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wlad13 [49]
1 year ago
10

Potassium carbonate (K2CO3) has one potassium atom bonded to carbon, and that carbon is also bonded to three oxygens. Their elec

tronegativities are shown in the list:
K 0.82
C 2.55
O 3.44
How would you describe the bonds (K-C and C-O) in this molecule?
A: K-C is polar covalent, and C-O is nonpolar covalent.
B: K-C is ionic, and C-O is nonpolar covalent.
C: K-C is ionic, and C-O is polar covalent.
D: K-C is polar covalent, and C-O is polar covalent.
Chemistry
1 answer:
dybincka [34]1 year ago
3 0

Answer:

I'm not all the way sure but I think it is C

Explanation:

The reason is because an ionic bond essentially donates an electron to the other atom participating in the bond. Also, the C-O I think is polar since it is evenly distributed.

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
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<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

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The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

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