<span>On the Periodic Table metals are found on the left of the table and will be positive, while non-metals are found on the right of the table and will be negative. </span>
Answer:
Carbon : chemical symbol is "C"
Chlorine: chemical symbol is"Cl"
Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
Answer:
Silver Acetate would be the Limiting Reagent.
Explanation:
The balance chemical equation for the given double displacement reaction is as;
HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂
Step 1: <u>Calculate Moles of Starting Materials:</u>
Moles of HCl:
Moles = Mass / M.Mass
Moles = 72.9 g / 36.46
Moles = 1.99 moles
Moles of AgC₂H₃O₂:
Moles = 150 g / 166.91 g/mol
Moles = 0.898 moles
Step 2: <u>Find out Limiting reagent as:</u>
According to balance chemical equation.
1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂
So,
1.99 moles of HCl will react with = X moles of AgC₂H₃O₂
Solving for X,
X = 1.99 mol × 1 mol / 1 mol
X = 1.99 mol of AgC₂H₃O₂
Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.
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