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NARA [144]
3 years ago
10

Which of the following statements is true?

Chemistry
1 answer:
natita [175]3 years ago
5 0

Answer:

b

Explanation:

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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

From the reaction, we conclude that

As, 1 mole of Ba(OH)_2 react to give 2 mole of H_2O

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of H_2O

Now we have to calculate the change in enthalpy of the reaction.

\Delta H_{rxn}=-\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

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

\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

3 0
3 years ago
Write short notes on different planet of the solar system​
marishachu [46]
Should I eat my chicken nuggets if my throat is hurting
7 0
3 years ago
Can an intermediate appear as a reactant in the first step of a reaction mechanism?.
saveliy_v [14]

Answer:

No

Explanation:

An intermetiate in a chemical reaction is something that is formed and reused in a reaction.

8 0
2 years ago
a 108ml sampl of gas has a mass of 77.96mg at a pressure of 1140mmhg and temperature of 183°c .what is the molar mass of the gas
madam [21]

Answer:

146.85 g/mol

Explanation:

PV=nRT

n=mass/molar mass

covert from mmhg to atm = 0.184 atm

convert from ml to L= 0.108 L

convert from degree C to K= 456.15 K

convert from mg to g= 0.07796g

then rearrange the formula:

n=PV/RT

=(0.184)(0.108)/(0.08206)(456.15)

n= 5.308*10^(-4)

rearrange the n formula interms of molar mass:

Molar mass= mass/n

=0.07796/(5.308*10^-4)

molar mass= 146.85g/mol

3 0
3 years ago
Read 2 more answers
How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid
Leno4ka [110]

Answer:

29.3 g HClO₄

Explanation:

We have 39.1 grams of 74.9 wt% aqueous perchloric acid solution, that is, there are 74.9 grams of perchloric acid in 100 grams of perchloric acid solution. The mass, in grams, of perchloric acid contained in 39.1 grams of perchloric acid solution is:

39.1 g Solution × (74.9 g HClO₄/100 g Solution) = 29.3 g HClO₄

5 0
3 years ago
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