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castortr0y [4]
3 years ago
6

What is the enthalpy of formation (?H?f) of NaHCO3(s) from its constituent elements.

Chemistry
2 answers:
vova2212 [387]3 years ago
7 0

Answer:

Enthalpy of formation = -947.68KJ/mol

Explanation:

Enthalpy of formation is the heat change when one mole of a substance is formed from its element in its standard states and in standard conditions of temperature and pressure. it may be positive or negative, if positive, it is an endothermic reaction where the heat content of the product is greater than that of the reactants, and if negative, it is exothermic reaction - where the heat content of the reactants is greater than the products. the enthalpy of formation is measured in KiloJoule/Moles (KJ/Mole).

From the value of the enthalpy of formation of NaHCO3, it shows that the reaction is exothermic, that is the formation of NaHCO3 from its constituents elements. As such, the heat content of the reactants is greater than the products.

The step by step explanation is shown in the attachment.

Shtirlitz [24]3 years ago
3 0

Answer:

-947.6 kJ/mol.

Explanation:

The constituent elements combined to result in NaHCO3(s)

NaHCO3(s) → (1/2).Na2CO3(s) + (1/2).CO2(g) + (1/2).H2O(g) : ΔHr = +64.6 kJ/mol

Enthalpy change of a reaction can be measured and given the symbol,  . It is equal to the difference in enthalpy between reactants and products

From this equation we deduct that water is produced as a vapour, hence the reaction is endotermic and ΔHr is positive.

Therefore, To find the enthalpy of formation of NaHCO3 from this equation, we use the general equation which relates the enthalpy change ΔHr of the reaction to the enthalpies ΔHf of formation for the reactants and products.

ΔHr = ∑ΔH(products) - ∑ΔH(reagents)

where the  

ΔH is equal to the change in the potential energy of enthalpy of the chemical bond.

In ideal test scenarios, these values should be supplied, as you will not be able to look them up. Here, I've supplied values here.

ΔHf : Na2CO3(s) = -1130.7, CO2(g) = -393.5, H2O(g) = -241.8 kJ/mol

Using these figures (you may wish to check them in Wikipedia or elsewhere), we get

+64.6 = [(1/2)*(-1130.7) + (1/2)*(-393.5) + (1/2)*(-241.8)] - ΔHf(NaHCO3)

With the result that the enthalpy of formation of NaHCO3 is obtained as -947.6 kJ/mol.

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What is the systematic name of the following compound?<br> Li2S
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Answer:

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6 0
3 years ago
1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%
Tatiana [17]

Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

3 0
3 years ago
13. How many moles of sodium hydroxide are needed to neutralize 50grams of sulfuric acid?
Phoenix [80]

Answer:

1.02mole

Explanation:

The reaction equation is given as:

     2NaOH  +  H₂SO₄ →  Na₂SO₄  + 2H₂O

Given:

Mass of H₂SO₄  = 50g

Unknown:

Number of moles of NaOH = ?

Solution:

To solve this problem, we first find the number of moles of the acid given;

  Number of moles  = \frac{mass}{molar mass}

Molar mass of H₂SO₄ = 2(1) + 32 + 4(16)  = 98g/mol

Now;

   Number of moles = \frac{50}{98}   = 0.51mole

From the balanced reaction equation:

       1 mole of H₂SO₄ will be neutralized by 2 mole of NaOH

     0.51 mole of H₂SO₄ will be neutralized by 2 x 0.51  = 1.02mole of NaOH

6 0
2 years ago
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Trava [24]

Answer:

Section Three

1. D

2. B

3. A

4. G

5. C

6. F

7. E

Section Four

1. F

2. E

3. B

4. A

5. C

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The nonmetals in Groups 5A, 6A, and 7A...
antiseptic1488 [7]

Answer:

The nonmetals in Groups 15, 16, and 17 form ions with charges of 3-, 2-, and 1-, respectively.

Explanation:

The elements of Group 18 (the noble gases) have a complete valence shell of eight electrons.

It is <em>easier</em> for the elements of Groups 15 to 17 <em>to</em> <em>gain</em> three, two, or one electron(s) to get a complete valence shell <em>than it is to lose</em> five, six, or seven valence electrons.

Thus, they form <em>negative ions</em> with charges of 3-, 2-, and 1-, respectively.

The charges <em>do not correspond</em> to the Group numbers of 15 to 17 or the old (pre-1990) Group numbers of 5A to 7A.

6 0
3 years ago
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