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ladessa [460]
3 years ago
13

A certain AB4 molecule has a square-planar molecular geometry.

Chemistry
1 answer:
harina [27]3 years ago
3 0

Answer:  b). The B−A−B angles between neighboring B atoms are 90∘.

Explanation:  The square planar geometry is the geometry in which the four groups are connected through an angle of 90∘ and all 5 atoms are in one plane.

It is not necessary for the molecule to have two non bonding pair of electrons on atom A but if it does have , then its parent geometry will be distorted octahedral. And if it doesn't have this non bonding pair of electrons then the geometry would be the square planar.

Thus the molecule has four electron domains about the central atom A if the parent geometry is square planar otherwise it would have 8 electron domain in distorted octahedral geometry.

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__Mg + __Fe2O3 -> __MgO + __Fe
Anika [276]

Answer:

Your correct answer is A. 1,2,1,2

Explanation:

Please mark brainliest!

7 0
3 years ago
Which of the following is a galvanic cell?
sladkih [1.3K]

C. Aluminum (Al) oxidized, zinc (Zn) reduced

<h3>Further explanation</h3>

Given

Metals that undergo oxidation and reduction

Required

A galvanic cell

Solution

The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

\large {\boxed {\bold {E ^ ocell = E ^ ocatode -E ^ oanode}}}

or:  

E ° cell = E ° reduction-E ° oxidation  

For the reaction to occur spontaneously (so that it E cell is positive), the  E° anode must be less than the E°cathode

If we look at the voltaic series:

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>

The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)

<em />

From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.

7 0
3 years ago
Read 2 more answers
2. _____ is the distance and direction of an object’s change in position from the starting   a.  Speed b.  Displacement   c.  Ve
Llana [10]
Displacement is the distance and direction of an object's change in position from the starting.
Hence option B is correct.

Hope this helps!
3 0
3 years ago
What is the formula to find the area of circle​
Advocard [28]

Answer:

A= π * r *2

Explanation:

6 0
3 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
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