Answer : The molar mass of the unknown gas will be 79.7 g/mol
Explanation : To solve this question we can use graham's law;
Now we can use nitrogen as the gas number 2, which travels faster than gas 1;
So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1
We can compare the rates of both the gases;
So here, Rate of gas 2 / Rate of gas 1 =
Now, 1.687 = square root [
]
When we square both the sides we get;
2.845 = (molar mass 1) / (28.01 g/mol N2)
On rearranging, we get,
2.845 X (28.01 g/mol N2) = Molar mass 1
So the molar mass of unknown gas will be = 79.7 g/mol
To answer this problem, we use Hess' Law to calculate the overall enthalpy of the reactions. The goal is to add all the reactions such that the final reaction is C<span>5H12 (g) + 8O2 (g) → 5CO2 (g) + 6H2O (l) through cancellation adn multiplication. The first equation is multiplied by 5, the second one is multiplied by 6 and the third one is reversed. The final answer is -3538 J or -3.54 x10^3 kJ.</span>
<h2><em>Solid. Because molecules densely packed it has a high resistance to flow thereby lower kinetic energy.
</em></h2>
Answer:
The equation to show the the correct form to show the standard molar enthalpy of formation:
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Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Given, that 1 mole of
gas and 1 mole of
liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

Divide the equation by 2.

The equation to show the the correct form to show the standard molar enthalpy of formation:

Camels lose less water through their urine and feces than many other mammals. Their kidneys concentrate water heavily, leading to salty urine. The intestines also reabsorb water from intestinal material as it is digested, so dry feces are produced.