Use the information below, to write a word equation, then change it into a balanced symbol equation.

While a certain isotope decays, it emits photons. What kind of decay is happening?

The correct answer is Gamma decay. It happens after beta and alpha decay because what's left after those two can enter a new process of gamma decay. This releases gamma rays which is a more complex term for the photons that you mentioned before. These rays can be dangerous for humans so care not to get caught in them.

6 0

2 years ago

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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago

Do the number of protons in an atom of an element always equals its atomic weight.

Elenna [48]

No. The number of a protons is not equal to it's atomic weight, instead it is equal to the 'atomic number'

5 0

3 years ago

What is the percent of S in

ollegr [7]

The percentage of the sulfur (S) in the compound CuSO₄ is 20.1 %.

<h3>What is the mass percentage?</h3>

The percentage of an element in a compound can be determined as the number of parts by mass of that element present in 100 parts by mass of the given compound.

First, calculate the molecular mass of the given compound by the addition of the atomic masses of all the present elements in the molecular formula. Then, the percentage of the elements can be determined by dividing the total mass of the element by the molar mass of the compound multiplied by 100.

Given, the atomic mass of copper, sulfur, and oxygen is 63.55 g, 32.07 g, and 16.0g respectively.

The molecular mass of CuSO₄ = 63.55 + 32.07 + 4(16.0) = 159.62 g

The mass percentage of the sulphur = (32.07/159.62) × 100 = 20.1 %

Therefore, the mass percentage of the sulfur is equal to 20.1 %.

Learn more about the mass percentage, here:

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7 0

10 months ago

Monica wrote the following observations about stars: "I observed a pattern of five stars at 8:00 p.m. and again at 9:00 p.m. The

stiv31 [10]

The answer would be b