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maxonik [38]
3 years ago
13

Balance all the equations

Chemistry
1 answer:
siniylev [52]3 years ago
5 0

Answer:

Balancing Chemical Equations

Balance the equations below:

1) __1__ N2 + __3__ H2      >       _2__ NH3

2) __2__ KClO3  >⮴ __2__ KCl + __3__ O2

3) __2__ NaCl + __1__ F2 ⮴ __2__ NaF + __1__ Cl2

4) __2__ H2 + __1__ O2 ⮴ __2__ H2O

5) __1__ Pb(OH)2 + __2__ HCl ⮴ __2__ H2O + __1__ PbCl2

6)__2__ AlBr3 + __3__ K2SO4 ⮴ __6__ KBr + __1__ Al2(SO4)3

7) __1__ CH4 + __2__ O2 ⮴ __1__ CO2 + __2__ H2O

8) _1___ C3H8 + _5___ O2 ⮴ __3__ CO2 + __4__ H2O

9) __2__ C8H18 + __25__ O2 ⮴ __16__ CO2 + __18__ H2O

10) __1__ FeCl3 + __3__ NaOH ⮴ __1__ Fe(OH)3 + __3__NaCl

11) __4__ P + __5__O2 ⮴ __2__P2O5

12) __2__ Na + __2__ H2O ⮴ __2__ NaOH + __1__H2

13) __2__ Ag2O ⮴ __4__ Ag + __1__O2

14) __1__ S8 + __12__O2 ⮴ __8__ SO3

15) _6___ CO2 + _6___ H2O ⮴ __1__ C6H12O6 + __6__O2

16) ___1_ K + __1__ MgBr ⮴ __1__ KBr + __1__ Mg

17) __2__ HCl + __1__ CaCO3 ⮴ __1__ CaCl2 + __1__H2O + __1__ CO2

18) __1__ HNO3 + __1__ NaHCO3 ⮴ __1__ NaNO3 + __1__ H2O + _1___ CO2

19) _2___ H2O + __1__ O2 ⮴ __2__ H2O2

20) __2__ NaBr + __1__ CaF2 ⮴ __2__ NaF + __1__ CaBr2

Explanation:

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Please explain:) I would really appreciate a step by step explanation if possible.
UNO [17]

The enthalpy change of the reaction, ΔH = -311 kJ

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

<h3>What is the enthalpy change for the reduction of ethyne to form ethane?</h3>

The enthalpy change for the reaction is obtained from the summation of the enthalpies of the reactions of the intermediate steps according to Hess's law.

The equation of the reaction is given below:

  • C₂H₂ + 2 H₂ → C₂H₆

The enthalpy of the reaction, ΔH = ΔH₁ + 2ΔH₂ + (-ΔH₃)

ΔH = {(-1299) + (2 * -286) + (1560)}Kj

ΔH = -311 kJ

The equation for the methanation reaction is given below:

3 H₂O + CO → CH₄ + H₂O

The enthalpy for the methanation reaction is as follows:

ΔH = 1.5ΔH₁ + 0.5*(-ΔH₂) + ΔH₃ + -ΔH₄

ΔH = (-483.6 * 1.5) + (0.5 * 221.0) + (-802.7) + (393.5)

ΔH = -1024.1 kJ/mol

Molar mass of CO = 28 /mol

Enthalpy change involved in the reaction of 300 g of CO = 300/28 * -1024.1 kJ/mol

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

In conclusion, the enthalpy changes are calculated from the enthalpy values of the  intermediate reactions.

Learn more about enthalpy changes at: brainly.com/question/26991394

#SPJ1

7 0
2 years ago
Question 10 (1 point)
Vinil7 [7]

Answer:

A,C,D,B

Explanation:

1killometer=1000m

1mm=0.001m

1cm=0.01m

base unit of length is meter

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Answer:

42%

Explanation:

i don't know if it's right but...

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Explanation: I did this and that’s the answer

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