Answer:
Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.
Explanation:
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Answer:
- <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>
Explanation:
STP stands for standard pressure and temperature.
The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:
- Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).
- After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).
Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.
With the later definition, the volume of a mol of gas at STP is 22.7 liter.
I will use the traditional measure of 22.4 liter per mole of gas.
<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>
- n = mass in grams / molar mass
- Atomic mass of nitrogen: 14.0 g/mol
- Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
- n = 14.0 g / 28.0 g/mol = 0.500 mol
<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>
- 22.4 liter / mol = x / 0.500 mol
- Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.
<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.
Answer:
3,855.532 grams
Explanation:
1 pound = 453.592 grams
8.50 = ? grams
--> 8.50 * 453.592 = 3,855.532 grams.
Answer:
The most abundant isotope is 1.007 amu.
Explanation:
Given data:
Average atomic mass = 1.008 amu
Mass of first isotope = 1.007 amu
Mass of 2nd isotope = 2.014 amu
Most abundant isotope = ?
Solution:
First of all we will set the fraction for both isotopes
X for the isotopes having mass 2.014 amu
1-x for isotopes having mass 1.007 amu
The average atomic mass is 1.008 amu
we will use the following equation,
2.014x + 1.007 (1-x) = 1.008
2.014x + 1.007 - 1.007 x = 1.008
2.014x - 1.007x = 1.008 - 1.007
1.007 x = 0.001
x= 0.001/ 1.007
x= 0.0009
0.0009 × 100 = 0.09 %
0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.
now we will calculate the abundance of second isotope.
(1-x)
1-0.0009 = 0.9991
0.9991 × 100= 99.91%