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Lisa [10]
3 years ago
12

Which reaction occurs spontaneously?

Chemistry
2 answers:
Anna007 [38]3 years ago
7 0
The reaction occurs spontaneously is <span>Cl2(g) + 2NaBr(aq)-->Br2(l) + 2NaCl(aq). The answer is number 1. The rest of teh choices do not asnwer the question above.</span>
WARRIOR [948]3 years ago
4 0

The answer is: (1) Cl2(g) + 2NaBr(aq)-->Br2(l) + 2NaCl(aq).

In this chemical reaction chlorine change oxidation number from 0 to -1 (reduction) and bromine change oxidation number from -1 to 0 (oxidation).

Chlorine is stronger oxidation reagent than bromine.

In VIIA or group 17 (halogen elements) there are six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At) and tennessin(Ts).

They have high electronegativity because they have seven valence electrons in their outermost energy level, so they can gain an electron to have the octet rule.  

Going down in the group, element are weaker oxidazing reagent and less reactive.

Electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

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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
ryzh [129]

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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