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3 years ago

When Iodine-131 emits a β particle (beta particle), what nuclide is produced? *

Chemistry

2 answers:

k0ka [10]3 years ago

5 0

When Iodine-131 emits a β particle will produce Xe-131

<h3>Further explanation </h3>

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,

alpha α particles ₂He⁴
beta β ₋₁e⁰ particles
gamma particles ₀γ⁰
positron particles ₁e⁰
neutron ₀n¹

So for reaction Iodine-131 :

$\tt _{53}^{131}I\rightarrow _{-1}^0\beta +_{54}^{131}Xe$

erastovalidia [21]3 years ago

5 0

Question, Answer, and Explanation are within the picture of files:

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Calculate the enthalpy of reaction for 2CO + O2 → 2CO2. given the following bond energies:

kotegsom [21]

Answer : The enthalpy change for the reaction is 1043 kJ/mol.

Explanation :

The given chemical reaction is:

$2CO+O_2\rightarrow 2CO_2$

As we know that:

The enthalpy change of reaction = E(bonds broken) - E(bonds formed)

$\Delta H=[(2\times B.E_{C\equiv O})+(1\times B.E_{O\equiv O})]-[2\times B.E_{C=O}]$

Given:

$B.E_{C\equiv O}$ = 1074 kJ/mol

$B.E_{O\equiv O}$ = 499 kJ/mol

$B.E_{C=O}$ = 802 kJ/mol

Now put all the given values in the above expression, we get:

$\Delta H=[(2\times 1074kJ/mol)+(1\times 499kJ/mol)]-[2\times 802kJ/mol]$

$\Delta H=1043kJ/mol$

Therefore, the enthalpy change for the reaction is 1043 kJ/mol.

7 0

3 years ago

1) Calculate the empirical formula of a compound with 36g of carbon and 96g of oxygen.

grandymaker [24]

Answer:

C4H6

Explanation:

See attached table

Convert each of the masses into moles by dividing the mass by the molar mass of that element. That yields 3.83 moles of C and 6 moles of O. I rounded up the C to 4 moles to result in an empirical formula of C4H6

5 0

2 years ago

An aqueous solution of methylamine (ch3nh2) has a ph of 10.68. how many grams of methylamine are there in 100.0 ml of the soluti

ruslelena [56]

Answer:

3.4 mg of methylamine

Explanation:

To do this, we need to write the overall reaction of the methylamine in solution. This is because all aqueous solution has a pH, and this means that the solutions can be dissociated into it's respective ions. For the case of the methylamine:

CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻ Kb = 3.7x10⁻⁴

Now, we want to know how many grams of methylamine we have in 100 mL of this solution. This is actually pretty easy to solve, we just need to write an ICE chart, and from there, calculate the initial concentration of the methylamine. Then, we can calculate the moles and finally the mass.

First, let's write the ICE chart.

CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻ Kb = 3.7x10⁻⁴

i) x 0 0

e) x - y y y

Now, let's write the expression for the Kb:

Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]

We can get the concentrations of the products, because we already know the value of the pH. from there, we calculate the value of pOH and then, the OH⁻:

The pOH:

pOH = 14 - pH

pOH = 14 - 10.68 = 3.32

The [OH⁻]:

[OH⁻] = 10^(-pOH)

[OH⁻] = 10^(-3.32) = 4.79x10⁻⁴ M

With this concentration, we replace it in the expression of Kb, and then, solve for the concentration of methylamine:

3.7*10⁻⁴ = (4.79*10⁻⁴)² / x - 4.79*10⁻⁴

3.7*10⁻⁴(x - 4.79*10⁻⁴) = 2.29*10⁻⁷

3.7*10⁻⁴x - 1.77*10⁻⁷ = 2.29*10⁻⁷

x = 2.29*10⁻⁷ + 1.77*10⁻⁷ / 3.7*10⁻⁴

x = [CH₃NH₂] = 1.097*10⁻³ M

With this concentration, we calculate the moles in 100 mL:

n = 1.097x10⁻³ * 0.100 = 1.097x10⁻⁴ moles

Finally to get the mass, we need to molar mass of methylamine which is 31.05 g/mol so the mass:

m = 1.097x10⁻⁴ * 31.05

<h2>m = 0.0034 g or 3.4 mg of Methylamine</h2>

3 0

3 years ago

Which statement about anabolism is true?

skelet666 [1.2K]

Answer:

i would say B? Im sorry if im wrong

Explanation:

i think that because Catabolic reactions release energy, while anabolic reactions use up energy. Anabolism is the opposite of catabolism.

8 0

2 years ago

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Which of these is a chemical property?

Katyanochek1 [597]

Yes, refer to the previous answer.

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3 years ago

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