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maw [93]
3 years ago
11

Sodium carbonate reacts with silver nitrate according to the following balanced equation: Na2CO3 (s) + 2 AgNO3 (aq) → Ag2CO3 (s)

+ 2 NaNO3 (9) If 3.60 g of Na2CO3 is allowed to react with 5.14 g of AgNO3, what mass of Na2CO3 will remain at the end of the reaction?
a) 2.01 g
b) 1.54 g
c) 0.423 g
d) 0.0150 g
e) 0g
Chemistry
1 answer:
klemol [59]3 years ago
8 0

Answer:

a) 2.01 g

Explanation:

  • Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃

First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:

  • 0.0302 mol AgNO₃ * \frac{1molNa_2CO_3}{2molAgNO_3}  = 0.0151 mol Na₂CO₃

So the remaining Na₂CO₃ moles are:

  • 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃

Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:

  • 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃

The closest answer is option a).

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Why a neutral atom has the same number of protons and electrons
Alex777 [14]

Explanation:

When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral. ... Chemically, we say that the atoms have formed bonds.

5 0
3 years ago
2Mg + O2à MgO For this unbalanced chemical equation, what is the coefficient for oxygen when the equation is balanced? A. 1 B. 2
Vikki [24]

Answer:

The answer to your question is the letter A. 1

Explanation:

Unbalanced chemical reaction

                      Mg  +  O₂  ⇒  MgO

               Reactants    Elements    Products

                       1            Magnesium       1

                       2           Oxygen              1

Balanced chemical reaction

                      2Mg  +  O₂  ⇒  2MgO

              Reactants    Elements    Products

                       2           Magnesium       2

                       2           Oxygen             2

Conclusion

The coefficients of the balanced equation are 2, 1, 2

7 0
3 years ago
Read 2 more answers
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

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Otrada [13]

Answer:

Option 3. The catalyst does not affect the enthalpy change (\Delta H_\text{rxn}) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction (\Delta H_\text{rxn}) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change \Delta H_\text{rxn} of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction \Delta H_\text{rxn} measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

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aliya0001 [1]
I have attached the answer. hopefully, i read the problems correctly. let me know if I did not.

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