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3 years ago

Uranium-232 has a half-life of 68.8 years. After 344.0 years, how much uranium-232 will remain from a 375.0-g sample?

Chemistry

1 answer:

N76 [4]3 years ago

8 0

Uranium-232 will remain 11.72 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

t 1/2 = half-life = 68.8 years

T = duration of decay = 344 years

No= initial sample = 375 g

Uranium-232 remain :

$\tt Nt=375.\dfrac{1}{2}^{344/68.8}\\\\Nt=375.\dfrac{1}{2}^5\\\\Nt=11.72~g$

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What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9

vladimir2022 [97]

<u>Answer:</u> The pH of resulting solution is 9.08

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ........(1)

<u>For HCl:</u>

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

$0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol$

<u>For ammonia:</u>

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

$0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol$

The chemical reaction for hydrochloric acid and ammonia follows the equation:

$HCl+NH_3\rightarrow NH_4Cl$

Initial: 0.006 0.01

Final: - 0.004 0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L (Conversion factor: 1 L = 1000 mL)

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})$

We are given:

$pK_b$ = negative logarithm of base dissociation constant of ammonia = $-\log (1.8\times 10^{-5})=4.74$

$[NH_4Cl]=\frac{0.006}{0.035}$

$[NH_3]=\frac{0.004}{0.035}$

pOH = ?

Putting values in above equation, we get:

$pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.92=9.08$

Hence, the pH of the solution is 9.08

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3 years ago

THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.

Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

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3 years ago

Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if 2 moles of pentane (C5H12) were supplied, and an unlimited amount

Kamila [148]

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Explanation:

We have the following combustion of pentane (C₅H₁₂):

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Knowing the chemical reaction we devise the following reasoning:

if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O

then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O

X = (2 × 6) / 1 = 12 moles of water H₂O

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combustion of organic compounds

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3 years ago

You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca

vova2212 [387]

The equation to be used are:

PM = ρRT
PV = nRT
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The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
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3 years ago

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Elena-2011 [213]

Answer:

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3 years ago