Which of the following reactions involves two or more substances combining to form a new compound?

The reaction 2a → a2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if

katovenus [111]

The integrated rate law for a second-order reaction is given by:

$\frac{1}{[A]t} = \frac{1}{[A]0} + kt$

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

<span>k =</span> the rate constant for the reaction

<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min

Hence, $\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)$

<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>

<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M

7 0

3 years ago

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If one benzene molecule is placed in water, the total entropy of the water + benzene system:___________

valkas [14]

Answer:

d) increases

Explanation:

Benzene is an aromatic hydrocarbon which is obtained from the destructive distillation of coal. It is a colourless volatile liquid with a sweet smell. It boils ar 80° C (353 K) and freezes at 5°C (2278 K). It is insoluble in water but mixes in all proportions with ethanol, ethoxyethane and methylbenzene. The reason which benzene is insoluble in water is that benzene is a non-polar compound and water is polar, meanwhile only "like dissolves like". So, when the benzene molecule is placed in water. There will be distortion and disturbance between the benzene molecule and the water. Thus, the particle of each molecule will be distant from each other. This state results to change in the entropy of the system as the entropy of the system increases.

3 0

3 years ago

8.0 mol AgNO3 reacts with 5.0 mol Zn in

Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgNO₃: 2 moles
Zn: 1 mole
Ag: 2 moles
Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

$amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}$

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

$amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}$

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

#SPJ1

4 0

2 years ago

Please help me with the questions?

weqwewe [10]

Answer:

1. Forecast

2. Cold front

3. Shelf Cloud

4. Barometer

5. Pressure

6 Prevailing Westerlies

7. Thermometer

8. Weather

6 0

3 years ago

The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c

olchik [2.2K]

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

7 0

3 years ago