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2 years ago

Name the element described in each of the following:

Chemistry

1 answer:

musickatia [10]2 years ago

6 0

Period 4 transition element that forms 2+ ion with a half‐filled d sub level is Manganese (Mn)

What is the half-filled d sub-level?

Transition metals are an interesting and challenging group of elements. They have perplexing patterns of electron distribution that don’t always follow the electron-filling rules. Predicting how they will form ions is also not always obvious.

Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration. In addition, the majority of transition metals are capable of adopting ions with different charges. Iron, which forms either the Fe2+ or Fe3+ ions, loses electrons as shown below.

Some transition metals that have relatively few d electrons may attain a noble-gas electron configuration. Scandium is an example. Others may attain configurations with a full d sublevel, such as zinc and copper.

to know more about half-filled d sub-level

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Given the following, determine ΔG°f at 298 K for SnO. Sn(s) + SnO2(s) → 2SnO(s) ; ΔG° = 12.0 kJ at 298K

Virty [35]

Answer:

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

Explanation:

$Sn(s) + SnO_2(s)\rightarrow 2SnO(s), \Delta G_{f}^{o} = 12.0 kJ$

$\Delta G_{f,SnO_2}^{o}= -515.8kJ/mol$

$\Delta G_{f,Sn}^{o}= 0 kJ/mol$

$\Delta G_{f,SnO}^{o}=?$

$\Delta G_{f}^{o}=\sum ((\Delta G_{f}^{o})_ {products})-\sum ((\Delta G_{f}^{o})_ {reactants})$

$12.0 kJ=(2 mol\times \Delta G_{f,SnO}^{o})-(1mol\times 0 kJ/mol+1 mol\times -515.8 kJ/mol)$

$\Delta G_{f,SnO}^{o}=-251.9 kJ/mol$

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

3 0

3 years ago

One atom of silicon can properly be combined in a compound with

julia-pushkina [17]

Answer:

C. two atoms of oxygen.

Explanation:

Step 1: Data given

Silicon has 14 electrons

Silicon is part of Group IV, all the elements there have 4 valence electrons.

It can form a compound when 4 valence electrons bind with the 4 valence elctrons of silicon

A. four atoms of calcium.

Calcium has 2 valence elctrons. 4 atoms of calcium cannot bind on 1 atom of silicon since there are only 4 valence electrons.

B. one atom of chlorine.

1 atom of chlorine has 7 valence electrons. Chlorine can bind with an atom with 1 valence electron. Since silicon has 4 valence electrons, they will not bind.

Silicon can bind with 4 atoms of chlorine to form SiCl4

C. two atoms of oxygen.

Oxygen has 6 valence electrons, this means oxygen can bind with an element with 2 valence electrons.

Since silicon has 4 valence electrons, it can bind with 2 atoms of oxygen to form SiO2 (silicon dioxide).

D. three atoms of hydrogen.

Hydrogen has 1 valence electron. 1 hydrogen atom can bind with an element that has 7 valence electrons.

Three atoms of hydrogen can bind with an element that has 5 valence electrons.

Silicon will not bind with 3 atoms of hydrogen ( but can bind with 4 atoms of hydrogen)

5 0

3 years ago

A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104 $[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0

3 years ago

The crust and upper mantle make up Earth's ____.

Alekssandra [29.7K]

Answer:

a. the lithosphere

Explanation:

google

7 0

2 years ago

How many milliners of hydrogen gas ar produced by the reaction 0.020 moles of magnesium with excess of hydrochloride acid at sto

jeyben [28]

Answer:- 448 mL of hydrogen gas are formed.

Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:

$Mg+2HCl\rightarrow MgCl_2+H_2(g)$

There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.

moles of Hydrogen gas formed = 0.020 mol

At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.

$0.02mol(\frac{22.4L}{1mol})$

= 0.448 L

They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL

$0.448L(\frac{1000mL}{1L})$

= 448 mL

So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.

5 0

3 years ago