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2 years ago

Emily spills concentrated sodium hydroxide solution on her lab bench. What she should do first?

Chemistry

1 answer:

blagie [28]2 years ago

4 0

<h2>Answer:</h2>

She should notify her teacher.

<h2>Explanation:</h2>

Sodium Hydroxide is very corrosive in nature and it causes Redness, Pain, Serious skin burns and Blisters. Sodium hydroxide does not produce systemic toxicity, but is very CORROSIVE and can cause severe burns in all tissues that it comes in contact with. Therefore she should not touch it and she must notify her teacher as soon as possible.

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Explanation:

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3 years ago

You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft

olya-2409 [2.1K]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

$Mass=Density\times Volume$

Volume of solution = Volume of HCl + Volume of $Ba(OH)_2$

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

$Mass=1g/mL\times 93.1mL=93.1g$

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat released = ?

m = mass = 93.1 g

$C_p$ = specific heat capacity of water = $4.184J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = $29.83^oC$

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

$Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC$

$Q=1881.43J=1.88kJ$ (1 kJ = 1000 J)

Now we have to calculate the moles of $Ba(OH)_2$ and HCl.

$\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}$

$\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}$

$\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $Ba(OH)_2$ react with 2 mole of $HCl$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $HCl$

From this we conclude that, $HCl$ is an excess reagent because the given moles are greater than the required moles and $Ba(OH)_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$