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2 months ago

Fluorine-18 is a positron emitter used in pet scans. write a balanced nuclear equation for the reaction.

Chemistry

2 answers:

lorasvet [2.7K]2 months ago

8 0

Nuclear reaction: ¹⁸F → ¹⁸O + e⁺ (positron) + ve (electron neutrino).

Beta decay refers to a radioactive process where a nucleus emits a beta particle along with a neutrino.

There are two variants of beta decay: beta minus and beta plus.

In beta plus decay, a proton transforms into a neutron, a positron, and an electron neutrino, resulting in no change in the mass number. In this process, fluorine (with an atomic number of 9) converts into oxygen (atomic number 8) because it loses one proton.

In beta minus decay, a neutron is transformed into a proton, releasing an electron and an electron antineutrino.

VMariaS [2.9K]2 months ago

5 0

<span>Due to constraints in typography, I will describe the equation instead of providing it in writing. Crude representation. 18 18 0 F --> O + e 9 8 1 In detail, each of the three components includes both a left superscript and a left subscript, differing from the standard placement on the right side that is typically used. The equation depicts F, with a left superscript of 18 and a left subscript of 9, representing fluorine with an atomic weight of 18 and 9 protons. Followed by a right arrow showing the reaction's direction. This is followed by the letter O with a left superscript of 18 and a left subscript of 8, indicating oxygen with an atomic weight of 18 and 8 protons. Then a plus sign appears to indicate addition. Lastly, either the lowercase letter "e" or the uppercase Greek character beta, with a left superscript of 0 and a left subscript of 1 or +1, denotes the emission of a positron, which has a positive charge and an atomic weight of 0.</span>

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it takes 151 kJ/mol to break an iodine-iodine single bond. calculate the maximum wavelength of light for which an iodine-iodine

alisha [2963]

Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ

Where h is Planck's constant = 6.626 × 10⁻³⁴ js

c is the speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ

λ = hc / E

λ = (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

λ = 7.92 × 10⁻⁷ m

6 0

2 months ago

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

KiRa [2933]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × $\frac{1mol}{159,609g}$ = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × $\frac{1mol}{18,02g}$ = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0

2 months ago

What is the density (in g/L) of a gas with a molar mass of 16.01 g/mol at 1.75 ATM and 337 K?

Anarel [2989]

To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature.

P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k

Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L

8 0

1 month ago

Read 2 more answers

Compared to a 26-gram sample of NaCl(s) at STP, a 52-gram sample of NaCl(s) at STP has(1) a different density(2) a different gra

KiRa [2933]

Choice (3) is correct: they exhibit the same chemical properties. Both samples are under identical conditions, so their densities remain unchanged. Gram-formula mass is an intrinsic property of the substance and does not vary. The volume would change if the mass changes.

6 0

1 month ago

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

Anarel [2989]

Answer:

The amount of calcium sulfate that precipitates is 6.14 grams.

Explanation:

Step 1: Provided data

We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−

The Ksp for CaSO4 is 2.40×10^−5.

Step 2: Determine moles of Ca^2+

Moles of Ca^2+ = Molarity of Ca^2+ * Volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles of Ca^2+ = 0.05 moles

Step 3: Determine moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Compute total volume

Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Compute Q

Q = [Ca2+] [SO42-]

[Ca2+] = 0.050 M and [SO42-]

Qsp = (0.050)(0.050) = 0.0025 >> Ksp

This indicates that precipitation will take place.

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M (molar solubility)

Step 7: Determine total dissolved CaSO4

Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g

Step 8: Calculate initial mass of CaSO4

Initial moles of CaSO4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate precipitate mass

6.807 - 0.667 = 6.14 grams.

The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.

5 0

2 months ago