Is a Big Mac an element, compound, or mixture?

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

6 0

3 years ago

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What is the atomic number and name of an element that has 15 protons in its nucleus?

just olya [345]

Answer:

phosphorus.

Explanation:

The atomic number of phosphorus is 15 so the protons will be 15.

4 0

3 years ago

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Two liquids, a and b are immiscible. liquid a has a density of 0.89 g/ml. liquid b has a density of 0.72 g/ml. what would you ex

Burka [1]

In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.

The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.

5 0

3 years ago

Use the IUPAC nomenclature rules to give the name for this compound - Al2S3.

jonny [76]

Aluminium Sulfide
According to rules the positive specie is named first and the negative specie is named last.

6 0

3 years ago

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The average atomic mass of Eu is 151.96 amu. There are only two naturally occurring isotopes of europium, Eu with a mass of 151.

earnstyle [38]

Answer:

The percentage abundance of Eu isotopes are 52 % and 48 % .

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope,:

% = x %

Mass = 151.0 amu

For second isotope :

% = 100 - x

Mass = 153.0 amu

Given, Average Mass = 151.96 amu

Thus,

$151.96=\frac{x}{100}\times {151.0}+\frac{100-x}{100}\times {153.0}$

Solving for x, we get that:

x = 52 %

<u>Thus percentage abundance of Eu isotopes are 52 % and 48 % .</u>

7 0

3 years ago