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1 year ago

Use each of the following reaction quotients to write the balanced equation:(b) Q = [NH₃]⁴[O₂]⁷ / [NO₂]⁴[H₂O]⁶

Chemistry

1 answer:

ira [324]1 year ago

7 0

4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g) is the balanced equation.

<h3>What is balanced equation?</h3>

If both the reactants and the products have the same number of atoms and total charge in each component of the equation, the equation describing the chemical reaction is said to be balanced. In other words, the mass and charge in the reaction are equal on both sides.

A chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulae, where the reactant entities are given on the left and the product entities on the right, with a plus sign between the entities in both the reactants and the products and an arrow that points towards the products, showing the direction of the reaction.

To learn more about balanced equation from the given link:

brainly.com/question/28297774

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If a molecule can hydrogen bond does it guarantee that it will have a higher boiling point than a molecule that cannot?

sleet_krkn [62]

A molecule that can h-bond will not always necessarily and does not have guarantee to have a higher boiling point than one than cannot have h-bond.
we can take an example of Pentan-2-one that cannot h-bond but instead of this it has a high boiling point that is 102.3 °C, while propan-1-ol can h-bond but it has a boiling point of 97.2°C, that is lower than the boiling point of Pentan-2-one.

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3 years ago

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2 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

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3 years ago

A scientific law is like a societal law because

natita [175]

A scientific law is like a societal law because they both are to be abided by their purposed way of doing things

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3 years ago

What is the effect of adding more water to the following reaction at equilibrium? CO2 + H2O H.CO.

LenaWriter [7]

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Explanation:

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3 years ago