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Ghella [55]
3 years ago
14

The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density

of acetic acid at 30.0 °C?
Chemistry
1 answer:
mash [69]3 years ago
7 0

Answer:

The density of acetic acid at 30°C = 1.0354_g/mL

Explanation:

specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)

Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)

= 0.9956 g/mL × 1.040

= 1.0354_g/mL

Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.

Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure

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Who was the first person to show strong empirical evidence for the existence of atoms?.
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Given what we know, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

<h3>Who was John Dalton?</h3>
  • He was a renowned scientist with knowledge in many fields.
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  • He proposed the atomic theory and carried out experiments to provide supporting evidence.

Therefore, given his proposal of the atomic theory and the experiments he carried out to provide evidence to support his claims, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

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How much heat (in Joules) is needed to raise the temperature of 257g of ethanol (cethanol=2.4 J/g°C) by 49.1°C?
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Answer:

Q = 30284.88 j

Explanation:

Given data:

Mass of ethanol = 257 g

Cp = 2.4 j/g.°C

Chnage in temperature = ΔT = 49.1°C

Heat required = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = 257 g× 2.4 j/g.°C × 49.1 °C

Q = 30284.88 j

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