Chemistry: The radioisotope that has the longest half-life is the best to use in powering planet and space exploration vehicles because they can travel farther. which radioisotope is the best to use?

The radioisotope that has the longest half-life is

bhhzd7499

14.01.2022

Uranium-238 is the radioisotope which is best to use.

Explanation:

Half-life : The time in which the concentration of a reactant is reduced to half of its original value is called half life period.

As it has been given that radioisotope that has the longest half-life is best to use as power source planet and space exploration vehicles:

Uranium 238 is with the highest value of half life which is equal to $4.5\times 10^9$ years.

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azzyla2003

02.12.2021

First, find the concentration of the NaCN solution in terms of molarity. The molar mass of NaCN is 49.0072 g/mol

Molarity = mol solute ÷ L solution

Molarity = (353 mg * (1 g / 1000 mg) * (1 / 49.0072 g/mol)) ÷ (500 mL * (1 L / 1000 mL))

Molarity = 0.0144 M NaCN

NaCN is a salt from a weak acid HCN and a strong base NaOH. The reaction is as follows:

HCN + NaOH --> NaCN + H2O

Upon hydrolysis, it becomes:

$CN^{-}$ + $H_{2}O$ --> HCN + $OH^{-}$

Initial: Amount of $CN^{-}$ is 0.0144 M, $H_{2}O$ is in excess, and the products are 0.

Change: This is unknown. Let this be x. For the $CN^{-}$ , it would be –x, while it would be +x for the products HCN and $OH^{-}$ .

Excess: This would be Initial + Change. Thus for $CN^{-}$ , it would be 0.0144 – x. For both products, it would be x for each.

The hydrolysis constant ( $K_{H}$ ) would be the amount of products over the reactants. Thus,

$K_{H}$ = (HCN)*( $OH^{-}$ ) / ( $CN^{-}$ ) = x2 / (0.0144 –x) --> equation 1

$K_{H}$ is also equal to the equilibrium constant of water ( $K_{W}$ = 1 x 10^-14) divided by $K_{A}$ .

$K_{H}$ = 1 x $10^{-14}$ ÷ 2.1 x $10^{-9}$ --> equation 2

Substituting equation 2 to equation 1:

$x^{2}$ ÷ (0.0144 –x) = 4.762 x $10^{-6}$

Solving for x,

X = 2.59 x $10^{-4}$

As mentioned before, x denotes the products HCN and $OH^{-}$ . Thus the amount of $OH^{-}$ is 2.59 x $10^{-4}$

To find pH,

pH = 14-(-log $OH^{-}$ )

pH = 14 – (-log 2.59 x $10^{-4}$ )

pH = 10.41

Thus, the answer is 10.41.

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