Clean? I’m pretty sure not sure what it means but.
Answer:
maybe, but id rather do automotive stuff, thats my second option.
Explanation:
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
Answer:
Carbon, germanium, tin and lead.
Explanation:
The silicon is belong to the carbon family. There are five elements in carbon family carbon, silicon, germanium, tin and lead. These five elements are present in same group i.e group fourteen. The elements present in same group have same number of valance electrons.
For example.
Carbon electronic configuration:
C₆ = [He] 2s² 2p²
Silicon electronic configuration:
Si₁₄ = [Ne] 3s² 3p²
Germanium electronic configuration:
Ge₃₂ = [Ar] 3d¹⁰ 4s² 4p²
Tin electronic configuration:
Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²
Lead electronic configuration:
Pb₈₂ = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
we can see that in case of all elements there are four valance electrons, which are equal to the valance electrons of silicon.
Answer:
Number of molecules = 1.8267×10^20
Explanation:
From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;
PV = nRT
where
P =pressure
V =volume
n = the number of moles
R is the gas constant equal to 0.0821 L·atm/mol·K
T is the absolute temperature
Given:
P = 6.75 atm;
T = 290.0 k,
; V = 1.07 cm³ = 0.001 L
( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)
n = 3.0335167*10^-4 moles
But there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 1.8267×10^20
