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3 years ago

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A student is studying the ways different elements are similar to one another. Diagrams of atoms from four different elements are

shown below. Nucleus Nucleus Atom 1 Atom 2 Nucleus Nucleus Atom 3 Atom 4

1 answer:

alisha [4.7K]3 years ago

6 0

Answer:

atom 1 and atom 2

Explanation:

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Tips for high school

musickatia [10]

Answer:

stoichiometry is hard but the rest is kinda easy if u pay attention

Explanation:

5 0

3 years ago

Read 2 more answers

The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the

deff fn [24]

Answer:

The rate constant of the reaction at 125˚ is $0.3115 \ \text{sec}^{-1}$ .

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

$k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}$

$\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}$

where $k$ is the rate constant, $E_{a}$ represents the activation energy of the chemical reaction, $R$ is the gas constant, $T$ is the temperature, and $A$ is the frequency factor.

The frequency factor, $A$ , is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be $k_{1}$ at an initial temperature $T_{1}$ and the rate constant $k_{2}$ at a final temperature $T_{2}$ , thus

$\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)$

$\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}$

Given that $E_{a} \ = \ 26.5 \ \ \text{kJ/mol}$ , $R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}$ , $T_{1} \ = \ \left(40 \ + \ 273\right) \ K$ , $T_{2} \ = \ \left(125 \ + \ 273\right) \ K$ , and $k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}$ , therefore,

$k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}$

8 0

2 years ago

Which of the following are common decomposers? (Select all that apply) fungi, ferns, bacteria, viruses, grasses

antiseptic1488 [7]

Fungi heijsiwhqibwudyenejidjd

4 0

3 years ago

Read 2 more answers

In synthesis reactions, there are always more _____________ than ________________.

nydimaria [60]

Answer:

A

Explanation:

always more reactants than products

4 0

3 years ago

Suppose you have a liquid solution containing two components, acetone and nonane. The boiling point of pure acetone is 56 C and

Tom [10]

Answer:

The boiling point is somewhere between 56 and 151 °C

Explanation:

Hello,

In this case, it is possible to compute it via rigorous methods in phase equilibrium by using for example a cubic equation of state to model the vapor phase and a suitable excess Gibbs free energy model for the liquid phase, nonetheless, it is an arduous task. In such a way, since the information about both acetone's and nonane's pure boiling points is given as well as acetone's mole fraction, which points out it is about a binary liquid solution, one could make up the boiling point is somewhere between 56 and 151 °C precising that it should be closer to 151 °C as the mixture is 90% nonane and 10% acetone.

Best regards.

4 0

3 years ago