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Romashka-Z-Leto [24]
3 years ago
9

An atom that has an atomic number of 38 and a mass number of 88 is an isotope of an atom that has An atom that has an atomic num

ber of 38 and a mass number of 88 is an isotope of an atom that has an atomic number of a. 38 and a mass number of 86. b. 50 neutrons and 38 protons. c. an atomic number of 39 and a mass number of 88. d. 50 protons and 38 neutrons.
Chemistry
1 answer:
gayaneshka [121]3 years ago
3 0

Answer: Option b. 50 neutrons and 38 protons.

Explanation:

Atomic number = proton number = 38

Mass number = proton number + Neutron

Mass number = 88

Proton = 38

Neutron number =?

Neutron number = Mass number — proton number

Neutron number = 88 —38 = 50

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The answer 
<span>the molar ratio for the following equation 
____C3H8 + ____O2 Imported Asset ____CO2 + ____ H2O

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4 0
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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
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Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

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Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

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Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

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⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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