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WITCHER [35]
3 years ago
5

Select the correct answer from each drop-down menu. At chemical equilibrium, the amount of because .

Chemistry
1 answer:
Elena-2011 [213]3 years ago
3 0

Answer:

The answer that completes the question are in BOLD:

At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.

Explanation:

In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.

The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.

When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.

N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.

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How to write half reaction for H2O2(aq) -> O2(g)
Ghella [55]
<span>The overall reaction is as follows:
H2O2 + 2ClO2 => 2ClO2- + O2 + 2H+ 
Half cell reactions:
</span><span>ClO2 + e- => ClO2- 
</span><span>H2O2 => O2 + 2H+ + 2e- </span>
7 0
2 years ago
The sources of errors in the kinetic experiment are:
Alexus [3.1K]

Answer: D) All of the above

Explanation:

The sources of errors in the kinetic experiment are depends upon the tools, due to some personal errors and also due to some methodological problem. As, experimental error during kinetic study is due to the difficulty to characterize the experimental errors depend on reaction conditions. The behavior of the experimental errors are depend on specific feature of the experiment systems.

4 0
3 years ago
HEEEEeEELP PLEasE! IS SCIENCE
Fynjy0 [20]
Gas and radio waves I’m pretty sure is correct
6 0
2 years ago
What product or ratio of properties remains constant (PxT, TxV, P/V, V/T, etx)?When you have decided, label the "constant?" colu
Naddika [18.5K]

Answer:

In this phenomenon we talk about ideal gases, that is why in these equations the constant is the number of moles and the constant R, which has a value of 0.082

Explanation:

The complete equation would have to be P x ​​V = n x R x T

where n is the number of moles, and if it is not clarified it is because they remain constant, as the question was worded.

On the other hand, the symbol R refers to the ideal gas constant, which declares that a gas behaves like an ideal gas during the reaction, and its value will always be the same, which is why it is called a constant. The value of R = 0.082.

The ideal gas model assumes that the volume of the molecule is zero and the particles do not interact with each other. Most real gases approach this constant within two significant figures, under pressure and temperature conditions sufficiently far from the liquefaction or sublimation point. The real gas equations of state are, in many cases, corrections to the previous one.

The universal constant of ideal gases is not a fundamental constant (therefore, choosing the temperature scale appropriately and using the number of particles, we can have R = 1, although this system of units is not very practical)

3 0
3 years ago
What is the pH of this solution?
Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M

Which is also:

[OH^-]=1.643x10^{-3}M

Thereafter we can compute the pOH first:

pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784

Finally, the pH turns out:

pH=14-2.784\\\\pH=11.216

Regards!

5 0
3 years ago
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