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Moles of Zn(NO3)2 in 131.50g of this substance.

Degger [83]

number of moles= mass / molecular mass
mass=131.50g
molecular mass of Zn(NO3)2 =189.39 g/mol
number of moles = 131.50 g/189.39 g/mol =0.69433 mol

3 0

4 years ago

Read 2 more answers

The acid-dissociation constant at 25.0 °c for hypochlorous acid (hclo) is 3.0 ⋅ 10−8. at equilibrium, the molarity of h3o+ in a

maks197457 [2]

Hypochlorous acid is a weak acid. The $K_{a}$ value for the dissociation of HOCl is $3.0 * 10^{-8}$

3 0

3 years ago

It didn't have science but it is a science question. The average, year-after-year conditions of temperature, precipitation, wind

larisa86 [58]

Hey there,

Question : The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its?

Answer : A, climate

Hope this helps :))

~Top♥

6 0

3 years ago

What is the molarity of a solution that contains 0.202 mol KCl in 7.98 L solution?

mr Goodwill [35]

Explanation:

molarity = no. of moles of solute/solution in litres

molarity =0.202/7.98

=0.025 M

8 0

3 years ago

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from

IgorLugansk [536]

The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g $S_8$ .

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of $S_8$ = 3.0 g

Molar mass of Ag = 107.8 g/mole

Molar mass of $S_8$ = 256 g/mole

Molar mass of $Ag_2S$ = 247.8 g/mole

First we have to calculate the moles of Ag and $S_8$ .

$\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles$

$\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)$

From the balanced reaction we conclude that

As, 16 mole of $Ag$ react with 1 mole of $S_8$

So, 0.0278 moles of $Ag$ react with $\frac{0.0278}{16}=0.00174$ moles of $S_8$

From this we conclude that, $S_8$ is an excess reagent because the given moles are greater than the required moles and $Ag$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag_2S$

From the reaction, we conclude that

As, 16 mole of $Ag$ react to give 8 mole of $Ag_2S$

So, 0.0278 moles of $Ag$ react to give $\frac{0.0278}{16}\times 8=0.0139$ moles of $Ag_2S$

Now we have to calculate the mass of $Ag_2S$

$\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S$

$\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g$

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

4 0

3 years ago