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pav-90 [236]
3 years ago
11

The equation, 2 H2(g) + O2(g) Imported Asset 2 H2O(l), represents an equation for a standard formation reaction.

Chemistry
2 answers:
disa [49]3 years ago
6 0

Answer : The given statement is false.

Explanation :

The given chemical reaction of formation of H_2O is:

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

Standard formation reaction :

In the standard formation reaction, one mole of substance always produced from its element that are present in their standard states.

Thus, the standard formation reaction of H_2O will be:

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)

Hence, the given statement is false.

VLD [36.1K]3 years ago
5 0
If this is asking if the equation is balanced, than the answer is true
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A 28.5 g piece of gold is heated and then allowed to cool. What is the change in temperature (°C) if the gold releases 0.106 kJ
Shkiper50 [21]

Considering the definition of calorimetry and sensible heat, the change in temperature if the gold releases 0.106 kJ of heat as it cools is 28.78°C.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change). It is calculated by the expression:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • Q= 0.106 kJ= 106 J (being 1 kJ=1000 J)
  • c= 25.4 \frac{J}{molC}
  • m= 28.5 g×\frac{1 mole}{196.967 g} = 0.145 moles (being 196.967 g/mole the molar mass of gold)
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Replacing:

106 J= 25.4 \frac{J}{molC}× 0.145 moles× ΔT

Solving:

ΔT=\frac{106 J}{25.4\frac{J}{molC}x0.145 moles}

ΔT= 28.78 C

The change in temperature if the gold releases 0.106 kJ of heat as it cools is 28.78°C.

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