All categories

1 month ago

Match the element or group to the rule assigning its oxidation state.

Chemistry

2 answers:

KiRa [2.9K]1 month ago

6 0

The correct sequence would be B, A, E, D, C. I hope this information assists you!

Alekssandra [3K]1 month ago

5 0

C. Isolated elements and atoms in gases have an oxidation state of 0. Elements in their free elemental state carry an oxidation number of zero. E. Elements in groups 1, 2, and 17, along with polyatomic ions, exhibit an ionic charge. Ions from Group 1, 2, and 17 are generated by alkali metals, alkaline metals, and halogens correspondingly, forming +1, +2, and -1 ions respectively. Polyatomic ions have charges of -1, -2, -3, like OH-, CO32-, and PO43-. B. Hydrogen carries an oxidation state of +1, but -1 when bonded to a diatomic metal. H generally has an oxidation state of +1. However, in metal hydrides like NaH (sodium hydride), it showcases an oxidation number of -1. A. Oxygen generally has an oxidation number of -2. D. Elements with varying oxidation numbers depend on other elements present in the compound.

You might be interested in

A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

3 0

1 month ago

Salt solutions can be __________ to give solid salts. What word completes this sentence?

alisha [2963]

Answer:

evaporated

Explanation:

Once the solution evaporates, only salt will remain, as the sole other component in the solution is water.

7 0

2 months ago

Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^

lorasvet [2795]

Fe 3+ + SCN- --> FeSCN 2+

.......Fe 3+.......SCN-.........FeSCN 2+ 
I.......0.04..........0.001.............. 
C........-x...............-x............. 
E.....0.04-x.....0.001-x...........x 

Keq = 203.4 = x / (0.04-x)(0.001-x) 
203.4 = x / (x^2 - 0.041x + 4x10^-5) 
203.4x^2 - 8.34x + 0.00094 = x 
203.4x^2 - 9.34x + 0.00094 = 0 
x = -0.0001M or 0.0458M 
therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+

5 0

2 months ago

Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a

lions [2927]

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

7 0

3 months ago

Read 2 more answers