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mart [117]
3 years ago
11

Which description of salt is a physical property? (2 points)

Chemistry
1 answer:
jarptica [38.1K]3 years ago
8 0

Product of mixing acids and bases describes salt is a physical property.

Product of mixing acids and bases

<u>Explanation:</u>

When an acid and a base are put together, they respond to kill the corrosive and base properties, creating a salt which portrays the physical property. The physical properties of table salt will be: Salt is a white cubic gem. At the point when the salt is unadulterated it clear.

It likewise shows up in white, dim or caramel shading relying on immaculateness. It is unscented yet has a solid salty taste. Fundamental salts contain the conjugate base of a feeble corrosive, so when they break down in the water, they respond with water to yield an answer with a pH more than 7.0.

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If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr
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The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be \frac{49}{40}=1.225 g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

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3 years ago
If 36.2g of Acetic Acid (HC2H302) was dissolved in 300. mL of water, what is the
uysha [10]

Answer:

2.01 M

Explanation:

Step 1: Calculate the moles of acetic acid (HC₂H₃O₂)

The molar mass of acetic acid is 60.05 g/mol. We will use this data to calculate the moles corresponding to 36.2 g of acetic acid.

36.2g \times \frac{1mol}{60.05g} = 0.603mol

Step 2: Convert the volume of solution to liters

We will use the relation 1000 mL = 1 L. We assume that the volume of solution is that of water (300 mL)

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Step 3: Calculate the molarity of the solution

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M = \frac{0.603mol}{0.300L} = 2.01 M

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