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zlopas [31]
3 years ago
9

Which of the following shows the balanced equation for the reaction that represents the decomposition of barium carbonate (BaCO3

)? (2 points) Select one: a. BaO + CO2yields BaCO3 b. 2BaO + CO2yields 2BaCO3 c. BaCO3yields BaO + CO2 d. 2BaCO3yields 2BaO + CO2
Chemistry
2 answers:
Inessa [10]3 years ago
5 0
BaCO3 = BaO + CO2, so the answer is C
Vedmedyk [2.9K]3 years ago
5 0

Answer : The correct option is, (c) BaCO_3(s)\rightarrow BaO(s)+CO_2(g)

Explanation :

Decomposition reaction : It is a type of chemical reaction in which one reactant breaks into two or more smaller compounds.

AB\rightarrow A+B

The given balanced decomposition reaction of barium carbonate is,

BaCO_3(s)\rightarrow BaO(s)+CO_2(g)

In this reaction, 1 mole of barium carbonate decomposes to yields 1 mole of barium oxide and 1 mole of carbon dioxide.

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Does the nucleus (protons and neutrons) weigh more than the electrons?
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4 0
2 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

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}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

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, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 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 = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

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 = 274 g

T_{final} = final temperature of water = 325.8 K

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

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

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
4 years ago
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