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2 years ago

Complete the equilibrium reactions which are pertinent to an aqueous solution of ag2co3. physical states are optional.

1 answer:

6 0

The equilibrium reactions pertinent to an aqueous solution of Ag2CO3 in which physical states are not considered is:

$\mathbf{Ag_2CO_{3 \ (s)} \rightleftharpoons 2Ag^{+}_{(aq)} + CO_3^{2-}}$

<h3>What are equilibrium reactions?</h3>

An equilibrium reaction is a chemical reaction process in which the reactants remain in a steady and stable condition both previously(reactant) and after the reaction is completed(product).

A reaction is considered to be at thermodynamic equilibrium when a reaction meets all three categories of equilibrium;

Equilibrium of temperature
Equilibrium in the chemical reaction.
Equilibrium with the surrounding factors like pressure.

The equilibrium reactions pertinent to an aqueous solution of Ag2CO3 is:

$\mathbf{Ag_2CO_{3 \ (s)} \rightleftharpoons 2Ag^{+}_{(aq)} + CO_3^{2-}}$

Learn more about equilibrium reactions here:

brainly.com/question/1619133

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The average speed between t = 1.0 s and t = 4.0 s will be 2.1+0.3 which is 2.4m/s

<h3>Velocity function </h3>

Given the equation for the position of the mouse expressed as:

$x(t)=pt^2+qt\\$

The velocity of the particle is expressed as:

$v(t)=2pt+q$

Given that p = 0. 40 m/s and q = -1. 10 m/s, the equation will become:

$v(t)=2(0.4)t-1.10\\v(t)=0.8t - 1.10$

If the time is <em>t = 1.0 s, the velocity will be:</em>

$v(1) = 0.8 - 1.10\\v(1) = -0.3m/s$

Similarly if t = 4.0ssec, hence:

$v(4) = 3.2 - 1.10\\v(4) = 2.1m/s$

The average speed between t = 1.0 s and t = 4.0 s will be 2.1+0.3 which is 2.4m/s

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Complete question

<em>A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.40 m/s2and q = -1.10 m/s . Determine the mouse's average speed between t = 1.0 s and t = 4.0 s. I have tried everything and the answer is not 0.40 m/s</em>

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Soon Yi loves to bake, and she is making flaky pastry. Soon Yi starts with a layer of dough 222 millimeters (\text{mm})(mm)left

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When the dough is folded, it increases by a constant factor. We can model the growth of the thickness using the exponential growth model

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Initial thickness, $T_0$ = 2mm

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$2(1+0.08)^n\geq 2.5\\2(1.08)^n\geq 2.5\\$Divide both sides by 2$\\\dfrac{2(1.08)^n}{2}\geq \dfrac{2.5}{2}\\\\1.08^n\geq 1.25\\\\$Change to logarithm form\\n \geq \log_{1.08}1.25\\\\n\geq \dfrac{\log 1.25}{\log 1.08} \\\\n\geq 2.9$

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