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3 years ago

As AgNO3 dissolves how does the temperature of the solution change

1 answer:

3 0

AGNO3 (s) ----> Ag^+ (aq) + NO3^- (aq) { inorganic reaction }

the dissolution process is endothermic ( absorb energy ) , the crystallization process is exothermic ( release energy )

the temperature will go down and feel cool as the solute dissolves
the rate of crystallization and rate of dissolution will be equal or the same at equilibrium. the rate of dissolution will initially increase , because the process needs to absorb energy to occur , by heating the solution energy is added and the solubility of the solute will decreases more silver nitrate .

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Which of the following correctly represents the transmutation in which a curium-242 nucleus is bombarded with an alpha particle

Alekssandra [29.7K]

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

Transmutation is defined as the process in which one chemical isotope gets converted to another chemical isotope. The number of protons or neutrons in the isotope gets changed.

The chemical equation for the reaction of curium-242 nucleus with alpha particle (helium nucleus) follows:

$_{96}^{242}\textrm{Cm}+_4^2\textrm{He}\rightarrow _{98}^{245}\textrm{Cf}+_0^1\textrm{n}$

The product formed in the nuclear reaction are californium-245 nucleus and a neutron particle.

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3 years ago

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

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3 years ago

Why does increasing concentration usually increase reaction rate?

Alex Ar [27]

The correct answer is C) There are more particle collision

With more particle collision, more reactions are created.

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The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole

Oksanka [162]

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Explanation:

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