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DiKsa [7]
3 years ago
7

Iron(II) is available to bond with chloride ion. How many of each type of ion will bond to form an ionic compound?

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

D) 1 iron(II), 2 chloride

Explanation:

Iron II chloride is the compound; FeCl2. It is formed as follows, ionically;

Fe^2+(aq) + 2Cl^-(aq) -----> FeCl2

The formation of one mole of FeCl2 involves the reaction one mole of iron and two moles of chloride ions. This means that in FeCl2, the ratio of iron to chlorine is 1:2 as seen above.

Therefore there is one iron II ion and two chloride ions in each mole of iron II chloride, hence the answer.

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Answer:

On this 40th anniversary of the moon landing, let's consider where these beliefs and urban legends come from.

Explanation:

im not sure.

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3 years ago
The volume of a bubble that starts at the bottom of a lake at 4.55°C increases by a factor of 8.00 as it rises to the surface wh
Masja [62]

Answer:

The depth of the lake is 67.164 meters.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas in bubble= ?

P_2 = final pressure of gas = 0.980 atm

V_1 = initial volume of gas = V

V_2 = final volume of gas = 8.00 × V

T_1 = initial temperature of gas = 4.55^oC=273.15+4.55=277.7 K

T_2 = final temperature of gas = 18.05^oC=273.15+18.05=291.2 K

Now put all the given values in the above equation, we get:

\frac{P_1\times V}{277.7 K}=\frac{0.980 atm\times 8.00\times V}{291.2 K}

P_1=7.476 atm

pressure of the gas in bubble initially is equal to the sum of final pressure and pressure exerted by water at depth h.

P_1=P_2+h\rho\times g

Where :

\rho = density of water = 1.00 g /cm^3=1000 g/m^3

g = acceleration due gravity = 9.8 m/s^2

7.476 atm=0.980 atm +h\rho\times g

6.496 atm=h\rho\times g

6.496 \times 101325 Pa=h\1000 g/m^3\times 9.8 m/s^2

h = 67.164 m

The depth of the lake is 67.164 meters.

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Answer:

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3 years ago
Determine the energy associated with an e- in hydrogen at n=2 to n=4
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Answer:

Kinetic energy of electron = -E₂ = 3.4 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

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