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Ivan
3 years ago
14

How can ionic bonds and naming be used in a real life example ?

Chemistry
1 answer:
Stells [14]3 years ago
8 0

Answer:

Melting snow more efficiently in winters, understanding the components of mineral water

Explanation:

Let's split this question into two parts. First of all, ionic bonds:

  • an example would be the application of the freezing point depression law. Remember that adding a solute to a specific solvent would decrease the freezing point of a solvent. This is the reason why we add ionic salts, NaCl, to snow in order to make it melt. Knowledge of the fact that 1 mol of NaCl, an ionic compound, dissociates into 2 mol of ions, sodium and chloride, yields us a van 't Hoff factor of 2 rather than 1 for non-electrolytes, molecular compounds. This means the same molality of ionic compounds would produce a twice larger decrease in the freezing point of a solvent;
  • an example for ionic naming is more trivial. Remember the difference between, say, calcium and calcium cation. Sometimes we may read that mineral water is full of calcium. Having chemical knowledge of ionic compound naming would lead us to a conclusion that this is wrong! Mineral water doesn't have any calcium in it, we don't see any metal in mineral water. However, mineral water contains calcium cations, Ca^{2+} and not Ca.
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Which bases can deprotonate acetylene? the pka values of the conjugate acids are given in parentheses. select all that apply. ch
Usimov [2.4K]

pKa is a value which is related to the acid dissociation constant Ka

pKa = -log Ka

i.e. Ka = 10^-pKa

The deprotonation reaction of acetylene is:

HC≡CH ↔ HC≡C⁻ + H⁺

pKa (HC≡CH) = 25

Solvents with pKa greater than 25 will deprotonate acetylene.

Ans: CH2=CH⁻ pka = 44 and CH3NH⁻ pka = 40  

4 0
4 years ago
Consider the following balanced equation for the following reaction:
Sidana [21]

<u>Answer:</u> The actual yield of the carbon dioxide is 47.48 grams

<u>Explanation:</u>

For the given balanced equation:

15O_2(g)+2C_6H_5COOH(aq.)\rightarrow 14CO_2(g)+6H_2O(l)

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Theoretical moles of carbon dioxide = 1.30 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

1.30mol=\frac{\text{Theoretical yield of carbon dioxide}}{44g/mol}\\\\\text{Theoretical yield of carbon dioxide}=(1.30mol\times 44g/mol)=57.2g

To calculate the theoretical yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Theoretical yield of carbon dioxide = 57.2 g

Percentage yield of carbon dioxide = 83.0 %

Putting values in above equation, we get:

83=\frac{\text{Actual yield of carbon dioxide}}{57.2}\times 100\\\\\text{Actual yield of carbon dioxide}=\frac{57.2\times 83}{100}=47.48g

Hence, the actual yield of the carbon dioxide is 47.48 grams

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3 years ago
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The temperature is the same, so the amount of kinetic energy will also be the same.

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