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3 years ago

Jenny places a strip of pH paper into a solution. When she removes the pH paper, it has turned yellow-green.

Chemistry

2 answers:

Ronch [10]3 years ago

4 0

Jenny puts the ph paper and lines it up PH SCALE to find how strong the solution.

vaieri [72.5K]3 years ago

4 0

Answer: compare the pH paper's color with a chart of colors and pH ranges

Explanation:

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When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so

zmey [24]

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $50ml+50ml=100ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

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

m = mass of water = 100 g

$T_{final}$ = final temperature of water = $27.5^0C$

$T_{initial}$ = initial temperature of metal = $21.0^0C$

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

$q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C$

$q=2719.6J=2.72kJ$

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of $HCl$ :

0.05 moles of $HCl$ releases heat = 2.72 KJ

1 mole of $HCl$ releases heat = $\frac{2.72}{0.05}\times 1=54.4KJ$

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

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