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ira [324]
2 years ago
13

During a reaction, the enthalpy of formation of an intermediate is 34 kJ/mol.

Chemistry
1 answer:
galina1969 [7]2 years ago
5 0

Since the enthalpy can be calculated from the heat of formation, the enthalpy is 136 kJ/mol.

<h3>What is enthalpy?</h3>

The enthalpy of a reaction is the heat that is lost or gained in that reaction. We know that the enthalpy can be calculated from the heat of  formation.

Thus, we can obtain the enthalpy of the reaction as 4 * 34 kJ/mol = 136 kJ/mol.

Learn more about enthalpy:brainly.com/question/13996238

#SPJ1

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3 years ago
Identify the product of radioactive decay and classify the given nuclear reactions accordingly.
marta [7]

Answer:  a)   _{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}

b)  _{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}

c)  _{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}

 d) _{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}

e) _{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}

Explanation:

A balanced nuclear equation is one in which the atomic number and mass number remains same on both sides of the equation i.e the number of protons and neutrons remain same.

General representation of an element is given as:  

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a) _{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}

b)  _{92}^{238}\textrm{U}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}

c)  _{94}^{242}\textrm{Pu}\rightarrow _{92}^{238}\textrm{Ra}+_{2}^{4}\textrm{He}

 d)   _{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}

e)   _{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}

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4 mol*40.08g/mol = 160.32 grams of Ca

please give thanks by clicking heart button :)

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