Answer:
The correct answer is - 3.012×10^23 molecules
Explanation:
The number of molecules is present in any mole of an element or compound are the same. To find the molecules you need to multiply the number of the moles of the compound by Avogadro's number:
Moles are calculated from the molecular weight, for AlF3
= 26.98 + 3×19
= 26.98 + 57
= 83.98
Then moles in 42 grams:
= 42/83.98
= 0.5001190 moles
Now the number of molecules = moles*Avogadro's number
= 0.500110×6.022×10^23
= 3.011×10^23 molecules
Here is the link to the answer:
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:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 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>
Answer:
Electrons
Explanation:
Because the nucleas is In the middle and the Electrons surround it
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
