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2 years ago

You may normally expect a _ reaction if a _ activation energy is required.

Chemistry

2 answers:

cestrela7 [59]2 years ago

6 0

Answer:

You may normally expect a slow reaction if a high activation energy is required.

Or, You may normally expect a fast reaction if a low activation energy is required.

Explanation:

Strike441 [17]2 years ago

4 0

Answer:

You may normally expect a slow reaction if a high activation energy is required.
Or, You may normally expect a fast reaction if a low activation energy is required.

Explanation:

To answer this question, we should define firstly the term of activation energy.

The activation energy is the minimum energy must the reacting molecules have to initiate the chemical reaction.

Also, we can define it as The energy barrier that the molecules must posses an amount of energy equal or more than the value of this barrier.

So, as the activation energy be of low value, the possibility that the reacting molecules can have this amount of energy to initiate and proceed the reaction increases and the reaction rate also increases.

Additionally, as the activation energy be of high value, the possibility that the reacting molecules can have this amount of energy to initiate and proceed the reaction decreases and the reaction rate also decreases.

Thus, we can answer this question as:

You may normally expect a slow reaction if a high activation energy is required.
You may normally expect a fast reaction if a low activation energy is required.

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What is the ratio of the diffusion rates of cl2 and o2? rate cl2 : o2 = 0.47 0.67 0.45 0.69 1.5?

matrenka [14]

The ratio of the diffusion rate of Cl2 and O2 is 1.5

calculation
rate of diffusion Cl2/ rate of diffusion O2 =√ molar mass of Cl2/ molar mass of O2

molar mass of Cl2 = 35.5 x2 = 71 g/mol
molar mass O2 = 16 x2 =32g/mol

that is rate of diffusion Cl2/ rate of diffusion of O2 =√ 71/ 32= 1.5

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2 years ago

What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

Veseljchak [2.6K]

Answer:

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

$CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol$

Mass of methane burnt = 530 g

Moles of methane burnt = $\frac{530 g}{16 g/mol}=33.125 mol$

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

$-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ$

$-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ$

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

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2 years ago

Avogadro’s number is a constant that helps scientists count atoms and molecules. It is approximately equal to 602,200,000,000,00

kondaur [170]

Avogrado's number = 6.022 × 10^23

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2 years ago

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Which letter indicates a subatomic particle with a mass of 5.489 x 10-4 amu

Ludmilka [50]

Answer:

electron

Explanation:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol = e⁻

Mass = 9.10938356×10⁻³¹ Kg

Mass in amu = 1/1838 = 5.4 × 10⁻⁴amu

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.

Symbol of proton= P⁺

Symbol of neutron= n⁰

Mass of proton=1.672623×10⁻²⁷ Kg

Mass of neutron=1.674929×10⁻²⁷Kg

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom

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2 years ago

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If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/

riadik2000 [5.3K]

Answer: The energy of the complex is $2.72\times 10^2kJ$

Explanation:

To calculate the energy of the complex, we use the equation given by Planck which is:

$\Delta E=\frac{N_Ahc}{\lambda}$

where,

$\lambda$ = Wavelength of the complex = $440nm=4.40\times 10^{-7}m$ (Conversion factor: $1m=10^9nm$ )

h = Planck's constant = $6.624\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

$\Delta E$ = energy of the complex

Putting values in above equation, we get:

$\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ$

Conversion factor used: 1 kJ = 1000 J

Hence, the energy of the complex is $2.72\times 10^2kJ$

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2 years ago

You may normally expect a _____ reaction if a _____ activation energy is required.

You may normally expect a _ reaction if a _ activation energy is required.