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3 years ago

According to atomic theory:

Chemistry

1 answer:

JulsSmile [24]3 years ago

7 0

Answer:

e. all of these

Explanation:

Let us check all the given options one by one:

a.The nucleus is positively charged.

Yes it is correct since nucleus contain protons and neutrons and protons are positively charged.

b.The nucleus contains both charged and uncharged particles.

Yes because protons are positively charged and neutrons are neutral in nature.

c.The electrons contribute very little to the total mass of the atom.

Yes we know all the mass of the atom is considered in center and mass of electron is negligible as compared to protons and neutrons.

d.The electrons are located in the atomic space outside the nucleus.

Yes, its a known fact.

e. All of these.

Since , all given options are correct .

Therefore , option e. is right .

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The initial reaction rate for the elementary reaction 2A + B → 4C was measured as a function of temperature when the concentrati

Artist 52 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776 $\frac{kJ}{mole}$

b) The frequency factor is 1.77 × $10^{18}$

c) The rate constant is 0.00033 $(\frac{dm^{3} }{mole} )^{2}$ $\frac{1}{s}$

Explanation:

From the question the elementary reaction for A and B is given as

2A + B → 4C

The rate equation the elementary reaction is

- $r_{A}$ = $k$ $[A]^{2}$ $[B]$

= $k[2]^{2}[1.5]$

= 6k

$k = \frac{-r_{A} }{6}$

When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

$k = Aexp(-\frac{E_{a} }{RT} )$

taking ln of both sides we have

$lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}$

Considering the graph for the rate constant $ln k$ and $(\frac{1}{T} )$ the slope from the equation is $-(\frac{E_{a} }{R})$ and the intercept is $ln A$

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of $ln k$ vs $(\frac{1}{T} )$ is shown on the fourth uploaded image

From the graph we can see that the slope is $-(\frac{E_{a} }{R} ) = - 15008$

Now we can obtain the activation energy $E_{a}$ by making it the subject in the equation also generally R which is the gas constant is $8.145 \frac{J}{kmole}$