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Kobotan [32]
3 years ago
14

Based on molecular orbital theory, the bond orders of the H—H bonds in H 2 , H 2 + , and H 2 - are _________, respectively

Chemistry
1 answer:
vodka [1.7K]3 years ago
4 0

Answer:

H2-1

H2+-1/2

H22- zero

Explanation:

Bond order= Bonding electrons-antibonding electrons/2

In H2, there are two bonding electrons and no antibonding electrons. In H2+ there is only one bonding electron and no antibonding electron while in H22- there are two bonding and two antibonding electrons respectively.

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Consider the reaction at 500 ° C 500°C . N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) K c = 0.061 N2(g)+3H2(g)↽−−⇀2NH3(g)Kc=0.06
Sav [38]

Answer:

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L  N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.  

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.  

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.  

4 0
3 years ago
Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

V = 38.75 L

3 0
3 years ago
You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second jar
melisa1 [442]

since both the jars are kept at the same temperature the vapor pressure will be same in both the cases.


3 0
2 years ago
Scientist can use trees to look at climates of the past. How? What information can they gather and how do they gather it? Explai
andrezito [222]

Explanation:

Scientist use trees a whole lot to look at climate of the past by examining tree rings.

These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.

Tree rings can be used to decipher the age of a tree.

  • These three rings can be used to interpret climatic patterns.
  • During a wet climate, the tree rings are more robust and bigger.
  • In a dry climate, the rings are thinner.
  • These alternating patterns can be used to decipher the climatic signatures in a tree.
  • Sometimes, it is possible to evaluate some certain isotopes that are useful in climatic studies.

learn more:

Climate change brainly.com/question/7824762

#learnwithBrainly

6 0
2 years ago
Atoms full valence electron shells are chemically stable. Most noble gases have eight valence electrons. So, one bottle for chem
kap26 [50]
Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.
6 0
2 years ago
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