<span>you have to find the differences of electronegativity for the atoms joined by the bond. if their diff is <0.4,
then it's non-polar
if it's between 0.4 to 1.7
then it's polar
1.7 and up it's ionic
Electronegativity values can be found on most periodic tables.</span>
Higher probability of loss. Chorionic villus sampling (CVS) and Amniocentesis (AC). The prenatal diagnosis technique can be done earlier in fetal development CVS (first trimester --> 10-13 weeks). AC (second trimester --> 16-20 weeks)
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Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
The answer is bohr hope this helps :)
The number of moles in each sample will be 0.391 moles, 30.7 moles, 0.456 moles, and 1350 moles
<h3>What is the number of moles?</h3>
The number of moles of a substance is the ratio of the mass of the substance to the molar mass.
In other words; mole = mass/molar mass.
Thus:
- moles of 18.0 g
= 18.0/46
= 0.391 moles
- moles of 1.35 kg
= 1350/44
= 30.7 moles
- moles of 46.1 g
= 46.1/101.1
= 0.456 moles
- moles of 191.8 kg
= 191800/142
= 1350 moles
More on the number of moles of substances can be found here: brainly.com/question/1445383
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