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2 years ago

Identify the false statement from the following.

1 answer:

7 0

The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.

<h3>What are the factors that affect London dispersion forces?</h3>

Generally, the factors which affects the London dispersion forces a dispersion force are as follows:

Shape of the molecules
Distance between molecules
Polarizability of the molecules

However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.

So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement

Learn more about London dispersion forces:

brainly.com/question/1454795

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Which two processes lead directly to the formation of both breccia and conglomerate

Luden [163]

Compaction and cementation

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3 years ago

In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

7 0

3 years ago

How many significant figures in 8400

IrinaK [193]

Answer:

there are two significant figures is the number 8400

Explanation:

5 0

2 years ago

Read 2 more answers

If the reduction reaction has a reduction potential of 0.1 V, and the oxidation reaction has a reduction potential of -0.4V, and

aleksley [76]

Answer : The value of ΔG expressed in terms of F is, -1 F

Explanation :

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

or,

$E^o=E^o_{reduction}-E^o_{oxidation}$

$E^o=(0.1V)-(-0.4V)=+0.5V$

Now we have to calculate the standard cell potential.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant

$E^o$ = standard e.m.f of cell = +0.5 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times F\times 0.5)$

$\Delta G^o=-1F$

Therefore, the value of ΔG expressed in terms of F is, -1 F

5 0

3 years ago

If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K and then I raise the pressu

mixas84 [53]

Answer:

The answer to your question is V2 = 29.6 l

Explanation:

Data

Pressure 1 = P1 = 12 atm

Volume 1 = V1 = 23 l

Temperature 1 = T1 = 200 °K

Pressure 2 = 14 atm

Volume 2 = V2 = =

Temperature 2 = T2 = 300°K

Process

1.- To solve this problem use the Combine gas law.

P1V1/T1 = P2V2/T2

-Solve for V2

V2 = P1V1T2 / T1P2

2.- Substitution

V2 = (12)(23)(300) / (200)(14)

3.- Simplification

V2 = 82800 / 2800

4.- Result

V2 = 29.6 l

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3 years ago