Answer:
Explanation:
Homogeneous mixture is a mixture in which the components of the mixture are in the same proportion throughout any sample extracted from the mixture while an heterogeneous mixture is a mixture in which the components of the mixture differ in term of proportion when different samples of the mixture are extracted and compared.
For example, a sandy water will have some parts (usually the bottom) of the mixture with more sand than other parts of the mixture, hence, it (sandy water) is a heterogeneous mixture. While salty and ocean water has it's salt dissolved in the same proportion throughout the water in the mixture, hence salty and/or ocean water is a homogeneous mixture.
Sandy water can be separated by filtration (i.e using a filter paper to separate the sand from the water when the mixture is poured over a filter paper) while salty and ocean water can be separated by distillation (i.e boiling of the mixture so the water molecules can boil and move through a tube as gas or steam into another container where they are cooled and converted back to liquid or water while leaving the solid salt component of the mixture in the boiling tube).
694,563,239 rounded to the nearest thousand is 694,563.
It's because the first digit from the right is for ones, second for tens, third for hundreds and fourth for thousands and that's the one that we should take a closer look at. You can round it either to 3 or 4, depends on the digit of hundreds. In this case 3239 is clearly closer to 3000 than 4000, that's why we round it to 694,563, not 694,564.
Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,
The expression for rate of reaction for the reactant :
![\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNH_3%3D-%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DO_2%3D-%5Cfrac%7B1%7D%7B7%7D%5Ctimes%20%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The expression for rate of reaction for the product :
![\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DNO_2%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
From this we conclude that, all the options are correct.
Answer:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
