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kolbaska11 [484]
3 years ago
6

55

Chemistry
2 answers:
nydimaria [60]3 years ago
8 0

Answer:

correct answer i C i think so

valentinak56 [21]3 years ago
4 0

Answer:

Safely throw away the liquids and get new ones.

Explanation:

Not A  because

you shouldn't smell them cause it could be toxic and hurt you

Not B

you should never guess which chemical are which cause if your not correct and use the wrong one you could have a chemical reaction

Not C

you cant mix random liquids it could make a chemical reaction to exploded ore worse create a hurtful gas

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Nitrogen has an atomic number of 7; therefore, it has _____ electrons in its outermost electron shell
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Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
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Answer : The correct option is, (A) -101.37 KJ

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}=1.6mole/L\times 0.087L=0.1392mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

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, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 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 = 87ml+87ml=174ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g

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 = 174 g

T_{final} = final temperature of water = 317.4 K

T_{initial} = initial temperature of metal = 298 K

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

q=174g\times 4.18J/g^oC\times (317.4-298)K

q=14110.008J=14.11KJ

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole

Therefore, the enthalpy of neutralization is, -101.37 KJ

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