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2 years ago

The boiling point of methanol is 64 7°C. Its melting point is -976°C. At room temperature (-25°C), methanol is in which state?

Chemistry

2 answers:

dsp732 years ago

5 0

Answer:

4.Liquid

Explanation:

marissa [1.9K]2 years ago

3 0

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.

The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.

At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults

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Vaselesa [24]

Answer:

The molar mass of the unknown acid is 386.8 g/mol

Explanation:

Step 1: Data given

Mass of the weak acid = 1.168 grams

volume of NaOH = 28.75 mL = 0.02875 L

Molarity of NaOH = 0.105 M

Since we only know 1 equivalence point, we suppose the acid is monoprotic

Step 2: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.105 M * 0.02875 L

Moles NaOH = 0.00302 moles

We need 0.00302 moles of weak acid to neutralize the NaOH

Step 3: Calculate molar mass of weak acid

Molar mass = mass / moles

Molar mass = 1.168 grams / 0.00302 moles

Molar mass = 386.8 g/mol

The molar mass of the unknown acid is 386.8 g/mol

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3 years ago

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

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To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

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Explanation:

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ValentinkaMS [17]

Answer:

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