Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?



volume of water added = final volume -initial volume
= 120-10
=110ml
Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M
Explanation :
Using neutralization law,

where,
= molar concentration of undiluted cognac = ?
= molar concentration of diluted cognac = 0.0844 M
= volume of undiluted cognac = 5.00 mL = 0.005 L
= volume of diluted cognac = 0.500 L
Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.


Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M
Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.

Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.

Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.

Let's write the reaction first.
HCl + H₂O ---> H₃O⁺ + Cl⁻
These reaction has two reactants, either the proton donor or the proton acceptor. Water is amphoteric, meaning it can act as an acid or base. Since HCl is an acid, then water in this reaction acts as a base.
1. The proton donor is HCl because it donates H+ to water which yields a hydronium ion, H₃O⁺.
2. The proton acceptor is water.
Step 1: Change density from g/mL to g/L;
0.807 g/mL = 807 g/L
Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume
Putting Values,
Mass = 807 g/L × 1 L
Mass = 807 g
Also,
Moles = Mass / M.mass
Putting values,
Moles = 807 g / 28 g.mol⁻¹
Moles = 28.82 moles
Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,
As,
P V = n R T
V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)
V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm
V = 704.76 L