Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where 
= rate constant at temp 
= rate constant at temp 
= activation energy
R= gas constant
= temperature = 
= temperature = 
![ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-85%5Ctimes%201000J%2Fmol%7D%7B8.314J%2FKmol%7D%5B%5Cfrac%7B1%7D%7B303%7D%20-%20%5Cfrac%7B1%7D%7B293%7D%5D)
Thus rate increases 3 times on raising the temperature from 20degree to 30 degree
Answer: -
The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.
Explanation: -
Temperature of the hydrogen gas first sample = 10 °C.
Temperature in kelvin scale of the first sample = 10 + 273 = 283 K
For the second sample, the temperature is 350 K.
Thus we see the second sample of the hydrogen gas more temperature than the first sample.
We know from the kinetic theory of gases that
The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.
So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.
Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:
45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present
Hope this answers the question. Have a nice day.
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Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble the data in one place.
2Mg + O₂ ⟶ 2MgO
n/mol: 2 5
Calculate the moles of MgO we can obtain from each reactant.
From Mg:
The molar ratio of MgO:Mg is 2:2
From O₂:
The molar ratio of MgO:O₂ is 2:1.
