What can we use to help us predict the results of reaction when we put metals into competition?

1 answer:

7 0

The activity series of metals as well as the electrode potential of metals can be used to compare the reactivity of metals.

<h3>What is used in comparing reactivity of metals?</h3>

The reactivity of metals can be compared using their electrode potentials which is a measures of the ability of the metal to donate electrons to another metal.

When comparing the reactivity of metals, the metal with the lesser negative electrode potential will be more reactive than another with a greater negative or positive electrode potential.

Therefore, the activity series of metals as well as the electrode potential of metals can be used to compare the reactivity of metals.

Learn more about activity series of metals at: brainly.com/question/17469010

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A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of

mina [271]

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.125

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.125=e^{-0.00440\times t}$

t = 473 year

8 0

3 years ago

On the basis of the information above, a buffer with a pH = 9 can best be made by using

telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for $\mathbf{H_3PO_4 , H_2PO^{-}_4 , HPO_4^{2-}}$ are $\mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \ 5\times 10^{-13}}$ respectively.

$\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}$

$\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}$

$\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}$

The reason while option D is the best answer is that, the value of pKa for both

$\mathbf{H_2PO^{-}_4 ,\ \& \ HPO_4^{2-}}$ lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.