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3 years ago

9

Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s

peed, which one is more likely to break through?

1 answer:

liubo4ka [24]3 years ago

7 0

Answer:

the heavy one

Explanation:

the heavy one because heavy things and break things and the light one can't

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When the equation Fe + Cl2 ----> FeCl3 is correctly balanced the coefficient for Cl2 is

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3 years ago

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3 years ago

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2 years ago

If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4

attashe74 [19]

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$ .

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of $SO_4^{2-}$ , it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

5 0

3 years ago

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