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sergey [27]
3 years ago
11

You carefully weigh out 20.00 g of CaCO3 powder and add it to 81.00 g of HCl solution. You notice bubbles as a reaction takes pl

ace. You then weigh the resulting solution and find that it has a mass of 92.80 g . The relevant equation is:
Chemistry
1 answer:
Zepler [3.9K]3 years ago
3 0

Answer:

The relevant equation is:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Explanation:

1 mol of calcium carbonate can react to 2 moles of Hydrochloric acid to produce 1 mol of water, 1 mol of calcium chloride and 1 mol of carbon dioxide.

The formed CO₂ is the reason why you noticed bubbles as the reaction took place

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Some potential energy was converted into thermal energy due to friction. The acceleration due to gravity became smaller as the box slid down the ramp.
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2 years ago
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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 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}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

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, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 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 = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

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 = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

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

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

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

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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The radioactive isotope I-131 is used in
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1.which of the following were found to destroy ozone in the upper atmosphere?
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Below are the answers:

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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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Convert the density 1.98 g/mL to the unit kg/m3.
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1.98 g/ml=1980 kg/m3 because 1 kg =1000g, 1m3=1000.000mL
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