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3 years ago

11

The formation of a solid when two liquids are mixed indicates a reaction.

1 answer:

Artyom0805 [142]3 years ago

5 0

The statement is true. The formation of a solid when two liquids are mixed indicates a reaction because a new substance is formed when the two liquid is mixed and it is evident by the precipitate observed. An example is the reaction of sodium iodide and lead nitrate yielding to lead iodide as the precipitate.

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How many grams of water are needed to react with 18.9 grams of li2o? given: li2o+h2oâ2lioh?

4vir4ik [10]

The balanced equation for the reaction is as follows;
Li₂O + H₂O ---> 2LiOH
Stoichiometry of Li₂O to H₂O is 1:1
Mass of Li₂O reacted - 18.9 g
Number of Li₂O moles reacted - 18.9 g / 30 g/mol = 0.63 mol
An equivalent amount of moles of water have reacted - 0.63 mol
mass of water required - 0.63 mol x 18 g/mol = 11.34 g
A mass of 11.34 g of water is required

4 0

3 years ago

If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a

strojnjashka [21]

Answer : The final temperature of the mixture is, $22.14^oC$

Explanation :

First we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g$

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.42J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol = 35.5 g

$m_2$ = mass of water = 45.0 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $8.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

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

$35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.14^oC$

Therefore, the final temperature of the mixture is, $22.14^oC$

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