The answer is "<span>An atomic nucleus is positively charged because it is composed of protons". An atomic nucleus actually contains nucleons which are made up of both protons and neutrons. Since neutrons are neutral or have no charge, the charge of an atomic nucleus mainly relies on the positive charge of the protons.</span>
Answer:
4.07 × 10⁹ mL
Explanation:
Step 1: Given data
Length of the pool (L): 60.0 m
Width of the pool (W): 35.0 m
Depth of water in the pool (D): 6.35 ft
Step 2: Convert "D" to m
We will use the relationship 1 m = 3.28 ft

Step 3: Calculate the volume of water (V)
We will use the following expression.

Step 4: Convert "V" to mL
We will use the relationship 1 m³ = 10⁶ mL.

Base on the question and in my further research and investigation about the said problem that states to identify the type of bonding in solid potassium, I would say that the bondings are metallic bonding and metallic. I hope you are satisfied with my answer and feel free to ask for more
Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ <em>pka = 4,74</em>
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
Solving:
1,82 =
<em>(1)</em>
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] <em>(2)</em>
Replacing (2) in (1)
<em>[CH₃COOH] = 3,55 mM</em>
And
<em>[CH₃COO⁻] = 6,45 mM</em>
Knowing that:
<em>CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O</em>
Having in the first 10mmol/L of CH₃COOH, you need to add <em>6,45 mmol/L of NaOH. </em>to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃<em>COOH </em>.
I hope it helps!