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GaryK [48]
3 years ago
6

For the following reaction in aqueous solution, identify all those species that will be spectator ions. Select all that apply.

Chemistry
1 answer:
uranmaximum [27]3 years ago
7 0

The reaction is incorrect. The correct reaction is: Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃. The options are:

A. Hg₂SO₄

B. Na₂SO₄

C. Na⁺

D. NO₃⁻

E. SO₄⁻²

F. Hg₂(NO₃)₂

G. NaNO₃

H. Hg⁺²

Answer:

C, D, E, H

Explanation:

The ions are formed after dissociation of the compound in the solution. When they're negatively charged, they're called anions, when they're negatively charged, they're called cations. If the ion is presented on both sides of the reaction, it is called a spectator ion.

Thus, the reaction given:

Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃

So, let's do the dissociation.

Na₂SO₄ is formed by the ions Na⁺ and SO₄⁻²;

Hg(NO₃)₂ is formed by the ions Hg⁺² and NO₃⁻;

HgSO₄ is formed by the ions Hg⁺² and SO₄⁻²;

NaNO₃ is formed by the ions Na⁺ and NO₃⁻.

Thus, the spectator ions are Na⁺, SO₄⁻², Hg⁺², and NO₃⁻.

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The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye
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Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

ln \frac{N}{N_0}  = - k*t ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

T _\frac{1}{2} = \frac{ln2}{k} ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of T_\frac{1}{2} in above formula (equation 2) :

28.8 yrs = \frac{ln 2}{ k }

Multiply both side by k

28.8 yrs * k = \frac{ln 2 }{k} * k

Dividing both side by 28.8 yrs

\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

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3 years ago
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