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dusya [7]
3 years ago
14

Balance the following chemical equations. 15. HgO + Cl2 ,HgCl + 02

Chemistry
1 answer:
Mamont248 [21]3 years ago
5 0

Answer:

ans is that

Explanation:

2hgo+cl2->2hgcl+o2

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Hess’s law
Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

<h3>What is Hess' law?</h3>

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

ΔH = 90 kJ

Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

Learn more about enthalpy and Hess law at: brainly.com/question/9328637

8 0
2 years ago
A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you
Anna71 [15]

Answer:

<h2>1.89 atm</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{1.23 \times 23}{15}  =  \frac{28.29}{15}  \\  = 1.886

We have the final answer as

<h3>1.89 atm</h3>

Hope this helps you

7 0
2 years ago
James exclaims to his chemistry teacher that a new element has been discovered that fits between Nickel and Copper on the period
myrzilka [38]

Aye you have the same class as me bruh I need help on some chemistry qustions

8 0
3 years ago
Read 2 more answers
AdiffusioncoupleofweldedAandBwasgivenadiffusionannealatelevatedtemperatureandthencooledtoroomtemperature.Chemicalanalysisofthesu
Murljashka [212]

Answer:

what

Explanation:

8 0
3 years ago
Select all of the following statements that are false about ΔGo and ΔG:a) If the reaction has a large negative ΔGo value, the re
OLEGan [10]

Answer:

a) If the reaction has a large negative ΔGo value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔGo and ΔG have the same magnitude, they just have opposite signs.

Explanation:

The fraction of the total heat energy if a system that does useful work is known as Gibb's free energy (G) and the change from the initial to final state is designated by \Delta G. It is observed that the values of \Delta G changes with experimental conditions such as temperature , pressure , concentration etc.

\Delta G^0 is the standard free energy change which is a balance of two natural tendencies of any system.

  1. Minimization of potential energy or enthalpic factor \Delta H^0

Maximization of disorderliness or entropic factor T\Delta S^0

Mathematically; \Delta G = \Delta H^0 - T\Delta S^0

Thus; from above mentioned, the statements that are true about ΔG⁰ and ΔG are:

ΔG⁰ and ΔG can have different values, they don't even have to have the same sign

For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point

If ΔG⁰ , measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.

while the false statements include:

a) If the reaction has a large negative ΔG⁰ value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔG⁰ and ΔG have the same magnitude, they just have opposite signs.

7 0
3 years ago
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