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3 years ago

Having an accurate model of an atom is important except

Chemistry

1 answer:

WITCHER [35]3 years ago

6 0

It can help support the explanation of the origin of the universe

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An acid is used to bring the pH of a solution from pH 9 to pH 6. How many times more acidic is the final solution than the initi

Molodets [167]

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5 0

2 years ago

Bohr's model was correct in assigning energy levels to electrons.

Jobisdone [24]

Answer:

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7 0

3 years ago

How many of grams of CuSO4 are in 475ml of a 2.0M aqueous solution

Anettt [7]

Answer:

151.63 g

Explanation:

We first get the number of moles;

Moles = Molarity × volume

= 2.0 × 0.475

= 0.95 moles

1 mole of CuSO4 = 159.609 g/mol

Therefore;

Mass = moles × molar mas

= 0.95 moles × 159.609 g/mol

= 151.63 g

8 0

4 years ago

Read 2 more answers

Rank the following compounds in order of decreasing boiling point: potassium fluoride (KF), acetylene (C2H2), and formaldehyde (

otez555 [7]

Answer: $KF$ > $CH_2O$ > $C_2H_2$

Explanation:

The order of boiling point depends upon the type of interactions present between the molecules.

Potassium fluoride (KF) is an ionic compound and the opposite ions are held together by strong electrostatic forces.

$C_2H_2$ is a covalent compound and the molecules are held together by weak van der Waals' forces.

Formaldehyde is a polar compound due to presence of polar carbonyl group. Hence dipole-dipole force is present between formaldehyde molecules.

Thus the decreasing order of boiling point is:

$KF$ > $CH_2O$ > $C_2H_2$

8 0

3 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0

3 years ago