairs of electrons
Triple bond: 3 pairs of electrons
Well the group fluorine is in is known as Halogens. Halogens all have seven valence electrons in their outer shell. They exist as diatomic molecules. Furthermore they readily combine with metals to form salts
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer:
The number of moles of xenon are 1.69 mol.
Explanation:
Given data:
Number of moles of xenon = ?
Volume of gas = 37.8 L
Temperature = 273 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
Now we will put the values in formula.
1 atm × 37.8 L = n × 0.0821 atm.L/ mol.K ×273 K
37.8 atm.L = n × 22.413 atm.L/ mol.
n = 37.8 atm.L / 22.413 atm.L/ mol.
n = 1.69 mol
The number of moles of xenon are 1.69.
The balanced chemical reaction is written as:
<span>3NO2 + H2O = 2HNO3 + NO
Assuming that the gases in this reaction are ideal gas, then we can use the conversion from L to moles which is 1 mol of ideal gas is equal to 22.4 L. We calculate as follows:
538 L NO2 ( 1 mol / 22.4L ) ( 1 mol NO / 3 mol NO2 ) ( 22.4 L / 1 mol ) = 179.33 L NO is produced</span>
