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3 years ago

Explain how to use an acid-base indicator to determine pH

2 answers:

7 0

Acid-base indicator changes color based on pH.

drop some in a solution n watch the color changes. different indicators show different colors at different pH. they usually have standard colors for comparison.

Gala2k [10]3 years ago

5 0

Answer:

Litmus paper

Explanation:

for this very example your gonna need a litmus paper

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C. physical; 100°C

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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

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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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3 years ago

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