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maks197457 [2]
2 years ago
11

In a closed system a reversible chemical reaction will reach a state of dynamic chemical when the rate of the forward reaction i

s to/than the rate of the reverse reaction.
SAT
1 answer:
vitfil [10]2 years ago
6 0

In a closed system a reversible chemical reaction will reach a state of dynamic chemical equilibrium   when the rate of the forward reaction is to be in or equal to the rate of the reverse reaction.

<h3>What occurs if a reaction reaches dynamic equilibrium in a closed system?</h3>

A closed chemical system is said to attain dynamic equilibrium if the rate of the forward reaction is said to have equals the rate of the backward reaction and also the level of concentrations of products and reactants are said to be the same.

Hence, In a closed system a reversible chemical reaction will reach a state of dynamic chemical equilibrium   when the rate of the forward reaction is to be in or equal to the rate of the reverse reaction.

Learn more about reverse reaction from

brainly.com/question/16614855

#SPJ1

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The conclusion which can be drawn with respect to the nature of an exchange-rate table in December 2013 is that: B. It would look different because exchange rate tables change constantly.

An exchange rate can be defined as a number that is used to denote or represent the financial value of a currency in comparison with another currency.

On a related note, the foreign exchange market is a market that is used for converting the currency of a particular country into the currency of another country. For example, the dollars (United States of America) can be converted into naira (Nigeria) in the foreign exchange market.

In the foreign exchange market, the exchange rate for various countries are typically shown or displayed by using the exchange-rate table, which changes constantly depending on the following factors:

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2 years ago
suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a
saul85 [17]

Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.

Explanation:

Let a Force pushes the hockey puck of mass m.

Then acceleration, a= \frac{F}{m}a=mF

From the equation of motion,

\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)

In the second case, when mass is 2m, then acceleration,

a'=\frac{F}{2m}a′=2mF

and t' is the time taken.

The final speed is v,

\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)

Hence, it would take two times the previous amount of time to push the pluck of double mass.

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