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2 years ago

Please help!!

Chemistry

1 answer:

xxTIMURxx [149]2 years ago

6 0

Answer:

In an experiment, a student transferred 4.50 mL of a liquid into a pre-weighed beaker (the weight of which was determined to be 35.986 g ).

Explanation:

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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

1 year ago

Gaseous ethane CH3CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. What is the

ololo11 [35]

Answer:

46.97g

Explanation:

Hello,

To find the theoretical yield, we must first of all get the equation of reaction right in order to know how the compounds combine together.

Equation of reaction;

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

From the equation of reaction,

2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂.

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Mass of CO₂ = 4 × 44 = 176g

Mass of C₂H₆ = 2 × 30 = 60g

Mass of O₂ = 7 × 32 = 224g

Therefore, 224g + 60g (O₂ + C₂H₆) = 176g of CO₂.

284g of reactants = 176g of product(CO₂)

(18 + 57.8)g of reactants = x g of products(CO₂)

X = (75.8 × 176) /284

X = 46.97g of CO₂.

Theoretical yield of CO₂ is 46.97g

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2 years ago

You start with an equimolar mixture of the gases SO₂ and O₂ in a container. The SO₂ and O₂ react to completion to form SO₃. If t

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Answer:

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Explanation:

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2 years ago

Substances change temperature at different rates because of their

murzikaleks [220]

Answer:

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2 years ago

Calculate the mass percent of HOCH 2CH 2OH in a solution made by dissolving 3.2 g of HOCH 2CH 2OH in 43.5g of water.

grin007 [14]

Answer:

$\%m/m=6.85\%$

Explanation:

Hello,

In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

$\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%$

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

$\%m/m=\frac{3.2g}{3.2g+43.5g} *100\%\\\\\%m/m=6.85\%$

Best regards.

5 0

2 years ago