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Ulleksa [173]
3 years ago
14

How many significant digits are in the measurements 14,433

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0
I believe there is 5. 
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Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2
pashok25 [27]

Answer : The value of \Delta H for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)     \Delta H_1=-571.6kJ/mole

Now we have to determine the value of \Delta H for the following reaction i.e,

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)    \Delta H_2=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=-(-571.6kJ/mole)

\Delta H_2=+571.6kJ/mole

Hence, the value of \Delta H for the reaction is +571.6 kJ/mole.

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ph ch3 h2nnh koh The Wolff-Kishner reaction involves the reaction of an aldehyde/ketone with hydrazine in the presence of KOH. T
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Ernest Shackleton's South! primarily uses which writing structure?
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Read 2 more answers
Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?
NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is 1.24\times 10^{-7}.

Explanation:

CO_2(g)\rightleftharpoons C(s)+O_2(g)

K_{goal}=?

K_{goal}=\frac{[C][O_2]}{[CO_2]}

2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)..[1]

K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}

2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)..[2]

K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}

CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)..[3]

K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}

[1] +  [2] + [3]

2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)

 ( on adding the equilibrium constant will get multiplied with each other)

K=K_1\times K_2\times K_3

K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}

K=1.53\times 10^{-14}

K=\frac{[C]^2[O_2]^2}{[CO_2]^2}

On comparing the K and K_{goal}:

K^2=K_{goal}

K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}

The value of the equilibrium constant for reaction asked is 1.24\times 10^{-7}.

4 0
3 years ago
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