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3 years ago

7

The relatively high boiling point of water is due to water having

1 answer:

Verizon [17]3 years ago

3 0

The answer should be hydrogen bonding. Water only has oxygen and hydrogen in it, which are both nonmetals, so you know the answer cannot be metallic or ionic. It also cannot be nonpolar because the electronegativity of the oxygens will make the molecule polar. You can also know it is hydrogen bonding because it can only take place when a hydrogen is attached to an oxygen, fluorine, or nitrogen. These bonds are very strong attractions, so the molecules are extremely hard to pull apart, creating a high boiling point. Hope that helps!

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NaHCO 3 and NaHSO 4 are acid salts. however aqueous solution of NaHCO 3 is basic while aqueous solution of NaHSO 4 is acidic. gi

posledela

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Explanation:

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3 years ago

2. Consider the reaction 2 Cg H18 (4) +250â (9) âº 16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1

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Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of $C_8H_{18}$ = 16.15 moles

First we have to calculate the moles of $H_2O$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react to give 18 moles of $H_2O$

So, 16.15 moles of $C_8H_{18}$ react to give $\frac{16.15}{2}\times 18=145.35$ moles of $H_2O$

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of $C_8H_{18}$ = 878 g

Molar mass of $C_8H_{18}$ = 114 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_8H_{18}$ .

$\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react with 25 moles of $O_2$

So, 7.702 moles of $C_8H_{18}$ react with $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

Now we have to calculate the mass of $O_2$ .

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g$

The mass of oxygen needed are 3080.8 grams.

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