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rjkz [21]
3 years ago
10

Natural gas, which you may assume to be pure methane, is burned in a furnace for space-heating purposes. (a) If twice the minimu

m of air is used for the combustion, and if the methane and air are initially at 20 C and the stack gases, including all of the water and vapor, are at 100 C, how much heat is liberated per mole of methane burned
Chemistry
1 answer:
Sergio [31]3 years ago
3 0

Answer:

no no no no no no no no no no no no no no

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Suppose that a 10-mL sample of a solution is to be tested for I− ion by addition of 1 drop (0.2 mL) of 0.13 M Pb(NO3)2.
Liono4ka [1.6K]

The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

<u>Given data : </u>

Volume of solution to be tested for I-ion = 10 mL

Volume of  Pb(NO₃)₂ = 0.2 mL

molarity of Pb(NO₃)₂ = 0.13 M

<h3>Determine the number of I that must be present </h3>

First step : calculate conc of PB²⁺ ions in the solution

conc of PB²⁺ ions =  ( molarity of Pb(NO₃)₂ * volume of  Pb(NO₃)₂ ) / ( total volume )

                              = ( 0.13 * 0.2 ) / ( 10 + 0.2 )

                              = ( 0.026 ) / ( 10.2 )  = 0.002549 M

Next step :<u> </u><u>determine the</u><u> molarity of  I</u>

using the dissociation reaction of PbI₂

PbI₂(s) ---> Pb²⁺ (aq)  + 2I (aq)

also;  Ksp = [ Pb²⁺ ] [ I ]²  ---- ( 1 )

From the question the given value of Ksp = 8.49 * 10⁻⁹

Therefore equation ( 1 ) becomes

8.49 * 10⁻⁹ = ( 0.002549 ) * [ I ]²

[ I ] = √ ( 8.49 * 10⁻⁹ ) / ( 0.002549 )

      = 0.0018 M

Final step :<u> Determine the minimum number of grams of I </u>

moles of I = molarity of I * total volume

                 = 0.0018 M * 10.2 mL

                 = 0.01836 * 10⁻³ mol

Hence we can conclude that The minimum number of grams of I that must be present is ; 0.01836 * 10⁻³ mol

Learn more about Pb(NO₃)₂ : brainly.com/question/25071409

8 0
2 years ago
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Identify whether the following activity on the table shown on the first uploaded image are examples of business level or  corporate level strategy

Answer:

The solution to this is shown on the second uploaded image

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The explanation is shown on the third and fourth  uploaded image

5 0
3 years ago
The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
4 years ago
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