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1 year ago

Item 4

Chemistry

1 answer:

Furkat [3]1 year ago

6 0

The statement that best describes a mixture is -

two or more substances that are mixed but can be separated back to their original form, maintaining their original properties.

Explanation:

A mixture is that form of matter in which two or more substances are simply mixed in any proportion and it can be separated back to their original form.

There are two types of Mixtures -

Homogeneous Mixtures
Heterogeneous Mixtures

In Homogeneous Mixtures, the composition is uniform throughout such as sugar in water , salt in water , sulphur in carbon disulphide , water and alcohol , whereas Heterogeneous Mixtures has non Uniform composition such as sand and salt , sugar and salt , wood , blood , water in soil etc.

Pure substance is a Homogeneous material which consists of a single type of particles atoms or molecules with definite set of properties.

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What is the mass of an atom that has 4 protons, 5 neutrons, and 4 electrons?

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The answer would be 9

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He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

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Which of the following affects cell potential?

Slav-nsk [51]

Answer:

the standard cell potential value

Explanation:

For every cell, we can calculate its standard electrode potential from the table of standard electrode potentials listed in many textbooks.

However, from Nernst's equation;

Ecell= E°cell - 0.0592/n log Q

Hence the standard cell potential (E°cell) affects the value of the calculated cell potential Ecell from Nernst's equation as stated above.

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2 years ago

The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 x 10-4 s -1 at

denpristay [2]

Answer:

$C_f=0.1755M$

Explanation:

Hello,

Based on the information and the units of the given data, the integrated rate law turns out into:

$\frac{dC}{dt}=-kC\\ C_f=C_0exp(-kt)\\C_f=0.25M*exp(-6.7x10^{-4}s^{-1}*8.8min*\frac{60s}{1min} )\\C_f=0.1755M$

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