Answer:
A radical is an atom, ion, or molecule that has unpaired electrons.
A compound is a combination of two or more elements.
Explanation:
In the Bronsted-Lowry acid-base definition, an acid is a molecule (or ion) that donates a proton and a base is a molecule (or ion) that accepts a proton.
Bronsted-Lowry theory of acid and bases took the Arrhenius definition one step further, as a substance no longer needed to be composed of hydrogen (H+) or hydroxide (OH-) ions in order to be classified as an acid or base. For example , consider the following chemical equation:
HCl (aq) +NH₃ (aq) → NH⁺₄ (aq) + Cl⁻(aq)
Here, hydrochloric acid (HCl) "donates" a proton (H+) to ammonia (NH3) which "accepts" it , forming a positively charged ammonium ion (NH4⁺) and a negatively charged chloride ion (Cl⁻). Therefore, HCl is a Bronsted-Lowry acid (donates a proton) while the ammonia is a Bronsted-Lowry base (accepts a proton). Also, Cl⁻ is called the conjugate base of the acid HCl and NH4⁺ is called the conjugate acid of the base NH3.
A Bronsted-Lowry acid is a proton (hydrogen ion) donor.
A Bronsted-Lowry base is a proton (hydrogen ion) acceptor.
In this theory, an acid is a substance that can release a proton (like in the Arrhenius theory) and a base is a substance that can accept a proton.
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Given what we know, we can confirm that an example is a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided.
<h3>What are examples of the situations given?</h3>
- A number that is a multiple of 10 is 40, since 10 times 4 equals 40.
- In order to get a product of 10, we can multiply two and five.
- To result in a quotient of 10, we can divide one hundred by ten.
<h3>What are non-examples of the situations given?</h3>
- one non-example of a multiple of 10 would be to multiply three and seven.
- A non-example of a product of 10 is to multiply the number fifty by twenty-five.
- To result in a non-example for a quotient of 10, we can divide the number fifteen by three.
Therefore, given the definition of an example as a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided, we can confirm that the ones listed above are correct.
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<h3>
Answer:</h3>
62.5 mL
<h3>
Explanation:</h3>
<u>We are given;</u>
- Molarity of the stock solution as 0.100 M
- Volume of the dilute solution as 250 mL
- Molarity of dilute solution as 0.0250 M
We are required to calculate the Volume of the stalk solution.
- Taking the volume and molarity of the stock solution to be V₁ and M₁ respectively, and volume and molarity of the dilute solution to be V₂ and M₂ respectively.
We are going to use the dilution formula;
- According to the dilution formula, M₁V₁ = M₂V₂
- Rearranging the formula;
V₁ = M₂V₂ ÷ M₁
= (0.025 M × 0.25 L) ÷ 0.100 M
= 0.0625 L
But, 1 L = 1000 mL
V₁ = 62.5 mL
Therefore, the volume of the stock solution is 62.5 mL
