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dalvyx [7]
3 years ago
8

“Na2SO3” represents sodium sulfite.What does the 4 mean in the formula 4Na2SO3?

Chemistry
2 answers:
Strike441 [17]3 years ago
6 0

The mass of substance which consists the equal number of particles as 12 grams of Carbon-12 is said to be mole.

For example:

The balanced chemical equation for production of water from hydrogen and oxygen is:

2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

That means 2 moles of H_2 reacts with 1 mole of O_2 to give 2 moles of  H_2O.

The given formula is 4Na_2SO_3.

Hence, 4 represents the number of moles in 4Na_2SO_3 formula that is 4 moles of Na_2SO_3.

The number of atoms of sodium, Na in 4Na_2SO_3 is:

4 \times 2 = 8

The number of atoms of sulfur, S in 4Na_2SO_3 is:

4 \times 1 = 4

The number of atoms of oxygen in, O in 4Na_2SO_3 is:

4 \times 3 = 12

Masteriza [31]3 years ago
3 0
You would find four put in front of the compound in an equation.  This is to indicate that there are four times more moles in the reaction than there is in the specie that has no number in front of it in the reaction.

  
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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
weqwewe [10]

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.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

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)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

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

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

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

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

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iVinArrow [24]

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Explanation:

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