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3 years ago

How did the Bohr model of the atom affect scientific thought?

Chemistry

2 answers:

Neporo4naja [7]3 years ago

6 0

Answer:

Its a.

Explanation:

Vadim26 [7]3 years ago

5 0

Answer:

C. Scientists accepted the model at first but later rejected it.

Explanation:

Scientists accepted the model at first because it explained the hydrogen emission spectrum.

However, with the development of quantum mechanics, scientists had to modify the model (not reject it).

Electrons still had specific energies, but they no longer travelled in fixed orbits.

Instead, electrons had a probability of being found in a given region of space.

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Given 7.20 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

julsineya [31]

The reaction between butanoic acid and ethanol is:

CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH3CH2 + H2O

Based on the reaction stoichiometry:

1 mole of butanoic acid forms 1 mole of ethyl butyrate

Now,

Molar mass of Butanoic acid = 88.0 g/mol

Given mass of butanoic acid = 7.20 g

Therefore, # moles of butanoic acid reacted = 7.20/88.0 = 0.0818 moles

# moles of ethyl butyrate formed = 0.0818 moles

Molar mass of ethyl butyrate = 116 g/mol

Mass of ethyl butyrate synthesized = 0.0818 * 116 = 9.49 g

8 0

3 years ago

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umka2103 [35]

Chicken has the most calories

7 0

3 years ago

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This infectious disease requires a special mask for protection

11111nata11111 [884]

Answer:

the black plague

Explanation:

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3 years ago

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A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

3 years ago

Calculate the standard entropy of vaporization of ethanol, C2H5OH, at 285.0 K, given that the molar heat capacity at constant pr

MArishka [77]

Answer: The standard entropy of vaporization of ethanol is 0.275 J/K

Explanation:

$C_2H_5OH(l)\rightleftharpoons C_2H_5OH(g)$

Using Gibbs Helmholtz equation:

$\Delta G=\Delta H-T\Delta S$

For a phase change, the reaction remains in equilibrium, thus $\Delta G=0$

$\Delta H=T\Delta S$

Given: Temperature = 285.0 K

$\Delta H=78.3J/mol$

Putting the values in the equation:

$78.3J=285.0K\times \Delta S$

$\Delta S=0.275J/K$

Thus the standard entropy of vaporization of ethanol is 0.275 J/K

4 0

3 years ago