Hey there!
Al + HCl → H₂ + AlCl₃
Balance Cl.
1 on the left, 3 on the right. Add a coefficient of 3 in front of HCl.
Al + 3HCl → H₂ + AlCl₃
Balance H.
3 on the left, 2 on the right. We have to start by multiplying everything else by 2.
2Al + 3HCl → 2H₂ + 2AlCl₃
Now we have 2 on the right and 4 on the left. Change the coefficient in front of HCl from 3 to 4.
2Al + 4HCl → 2H₂ + 2AlCl₃
Now, for Cl, we have 4 on the left and 6 on the right. Change the coefficient in front of HCl again from 4 to 6.
2Al + 6HCl → 2H₂ + 2AlCl₃
Now, our H is unbalanced again. 6 on the left, 4 on the right. Change the coefficient in front of H₂ from 2 to 3.
2Al + 6HCl → 3H₂ + 2AlCl₃
Balance Al.
2 on the left, 2 on the right. Already balanced.
Here is our final balanced equation:
2Al + 6HCl → 3H₂ + 2AlCl₃
Hope this helps!
im pretty sute the answer would be number 4.
mark brainliest :)
Answer:
5.00 mol Mg
10.0 mol Cl
40.0 mol O
Explanation:
Step 1: Given data
Moles of Mg(ClO₄)₂: 5.00 mol
Step 2: Calculate the number of moles of Mg
The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.
5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg
Step 3: Calculate the number of moles of Cl
The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.
5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl
Step 4: Calculate the number of moles of O
The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.
5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O
<span>C) <u>Colloids</u></span><span>
Colloids have small non-dissolved particles that flow around in the mixture. These particles do not settle over time. When a light is shined on colloids the scattering characteristic of the Tyndall effect are visable.</span>
