Answer:
See explanation
Explanation:
In looking at molecules to determine whether they are polar or not we have to look at two things basically;
i) presence of polar bonds
ii) geometry of the molecule
Now, we know that CCI2F2 is a tetrahedral molecule, but the molecule is not symmetrical. It has four polar bonds that are not all the same hence the molecule is polar.
In an electric field, polar molecules orient themselves in such a way that the positive ends of the molecule are being attracted to the negative plate while the negative ends of the molecules are attracted to the positive plate.
So the positive ends of CCI2F2 are oriented towards the negative plate of the field while the negative ends of CCI2F2 are oriented towards the positive ends of the field.
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
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Answer is: elements are always combined in the same proportion by mass.
Law of multiple proportions or Dalton's Law said that the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.
For example, nitrogen(I) oxide N₂O; m(N) : m(O) = 2·14 : 16 = 7 : 4.
Another example, water (H₂O) is made of two hydrogen atoms and one oxygen atom:
m(H) : m(O) = 2·1 : 16 = 1: 8.
Since one atmosphere is 29.9213 inches of mercury, we can simply find this by taking 29.9213 and multiplying that by 1.24. You end up with 37.102412 inches of mercury. I hope this helps you out, and good luck on the rest of your homework.
All of this involdes cell of different types
