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PSYCHO15rus [73]

3 years ago

12

The appropriate geometric "fixing" of two atoms which is necessary to start chemical rearrangement is called the:

2 answers:

Ivanshal [37]3 years ago

8 0

The answer is: activation complex.

Activated complex is collection of intermediate structures in a chemical reaction that forms while bonds are breaking and new bonds are forming.

The successful collisions must have enough energy (activation energy).

Chemical bonds are broken and new bonds are formed.

Particles are in constant, random motion and possess kinetic energy, molecules faster and have more collisions.

ahrayia [7]3 years ago

4 0

Hello there.

Question: <span>The appropriate geometric "fixing" of two atoms which is necessary to start chemical rearrangement is called the:

Answer: It is called activation complex.
</span><span>
Hope This Helps You!
Good Luck Studying ^-^</span>

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A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo

Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

Divide by 0.167 :

C : H = 1 : 3

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2 years ago

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cestrela7 [59]

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sergiy2304 [10]

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2 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

2 years ago

True or False: All material has the same density.

Burka [1]

Answer:

Different materials have different densities. So it is False

4 0

3 years ago

Read 2 more answers