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kap26 [50]
3 years ago
13

Which atomic model was proposed as a result of J. J. Thomson’s work?

Chemistry
2 answers:
lesya692 [45]3 years ago
4 0
The answer would be A) The plum pudding model.
victus00 [196]3 years ago
4 0

Answer: Option (a) is the correct answer.

Explanation:

J.J. Thomson is a renowned scientist who discovered the model of atom which is known as the plum pudding model.

Through this model Thomson stated that an atom consists of small particles. These small particles are positive and negative charges present inside an atom.

Whereas Democritus stated that an atom is indivisible and thus he proposed the indivisible atom model.

Ernest Rutherford discovered that an atom has lot of empty space and this space is nucleus. As he stated that positively charged particles are known as protons which are present inside a nucleus.

Rutherford's gold foil experiment discovered that an atom has nucleus in which most of the mass of an atom is concentrated.

Thus, we can conclude that the plum pudding model was proposed as a result of J. J. Thomson’s work.

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grandymaker [24]
There are 2 significant figures.
"Rule 3: Trailing zeros are significant if a decimal point is shown in the number, but may or may not be significant if no decimal point is shown."
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3 years ago
Can anyone think of some interesting facts about hydrogen? I know that there are plenty, I'm just drawing a blank right now. I'l
podryga [215]
hydrogen is the lightest element in the periodic table
7 0
3 years ago
Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical equation: ()zn
kompoz [17]

Based on the chemical equation, to balance the given equation we should use 2 as a coefficients of HCl on reacatnt side of the reaction.

<h3>What is balanced equation?</h3>

Balanced chemical equations are those equation in which each entities are present in same amount on reactant side as well as on the product side of the chemical reaction.

Given chemical reaction is:

Zn + HCl → ZnCl₂ + H₂

In the above reaction equation  is not balanced as number of chlorine and hydrogen atoms are not equal on both sides, so balanced equation will be:

Zn + 2HCl → ZnCl₂ + H₂

Hence we add 2 as a coefficient of HCl to balance the equation.

To know more about balance equation, visit the below link:

brainly.com/question/15355912

8 0
2 years ago
If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
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m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
2 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
2 years ago
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