Answer:
4552 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 55 mL
Molarity of stock solution (M₁) = 12 M
Molarity of diluted solution (M₂) = 0.145 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
12 × 55 = 0.145 × V₂
660 = 0.145 × V₂
Divide both side by 0.145
V₂ = 660 / 0.145
V₂ ≈ 4552 mL
Thus, the volume of the diluted solution is 4552 mL
Answer:
.0556 L
Explanation:
First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.
Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).
Finally, the answer will be .0556 L.
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Answer:
For any given element, ionization energy increases as subsequent electrons are removed. For example, the energy required to remove an electron from neutral chlorine is 1251 kJ/mol. ... An even sharper increase in ionization energy is witnessed when inner-shell, or core, electrons are removed.
Hope it helps :)
Answer:
Strong acid
Explanation:
An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
Hydroxonium ions are formed as a result of the chemical bonding between the oxygen of water molecules and the protons released by the acid due to its ionisation. This makes aqueous solution of acids conduct electricity.
A strong acid is one that ionizes almost completely. Examples are:
1. Hydrochloric acid
2. Tetraoxosulphate (VI) acid
3. Trioxonitrate (V) acid
4. Hydroiodic acid
5. Hydrobromic acid
Answer:
The final mass of sample is 1.3 g.
Explanation:
Given data:
Half life of H-3 = 12.32 years
Amount left for 15.0 years = 3.02 g
Final amount = ?
Solution:
First all we will calculate the decay constant.
t₁/₂ = ln² /k
t₁/₂ =12.32 years
12.32 y = ln² /k
k = ln²/12.32 y
k = 0.05626 y⁻¹
Now we will find the original amount:
ln (A°/A) = Kt
ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y
ln (3.02 g/ A) = 0.8439
3.02 g/ A = e⁰°⁸⁴³⁹
3.02 g/ A = 2.33
A = 3.02 g/ 2.33
A = 1.3 g
The final mass of sample is 1.3 g.
