<span>Microscopic foreign particles that droplets form on is called the hygroscopic nuclei. The particles are already present in the air in the form of dust, salt from seawater evaporation and combustion residue. Together the particles begin the forming process of droplets.</span>
<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆T
Heat released by anthracene= Heat absorbed by water
Heat absorbed by water =
mol of anthracene (MW=178,23 g/mol)
The enthalpy of combustion per mole of anthracene :
24.4 cm.
<h3>Explanation</h3>
HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.
The speed of a HCl or NH₃ molecule depends on its <em>kinetic energy</em>.
Where
- is the <em>kinetic energy</em> of the molecule,
- its mass, and
- the square of its speed.
Besides, the <em>kinetic theory</em> <em>of gases</em> suggests that for an ideal gas,
where its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the <em>average kinetic energy</em> of molecules shall be the same for <em>any ideal gas </em>at the same<em> temperature</em>. So is the case for HCl and NH₃
Where
- , , and the mass, speed, and kinetic energy of an HCl molecule;
- , , and the mass, speed, and kinetic energy of a NH₃ molecule.
The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their <em>molar mass</em>. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, . Therefore:
The <em>average </em>speed NH₃ molecules would be <em>if</em> the <em>average </em>speed of HCl molecules is 1.
Answer:
Oxygen is the limiting reactant.
Explanation:
Hello,
In this case, given the reaction:
Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:
Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:
In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.
Best regards.