Average atomic mass of an element is the sum of atomic mass of its isotopes multiplied by their respective percentage abundance.
There are four isotopes of element X:
1. 4.350 % with mass 49.94605 amu
2. 83.79% with mass 51.94051 amu
3. 9.5% with mass 52.94065 amu
4. 2.360% with mass 53.93888 amu
Thus, average atomic mass will be:
Or,
Therefore, average atomic mass of element X will be 51.99 amu (four significant figures)
Answer:
260 kPa
Explanation:
To answer this problem we can use <em>Gay-Lussac's law</em>, which states that at a constant volume (such as is the case with a rigid container):
Where in this case:
- T₂ = 273 °C ⇒ 273 + 273.16 = 546 K
We <u>input the data</u>:
- 340 kPa * 546 K = P₂ * 713 K
And <u>solve for P₂</u>:
We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound
C H O
mass 77.87 g 11.76 g <span>10.37 g
number of moles 77.87/12 11.76/1 10.37/16
moles = 6.48 = 11.76 =0.648
divide by least number of moles
6.48/0.648 11.76/0.648 0.648/0.648
= 10 =18.1 = 1
rounded off
C - 10 , H - 18 and O - 1
empirical formula - C</span>₁₀H₁₈O
mass of empirical unit = 12 x 10 + 1x 18 + 16 = 120 + 18 + 16 = 154
number of empirical units = molecular mass / mass of one empirical unit
= 154.25 / 154 = 1.00
Therefore molecular formula = C₁₀H₁₈O
Answer: density =3.377g/cm³
Explanation:
Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)
D= (M ×n) /(V×A)
M= 137g/mol
n= 2 (For BCC)
V=a³ , where a= 4r/√3
a= (4×222)/√3
a=512.69pm
a= 512.69×10^-10cm
V= ( 512.69×10^-10)^3
V= 1.3476×10^-22cm³
D= (137×2)/(1.3476×10^-22 × 6.02^23)
D= 3.377g/cm³
Therefore the density of barium is 3.377g/cm³
