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3 years ago

What is the mole fraction of NaOH in an aqueous solution that contains 15% NaOH by mass?

Chemistry

1 answer:

zhenek [66]3 years ago

6 0

Answer:

0.074

Explanation:

15% means that in 100 g of solution 15 g sodium hydroxide is present.

Mass of water = 100 - 15 = 85 g

Number of moles of sodium hydroxide:

Number of moles = 15 g/40 g/mol

Number of moles = 0.375 mol

Number of moles of water:

Number of moles = 85 g/18 g/mol

Number of moles = 4.7 mol

Moles fraction of NaOH:

moles of NaOH/ moles of solvent + moles of solute

0.375 mol/ 0.375 mol+4.7mol

0.375 mol / 5.075 mol

0.074

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As the temperature of water increases its density

fiasKO [112]

Answer:

Explanation:

The warmer the water, the more space it takes up, and the lower its density.

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2 years ago

Read 2 more answers

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

The reaction of pyrrole with bromine forms predominantly __________. View Available Hint(s) The reaction of pyrrole with bromine

xenn [34]

Answer:

a) 2-bromopyrrole

Explanation:

Our options for this questions are:

a) 2-bromopyrrole

b) 2,3-dibromopyrrole

c) N-bromopyrrole

d) 3-bromopyrrole

To understand how the reaction works we have to start with the resonance structures. (Figure 1), on these structures, we will obtain a negative charge on carbon 2 in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.

The second step is to check how the mechanism take place. An electrophile is generated by the $Br_2$ and $FeBr_3$ . This electrophile can be attacked by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).