Answer:
Explanation:
The molar mass is the mass of a substance in grams per mole.
To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.
The compound given is:
The compound has 1 Ca (calcium) and 2 Cl (chlorine).
Mass of Calcium
- The molar mass of calcium is 40.08 g/mol
- There is only one atom of Calcium in CaCl₂, so the number above is what we will use.
Mass of Chlorine
- The molar mass of chlorine is 35.45 g/mol
- There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
- 35.45 * 2= 70.9 g/mol
Molar Mass of CaCl₂
- Now, to find the molar mass, add the molar mass of 1 calcium and 2 chlorine.
- 40.08 g/mol + 70.9 g/mol =110.98 g/mol
The molar mass of CaCl₂ is <u>110.98 grams per mole. </u>
Answer:
Explanation:
By the kinetic molecular theory (particle model), all matter consists of particles, there are spaces between the particles, the particles are in constant random motion, and there are forces of attraction and repulsion between the particles.
Furthermore, temperature is defined to be a measure of the average kinetic energy of the particles.
Evaporation is a change of phase from liquid to gas explained as follows :
When particles in the liquid phase are heated, they gain kinetic energy and move faster and further apart. Eventually they have enough energy to escape the forces of attraction holding them together in the liquid phase and they move very fast and far from each other and exist in the gaseous phase.
Answer:
Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules). The number of atoms before and after the chemical change is the same but the number of molecules will change.
Explanation:
Answer:
Explanation:
For a first order reaction the rate law is:
![v=\frac{-d[A]}{[A]}=k[A]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B-d%5BA%5D%7D%7B%5BA%5D%7D%3Dk%5BA%5D)
Integranting both sides of the equation we get:
![\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%7D%20%5C%2C%20dx%20%3D-k%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt)
where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.
From that integral we get the integrated rate law:
![ln\frac{[A]}{[A]_{0} } =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_%7B0%7D%20%7D%20%3D-kt)
![[A]=[A]_{0}e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_%7B0%7De%5E%7B-kt%7D)
![ln[A]=ln[A]_{0} -kt](https://tex.z-dn.net/?f=ln%5BA%5D%3Dln%5BA%5D_%7B0%7D%20-kt)
![k=\frac{ln[A]_{0}-ln[A]}{t}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%5BA%5D_%7B0%7D-ln%5BA%5D%7D%7Bt%7D)
therefore k is
Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:
a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:
Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:
c) The leftover is computed as follows:
NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
