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3 years ago

Which of the following would not be defined as an acid A.KOH, B. HNO3, D. H3PO4 and why

Chemistry

1 answer:

tigry1 [53]3 years ago

3 0

B. HNO3 because it is copper and can not move the place of hydrogen's atom from the acid

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What happens to weather patterns when a cold front approaches? Answer in complete sentences. Its timed!! YOU WILL GET BRAINLIEST

Kisachek [45]

Answer:

When a cold front approaches there can be a sudden drop in temperature, strong wind gusts, and heavy rain and thunderstorms.

Explanation:

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3 years ago

Which of the following people would most often use chemistry in their work?

aleksklad [387]

Answer:

I would suggest the answer D: A paint manufacturer

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Crank

Answer:

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3 years ago

Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol

DENIUS [597]

Answer:

methyl tert-butyl ether is produced as a product

Explanation:

Nucleophilic substitution in solvolysis reaction of tert-butyl bromide goes through $S_{N}1$ mechanism.
Because Br is a good leaving group due to it's large size and high polarizability and methanol is an weak nucleophile.
In the first step, tert-butyl cation is produced. In the second step, methanol attacks the carbocation and produce methyl tert-butyl ether after deprotonation.
Reaction and structure of product has been shown below.

4 0

3 years ago

Calculate the freezing point and boiling point of a solution containing 14.2 gg of naphthalene (C10H8)(C10H8) in 111.0 mLmL of b

krek1111 [17]

Answer:

The freezing point of solution = -0.34 °C

The boiling point of solution = 82.98 °C

Explanation:

Step 1: Data given

Mass of naphthalene = 14.2 grams

Molar mass naphthalne = 128.17 g/mol

Volume of benzene = 111.0 mL

Density of benzene = 0.877 g/mL

Kf(benzene)=5.12°C/m

Freezing point benzene = 5.5 °C

Kb(benzene)=2.53°C/m

Boiling point benzene = 80.1 °C

Step 2: Calculate mass of benzene

Mass benzene = density * volume

Mass benzene = 0.877 g/mL * 111.0 mL

Mass benzene = 97.3 grams

Step 3: Calculate moles naphthalene

Moles naphthalene = mass naphthalene / molar mass napthalene

Moles napthalene = 14.2 grams / 128.17 g/mol

Moles naphthalene = 0.111 moles

Step 4: Calculate molality

Molality = moles naphthalene / mass benzene

Molality = 0.111 moles / 0.0973 kg

Molality = 1.14 molal

Step 5: Calculate freezing point of a solution

ΔT = i*kf*m

ΔT = 1 * 5.12 °C/m * 1.14 m

ΔT = 5.84 °C

ΔT = T(pure solvent) − T(solution)

The freezing point of solution = T pure - ΔT

5.5 - 5.84 = -0.34 °C

The freezing point is -0.34 °C

Step 6: Calculate boiling point of a solution

ΔT = i*kb*m

ΔT = 1 * 2.53 °C/m * 1.14 m

ΔT = 2.88 °C

ΔT = Tb (solution) - Tb (pure solvent)

The boiling point of solution = T pure + ΔT

The boiling point of solution = 80.1 °C + 2.88

The boiling point of solution = 82.98 °C

6 0

3 years ago