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maksim [4K]
3 years ago
8

Write the isotope notation for an element with 5 protons and 6 neutrons.

Chemistry
1 answer:
Korolek [52]3 years ago
4 0

The isotope notation : \large {{{11} \atop {5}} \right X}

<h3> Further explanation </h3>

Given

53 protons and 20 neutron

Required

The isotope notation

Solution

Isotopes are elements that have the same atomic number with a different mass number

The following element notation,

 \large {{{A} \atop {Z}} \right X}

X = symbol of element

A = mass number

   = number of protons + number of neutrons

Z = atomic number

   = number of protons = number of electrons, on neutral elements

So the mass number of element = 5 + 6 11

Atomic number = 5

The symbol :

\large {{{11} \atop {5}} \right X}

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Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO
Irina-Kira [14]

Answer : The enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of H_2SO_4 will be,

S+H_2+2O_2\rightarrow H_2SO_4    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) S+O_2\rightarrow SO_2     \Delta H_1=-298.2

(2) SO_2+\frac{1}{2}O_2\rightarrow SO_3    \Delta H_2=-98.2

(3) SO_3+H_2O\rightarrow H_2SO_4    \Delta H_3=-130.2

(4) H_2+\frac{1}{2}O_2\rightarrow H_2O    \Delta H_4=-285.8

Now adding all the equations, we get the expression for enthalpy of formation of H_2SO_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)

\Delta H=-812.4kJ/mol

Therefore, the enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

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A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?
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Answer:

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Explanation:

Given data:

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Mass of CuSO₄ = 3.28 g

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Solution:

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2Al + 3CuSO₄   →   Al₂ (SO₄)₃ + 3Cu

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now we will compare the moles of reactant with product.

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                 2          :             1

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                Al           :            Cu

                 2            :              3

               0.05         :            3/2×0.05 = 0.075 mol

         CuSO₄           :           Al₂ (SO₄)₃

                3             :             1

               0.02         :          1/3×0.02=0.007 mol

         CuSO₄           :            Cu

               3               :              3

               0.02         :              0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

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