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3 years ago

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3. Which part of Thomson's plum pudding model was different from Rutherford's model?

1 answer:

horsena [70]3 years ago

8 0

Answer:

The difference between the Thompson's plum pudding model and the Rutherford atom model is the location of the electrons (option a).

Explanation:

While Thompson compared his atom to a plum pudding, where the electrons floated freely in the pudding, Rutherford established the arrangement of the electrons in orbitals, which were found around the atomic nucleus like the planets around the sun.

Rutherford's findings also established the existence of a small, positively charged nucleus.

<em>Thompson and Rutherford models did not differentiate between the charges of electrons and protons , overall charges or overall size of the atom.</em>

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commercial cold packs consist of solid NH4NO3 and water. In a coffee-cup calorimeter, 5.60g NH4NO3 is dissolved in 100g of water

fgiga [73]

Answer:

-1.37 kJ/mol

Explanation:

The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$ is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 5.60 g

Specific heat = 4.18 J/g°C

$\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C$

So,

$\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J$

Negative sign signifies loss of heat.

Also, 1 J = 0.001 kJ

So,

$\Delta H=-0.096\ kJ$

Also,

Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{5.60\ g}{80.043 \ g/mol}$

$Moles= 0.06996\ mol$

Thus, $\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol$

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4 years ago

Why did engineers need to design a sunshade for Mercury Messenger?

zloy xaker [14]

The answer is so it didn't burn.

8 0

3 years ago

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You are given a solution containing a pair of enantiomers (a and b). careful measurements show that the solution contains 77% a

san4es73 [151]

The ee solution would be 55%

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4 years ago

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Consider the reaction 4FeS2 + 11O2 → 2Fe2O3 + 8SO2. If 8 moles of FeS2 react with 15 moles of O2, what is the limiting reactant?

omeli [17]

Answer:

O2

Explanation:

for find the limiting reactant you must calculate the moles of the reactants from the amount that you have and from the MM:

MM FeS2 = 120n = 26.2g / 120g/mol = 0,218 mol

MM O2 = 32n = 5,44g/32g/mol = 0,17 mol

The limiting reactant is

O2

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3 years ago

Which half-reaction correctly represents reduction? *

nasty-shy [4]

Answer:

C

Explanation:

Reduction can be seen through addition (gaining) of electrons, addition of Hydrogen or removal of Oxygen

Mn7+ is reduced to Mn4+ by the addition of 3 electrons.

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4 years ago