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3 years ago

How does bohrs model agree with Thomson model

2 answers:

8 0

Bohr's Model agree with Thompson's model because Bohr concluded that packets of energy are absorbed or emitted by the atoms when an electron changes shell. Thompson's conclusion states that atoms have small, negatively charged particles (referring to Electrons) as part of their internal structure.

schepotkina [342]3 years ago

7 0

Answers:

Bohr's model agrees with Thompson's model as both hold the fact that there are three sub=atomic particles which makes up an atom and there are equal number of protons and electron in a neutral atom

Explanation:

Bohr's model describes an atom as a planetary like shape in which the electrons revolve round the nucleus which is composed of protons and neutrons.

Thompson's model states an atom as a sphere which is composed of electrons surrounded by positively charged ions.

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A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form

liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.

6 0

3 years ago

Read 2 more answers

What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of e

Serjik [45]

Explanation:

According to Clausius-Claperyon equation,

$ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

The given data is as follows.

$T_{1} = 63.5^{o}C$ = (63.5 + 273) K

= 336.6 K

$T_{2} = 78^{o}C$ = (78 + 273) K

= 351 K

$P_{1}$ = 1 atm, $P_{2}$ = ?

Putting the given values into the above equation as follows.

$ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

$ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]$

$\Delta H$ = $\frac{0.559}{1.466 \times 10^{-4}} J/mol$

= $0.38131 \times 10^{4} J/mol$

= 3813.1 J/mol

Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.

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kolezko [41]

Answer:

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3 years ago

What element is in group 17 and period 2 of the periodic table?

Natali [406]

Answer:

Group 17 elements:

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What is the maximum kinetic energy of an ejected electron if silver metal is irradiated with 224 nm light? The

Genrish500 [490]

Answer:The maximum kinetic energy KEe of ejected electrons (photoelectrons) is given by KEe=hf−BE KE e = h f − BE , where hf is the photon energy and BE is the binding energy (or work function) of the electron to the particular material.

Explanation:

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