Answer:
The mass of the precipitate that AgCl is 3.5803 g.
Explanation:
a) To calculate the molarity of solution, we use the equation:

We are given:
Mass of solute (NaCl) = 1.46 g
Molar mass of sulfuric acid = 58.5 g/mol
Volume of solution = 

Putting values in above equation, we get:

0.09982 M is the concentration of the sodium chloride solution.
b) 
Moles of NaCl = 
according to reaction 1 mol of NaCl gives 1 mol of AgCl.
Then 0.02495 moles of NaCl will give:
of AgCl
Mass of 0.02495 moles of AgCl:

The mass of the precipitate that AgCl is 3.5803 g.
Explanation:
1st question answer true
2nd question low resistance
hope it helps
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!
Explanation:
the investigation lasts for 7 days !
hope this helps you.
Answer:
0.01932 L
Explanation:
First we <u>convert 105 mM to M</u>:
Next we <u>convert 552 mL to L</u>:
Then we use the following equation:
Where:
We<u> input the given data</u>:
- 3 M * V₁ = 0.105 M * 0.552 L
And <u>solve for V₁</u>: