The energy required to remove an electron from a neutral atom is known as its:

A solution of [FeSCN]^{2+} is found to have 37% transmittance at 447 nm. If the molar absorption coefficient (ε) is 4400 at λ_{m

alexandr1967 [171]

Answer:9.81×10^-5 M

Explanation: See attachments

8 0

4 years ago

The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o

choli [55]

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

4 0

3 years ago

Read 2 more answers

Really need help now please help!!

Zielflug [23.3K]

I'm sorry but the picture isn't clear enough

7 0

3 years ago

At a consent tempture the pressure on a 2.5 l ballon is increased from 2.4 atm to 5.8 atm. what is the new volume?

marissa [1.9K]

Answer:

1.034 L

Explanation:

P1 V1 = P2 V2

P1 V1 / P2 = V2

2.4 (2.5) / 5.8 = V2 = 1.034 L

8 0

2 years ago

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago