There are four substances, A, B C and D. A and C are pure substances. They cannot be broken down into anything simpler by chemic
al methods. B consists of A and D, which are not chemically bonded together. D is a pure substance. We can decompose D into A and C using chemical methods. How would we classify substance A, C, and D?
A: Substances A and C are elements, because they are pure substances that bond together to make substance D and cannot be broken using chemical processes . Substance D is a mixture, because it is a pure substance made up of A and C.
B: Substances A and C are compounds, because they are pure substances that bond together to make substance D and cannot be broken using chemical processes . Substance D is a mixture, because it is a pure substance made up of A and C.
C: Substances A and C are elements, because they are pure substances that bond together to make substance D and cannot be broken using chemical processes. Substance D is a compound, because it is a pure substance made up of A and C.
D: Substances A, C, and D, are elements. All pure substances are elements.
A: Substances A and C are elements, because they are pure substances that bond together to make substance D and cannot be broken using chemical processes . Substance D is a mixture, because it is a pure substance made up of A and C.
B: Substances A and C are compounds, because they are pure substances that bond together to make substance D and cannot be broken using chemical processes . Substance D is a mixture, because it is a pure substance made up of A and C.
C: Substances A and C are elements, because they are pure substances that bond together to make substance D and cannot be broken using chemical processes. Substance D is a compound, because it is a pure substance made up of A and C.
D: Substances A, C, and D, are elements. All pure substances are elements.
1 answer:
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Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.