humans are heating up the Earth's atmosphere is this an example of the law or theory support your answer

For the reaction 2NH3(g)↽−−⇀3H2(g)+N2(g) 2 NH 3 ( g ) ↽ − − ⇀ 3 H 2 ( g ) + N 2 ( g ) the equilibrium concentrations were found

forsale [732]

Answer:

0.324

Explanation:

The following data were obtained from the question:

Concentration of NH3, [NH3] = 0.25 M

Concentration of H2, [H2] = 0.3 M

Concentration of N2, [N2] = 0.75 M

Equilibrium constant (Kc) =.?

The balanced equation for the reaction is given below:

2NH3 <==> 3H2 + N2

The equilibrium constant, Kc for a given reaction is the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient. Thus, the equilibrium constant for the above reaction can be obtained as illustrated below:

Kc = [H2]³ [N2] / [NH3]²

Concentration of NH3, [NH3] = 0.25 M

Concentration of H2, [H2] = 0.3 M

Concentration of N2, [N2] = 0.75 M

Equilibrium constant (Kc) =.?

Kc = [H2]³ [N2] / [NH3]²

Kc = [0.3]³ × [0.75] / [0.25]²

Kc = (0.027 × 0.75) / 0.0625

Kc = 0.02025 / 0.0625

Kc = 0.324

Therefore, the equilibrium constant for the reaction is 0.324

4 0

3 years ago

A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:

hodyreva [135]

Answer:

-252.5 kJ/mol = ΔH H2O(g)

Explanation:

ΔH Fe2O3 = -825.5kJ/mol

ΔH H2 = 0kJ/mol

ΔH Fe = 0kJ/mol

Based on Hess's law, ΔH of a reaction is the sum of ΔH of products - ΔH of reactants. For the reaction:

Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(g)

ΔHr = 67.9kJ/mol = 3*ΔH H2O + 2*ΔHFe - (ΔH Fe2O3 + 3*Δ H2)

67.9kJ/mol = 3*ΔH H2O + 2*0kJ/mol - (ΔH -825.5kJ/mol + 3*Δ H2)

67.9 = 3*ΔH H2O(g) + 825.5kJ/mol

-757.6kJ/mol = 3*ΔH H2O(g)

6 0

3 years ago

A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil

Masteriza [31]

Answer : The value of $pK_a$ of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

$\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}$

$\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol$

Now we have to calculate the value of $pK_a$ of the weak acid.

The equilibrium chemical reaction is:

$HA+KOH\rightleftharpoons HK+H_2O$

Initial moles 0.25 0.03 0

At eqm. (0.25-0.03) 0.03 0.03

= 0.22

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HK]}{[HA]}$

Now put all the given values in this expression, we get:

$3.85=pK_a+\log (\frac{0.03}{0.22})$

$pK_a=4.72$

Therefore, the value of $pK_a$ of the weak acid is, 4.72

7 0

3 years ago

Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)

Digiron [165]

Answer: $=176.6kJmol^{-1}$

Explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

$\Delta H= {\text {sum of bond energies of reactants}}- {\text {sum of bond energies of products}}$

$\Delta H$ = {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

$\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} - {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}$

$=176.6kJmol^{-1}$

7 0

4 years ago

What is the minimum % relative error for a 0.5M NaOH solution?

scZoUnD [109]

Answer: 20%

Explanation:

Relative error = Absolute error/actual value

= 0.1/0.5

= 0.2

Hence,

Minimum % Relative error = 0.2 ×100%

= 20%

4 0

4 years ago

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humans are heating up the Earth's atmosphere is this an example of the law or theory support your answer ​

humans are heating up the Earth's atmosphere is this an example of the law or theory support your answer