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3 years ago

The molar mass of aniline is 93 g/mol. what is its molecular formula

Chemistry

1 answer:

Mashutka [201]3 years ago

4 0

Answer:

Molecular formula of aniline is C6H5NH2.

Explanation:

As we know, molecular mass can be calculated as

Molar mass = C6H5NH2

Molar mass = (6*12)+(1*7)+(28)

Molar mass = 93 g/mol

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Archy [21]

Answer:

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Explanation:

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3 years ago

The strong forces of attraction between the positive and negative regions of molecules are called

Marat540 [252]

They are called Dipole-dipole forces

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3 years ago

Compounds A and B are constitutional isomers (same molecular formula but

Naily [24]

Answer:

Trimethylacetaldehyde

Explanation:

For the unknow compound we have a molar mass of 86 g/mol. We have an even value for the mass, so the compound does not have nitrogen and we can have several posibilities:

A) $C_3H_2O_3$

B) $C_4H_6O_2$

C) $C_5H_1_0O$

D) $C_6H_1_4$

If we check the IR info a signal in 1730 cm-1 appears, this indicates that we have an oxo group (C=O). So, the D option can be discarded. The groups that can have the oxo group are: Carboxylic acids, Ketones and aldehydes.

We don have a signal in 3000 cm-1, so the carboxylic acid can be discarded. Now, is we check the info for the 1H NMR we only have 2 signals. If we only have 2 we will have a very symmetric compound.

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3 years ago

The average blood alcohol concentration (BAC) of eight male subjects was measured after consumption of 15 mL of ethanol (corresp

Jlenok [28]

Answer:

(a) 0.00587 mg/min

(b) - 0.00222 mg/min

Explanation:

The function that modeled the data is:

$C(t) = 0.0225te^{-0.0467t}$

Since you need to determine how rapidly the concentrations will vary (increase or decrease) at different times, this is the instantaneous speed or rate of change, you need to find the first derivative of C(t), C'(t)

The next steps show the use of the rule of chain to find C'(t):

$C(t) = 0.0225te^{-0.0467t}\\ \\ C'(t)=0.0225[e^{-0.0467t}-0.0467te^{-0.0467t}]\\ \\ C'(t)=0.0225e^{-0.0467t}-0.00105075e^{-0.00467t}t$

Now that you have the expression for the speed at which the concentration changes, you just need to substitute the values.

(a) How rapidly was the BAC increasing after 9 minutes? (mg/mL)/min

$C'(9)=0.0225e^{-0.0467(9)}-0.00105075(9)(e^{-0.0467(9)})}\\ \\ C'(t)=0.00857mg/min$

(b) How rapidly was it decreasing half an hour later (t = 30)? (mg/mL)/min

$C'(30)=0.0225e^{-0.0467(30)}-0.00105075(30)(e^{-0.0467(30)})}\\ \\ C'(t)=-0.00222mg/min$

6 0

3 years ago

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Gennadij [26K]

Answer:

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Explanation:

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3 years ago