It’s b and could I have the Brianliest plzzzz
1. Make a Prediction
2. Fill both beakers with water
3. Dissolve salt in one of the beakers
4. Place both in the freezer and observe
5. Write a report
(Always make the prediction first! That's a hypothesis!)
Answer:
The correct answer is 0.300 * 10^23 ions.
Explanation:
Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.
The moles of zinc chloride (ZnCl2) is,
= mass of zinc + 2 mass of chlorine
= 65.38 + 2 (35.45)
=65.38 + 70.90
= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)
Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine
Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8
= 3.537 g of chlorine
70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537
= 0.300 * 10^23 ions of chlorine.
I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.
After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.
Answer:
<h2>
The equilibrium constant Kc for this reaction is 19.4760</h2>
Explanation:
The volume of vessel used=
ml
Initial moles of NO=
moles
Initial moles of H2=
moles
Concentration of NO at equilibrium=
M

Moles of NO at equilibrium= 
=
moles
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)
<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles
<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles
∴
⇒
![Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH2O%5D%5E2%5BN2%5D%7D%7B%5BH2%5D%5E2%5BNO%5D%5E2%7D%20%28volume%20of%20vesselin%20litre%29)
<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles
⇒
⇒