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3 years ago

The energy required to remove the third electron from a dipositive ion is

Chemistry

2 answers:

jolli1 [7]3 years ago

5 0

The energy required is Ionization energy

const2013 [10]3 years ago

5 0

Third ionization energy

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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?

Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider $Mg(OH)_{2}$ as an insoluble substance, though it may be moderately soluble.

The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:

$Mg(NO_{3})_{2}_{(aq)}$ + $2NaOH_{(aq)}$ → $Mg(OH)_{2} {(s)}$ + $2NaNO_{3}_{(aq)}$

8 0

3 years ago

How do polar molecules respond when they are placed between two metal plates, one positively charged and one negatively charged?

kakasveta [241]

Gradpoint

The molecules align to the electric field

5 0

3 years ago

Read 2 more answers

indicate whether the entropy of the system increases or decreases. Mixing 10 mL of 90.0 °C water with 10 mL of 10 °C water. The

Ivanshal [37]

Answer:

When the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy

Explanation:

The answer that "the entropy will is increases" is correct as:

The water at 90° C i.e at higher temperature is mixed with the water at 10° C i.e the water at the lower temperature.

The water at lower temperature will have molecules with lower energy while the water with higher temperature will have molecules undergoing high thermal collisions. Thereby, when the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy.

Therefore, the answer is correct with respect to the water at lower temperature.

Meanwhile, for the water at higher temperature , the temperature of the system will decrease. Thus, the entropy of the water at higher level will decrease.

5 0

4 years ago

0.350 mol of a solid was dissolved in 260 mL of water at 21.2 oC. After the solid had fully dissolved, the final temperature of

Fittoniya [83]

Answer: Heat of the solution = mass water × specific heat water × change in temperature

mass water = 260ml (1.00g/ml ) = 260g

specific heat of water = c(water) = 4.184J/ g°C

Heat change of water = final temperature - initial temperature

= 26.5 - 21.2

= 5.3 °C

H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J

Molar heat = $\frac{5765J}{0.350mol}$

= 16473J/mol

Explanation: finding molar heat requires first to look at specific heat of water and the change of water temperature

7 0

3 years ago