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2 years ago

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How do I Balancing Chemical Equations

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Black_prince [1.1K]2 years ago

6 0

Answer:

what?

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The pH at the equivalence point will be a. less than 7.00 b. equal to 7.00 Allohalineup of a sto quivalence point of a titration

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Answer: c. greater than 7.00

Explanation: The equivalence point of a titration is when all the base is consumed by the acid. When a strong base and a strong acid react, the medium is neutralized because is produced water and salt (which won't suffer hydrolysis). How water's pH is 7, in this type of titration the pH of the equivalence point will be at pH=7. But on titration of a weak acid with a strong base, the reaction of the equivalence point produces water and the conjugate base of the acid. Because the acid is weak, their conjugate base will be strong and will suffer hydrolysis, producing hydroxyl ions, elevating the pH of the water and making it greater than 7.

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Which isotope is most commonly used in the radioactive dating of the remains of organic materials?

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(1) 14C also known as Carbon-14, is used to radio-date organic materials as well as earth materials.

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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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Using ethanol and sodium or potassium cyanide as the sources of the carbon atoms, along with any necessary inorganic reagents, s

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Answer:

Please see attachment

Explanation:

Please see attachment

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What is the main source of energy in stars?

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Answer:

nuclear fusion

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