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3 years ago

A 7.85 × 10-5 mol sample of copper-61 emits 1.47 × 1019 positrons in 90.0 minutes. what is the decay constant for copper-61

1 answer:

5 0

4.14x10^-3 per minute
First, calculate how many atoms of Cu-61 we initially started with by
multiplying the number of moles by Avogadro's number.

7.85x10^-5 * 6.0221409x10^23 = 4.7273806065x10^19
Now calculate how many atoms are left after 90.0 minutes by subtracting the
number of decays (as indicated by the positron emission) from the original
count.
4.7273806065x10^19 - 1.47x10^19 = 3.2573806065x10^19
Determine the percentage of Cu-61 left.
3.2573806065x10^19/4.7273806065x10^19 = 0.6890455577
The formula for decay is:
N = N0 e^(-Î»t)
where
N = amount left after time t
N0 = amount starting with at time 0
Î» = decay constant
t = time
Solving for Î»:
N = N0 e^(-Î»t)
N/N0 = e^(-Î»t)
ln(N/N0) = -Î»t
-ln(N/N0)/t = Î»
Now substitute the known values and solve:
-ln(N/N0)/t = Î»
-ln(0.6890455577)/90m = Î»
0.372447889/90m = Î»
0.372447889/90m = Î»
0.00413830987 1/m = Î»
Rounding to 3 significant figures gives 4.14x10^-3 per minute as the decay
constant.

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The presence of OH- in the solution shows that the solution is basic or alkaline.

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A base accepts hydrogen ion and the concentration of OH is always higher in base.

There is a presence of conjugate acid and conjugate base in the Bronsted and Lowry acid and base.

Conjugate acid is one which is formed when a base gained a proton.

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And from the Arrhenius base Theory, the base is one that dissociates in to water as OH-.

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3 years ago

Bridget is in science class. Her teacher gives her two unknown substances and asks her to determine their relative pH. She place

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Answer:

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Explanation:

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3 years ago

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

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Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

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3 years ago

g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint

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Answer:

The value is $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

The molar mass of $CaCO_3$ is $MW = 100.085 \ g/mol$

The total mass is $m_g = 12.3 \ g$

The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

The percentage of calcium in calcite is mathematically represented as

$C = \frac{40.07}{100.085} * 100$

$C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

$m= \frac{m_g}{5}$

$m= \frac{12.3}{5}$

$m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

$m_c = 2.46 * \frac{40.04}{100}$

$m_c = 0.985 \ g$

The uncertainty of mass of a single sample is mathematically represented as

$k = \frac{\Delta g }{5}$

$k = \frac{0.1 }{5}$

$k = 0.02\ g$

The uncertainty of mass of calcium in a single sample is mathematically represent

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$G = 0.00801 \ g$

Generally the average mass of calcium in each sample is

$L = 0.985 \pm 0.00801$

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Answer:

1. First

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