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Sphinxa [80]
3 years ago
11

Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )

Chemistry
1 answer:
Olenka [21]3 years ago
3 0

Answer:

\large \boxed{\text{-486 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                      2H₂   +   O₂ ⟶ 2H-O-H

Bonds:          2H-H    1O=O       4H-O

D/kJ·mol⁻¹:     436      498          464

\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.

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Comment Both propane and benzene are hydrocarbons. As a rule,
kozerog [31]

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

5 0
3 years ago
How many times greater is one gallon than one quart
horrorfan [7]
4, because there are 4 quarts in 1 gallon. 

Another way to think about it is 1 gallon= $1 and 1 quart= 25 cents 
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Hope this helped..
8 0
3 years ago
How many grams are in 3.14 moles of PI₃?
OverLord2011 [107]

Answer:

\boxed {\boxed {\sf 1290 \ g \ PI_3}}

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

  • Phosphorus (P): 30.973762 g/mol
  • Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

  • I₃: 126.9045 * 3=380.7135 g/mol
  • PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

We want to convert 3.14 moles to grams, so we multiply by that value.

3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \  mol \ PI_3}

The units of moles of PI₃ cancel.

3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }

1292.698 \ g\ PI_3

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

  • 1292.698

The 2 in the ones place tells us to leave the 9.

1290 \ g \ PI_3

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

4 0
3 years ago
SOMEOME HELLL PLS !!
Tpy6a [65]

Answer:

only 1

Explanation:

5 0
3 years ago
How many molecules of hydrogen gas are produced when 32.4 g of potassium are added to water?
Yuri [45]
Moles of K = 32.4/39 = 0.83 mole. According to stoichiometry, 2 moles of K produces 1mole of H2. Therefore, 0.83 mole of K produces = 0.83/2 = 0.415 moles of H2. Therefore number molecules of H2 = moles of H2 x 6.02 x 10^23 = 2.4983 x 10^23 molecules. Hope this helps!
4 0
3 years ago
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