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lora16 [44]
3 years ago
8

The formula in my book for carbonate is Co32- and I’m confused because carbon has 4 valence electrons and oxygen has 6 . How doe

s it end up being CO32-?? Please help
Chemistry
2 answers:
polet [3.4K]3 years ago
6 0
The total charge of the carbonate ion is -2. O2 in CO32- has a total charge of -6 (3 x -2). Hence, to find the charge of carbon -> x + -6 = -2. Therefore, carbon has a charge of +4.
This matches your theory stated whereby
1. Carbon has 4 valence electrons so in this case, it loses 4 electrons to obtain a stable noble octet config. & hence has a charge of +4.
2. Oxygen has 6 valence electrons so in this case, it gains 2 electrons & hence has a charge of -2.
balandron [24]3 years ago
4 0
Hello!!!

what exactly are you wanting to know? It isn’t clear from your question. Please advise and I will answer it for you ASAP. I think I know what you are wanting but I don’t want to assume.... thank you
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A simple equation relates the standard free‑energy change, ΔG∘′, to the change in reduction potential. ΔE0′. ΔG∘′ = −nFΔE0′ The
MA_775_DIABLO [31]

Answer :  The value of standard free energy is, -152.4 kJ/mol

Explanation :

The given balanced cell reaction is:

FADH_2+\frac{1}{2}O_2\rightarrow FAD+H_2O

The half reaction will be:

Reaction at anode (oxidation) : FADH_2\rightarrow FAD+2H^++2e^-     E^0_{Anode}=+0.03V

Reaction at cathode (reduction) : \frac{1}{2}O_2+2H^++2e^-\rightarrow H_2O     E^0_{Cathode}=+0.82V

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=(+0.82V)-(+0.03V)=+0.79V

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard free energy = ?

n = number of electrons transferred = 2

F = Faraday constant = 96.48kJ.mol^{-1}V^{-1}

E^o_{cell}  = standard electrode potential of the cell = 0.79 V

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

\Delta G^o=-(2)\times (96.48kJ.mol^{-1}V^{-1})\times (0.79V)

\Delta G^o=-152.4kJ/mol

Therefore, the value of standard free energy is, -152.4 kJ/mol

8 0
3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

4 0
2 years ago
There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen
alekssr [168]

Answer:

2.765amu is the contribution of the X-19 isotope to the weighted average

Explanation:

The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:

X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21

The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:

19.00amu * 0.1455 =

2.765amu is the contribution of the X-19 isotope to the weighted average

8 0
2 years ago
Is the continental crust less dense or more dense ?
zheka24 [161]

Answer:

It is less dense. It is also less dense than the oceanic crust.

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4 0
2 years ago
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iren2701 [21]
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
5 0
3 years ago
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