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3 years ago

How many moles are in 74.8 grams of NaCl?

Chemistry

1 answer:

Pie3 years ago

7 0

Answer:

1.28 mol NaCl

Explanation:

Find the mass of NaCl by adding the two elements atomic mass

Na: 22.99

Cl: 35.45

22.99+35.45=58.44 g

Now convert to moles

74.8 g / 58.44 g = 1.27994524298 mol

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Answer:

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Explanation:

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3 years ago

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq

Pavlova-9 [17]

Answer : The molarity of $I_3^-$ in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

$2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)$

First we have to calculate the moles of $Na_2S_2O_3$ .

$\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}$

$\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole$

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of $I_3^-$ .

From the balanced chemical reaction, we conclude that

As, 2 moles of $S_2O_2^{-3}$ react with 1 mole of $I_3^-$

So, 0.007696 moles of $S_2O_2^{-3}$ react with $\frac{0.007696}{2}=0.003848$ mole of $I_3^-$

The moles of $I_3^-$ = 0.003848 mole

Now we have to calculate the molarity of $I_3^-$ .

$\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}$

Now put all the given values in this formula, we get:

$\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M$

Therefore, the molarity of $I_3^-$ in the solution is, 0.128 M

8 0

3 years ago

17.g to mL<br> Density= 3.291g/mL

algol [13]

Density is defined as mass per unit volume
therefore density = mass ÷ volume

volume = mass÷density
volume in ml = 17 ÷ 3.291

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3 years ago

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Answer:

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3 years ago

What is light tell definition

swat32

Answer:

1..the natural agent that stimulates sight and makes things visible.

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"a shrewd light entered his eyes"

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3 years ago

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