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What molecular geometries are associated with AB3?
Trygonal pyramid
What is the GCF of a2b3 and ab3?
The GCF is ab3
Why don't all molecules with the general atomic formula AB3 have the same shape?
because the dipoles changes from different AB3 molecule and the change of the bonding electrons pairs and the lone electrons pairs. eg. BF3 has (3BP) the shape is trigonal planar PCl3 has (3BP and 1LP) the shape is trigonal pyramidal BrF3 has (3BP and 2Lp) the shape is T-shaped
What does a cubic yard of ab3 rock weight?
1000 pounds
A and b are prime numbers ab3 equals 54 find the values of a and b?
54 = 2 x 3 x 3 x 3 = 2 x 33. Thus if ab3 = 54 and a, b are prime, then a=2, b=3.
How do you factor a5 plus b5 equals?
a5+b5 = (a+b) (a4-a3b+a2b2-ab3+b4)
What is (ab3)4?
Assuming that you want (ab^3)^4, which is impossible to ask given the crap browser used by Answers, the solution is A^4b^12.
What is the qoutient when 32a 2 b 12 ab3 c is divided by 8 ab?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals", "squared", "cubed" etc. Please use "brackets" (or parentheses) because it is impossible to work out whether x plus y squared is x + y^2 of (x + y)^2.
Is AlN polar or non polar?
One can determine the polarity of AlBr3 by first drawing a simple Lewis structure. We know that Aluminum has three (3) valence electrons and Bromine has seven (7) valence electrons. To figure out the total number of valence electrons in the molecule before drawing the structure we do the following: V = 3 + (7 x 3) V = 24 electrons in AlBr3 Now we can draw a simple Lewis structure: Br - Al - Br l Br Each Bromine has one single bond with Aluminum and 3 lone pairs of electrons. If we calculate that each bond is two electrons and each Bromine has 6 electrons (3 lone pairs) then: 6 + 6 + 6 (from the Bromine atoms) + 2 + 2 + 2 (from the three bonds) = 24 electrons So, we know that our structure is correct. Now we check the VSEPR type. The letter 'A' represents the central atom (in our case, Aluminum), the letter 'B' represents the number of bonding atoms to the central element (Bromine) and the letter 'E' represents the number of lone electron pairs on the CENTRAL atom (Aluminum). From this we can tell that our VSEPR configuration is AB3 (no 'E' value as there are no lone pairs of electrons on Aluminum). Finally, we match our VSEPR configuration, AB3, with what VSEPR says about polarity. Because our molecule consists of a central atom surrounded by three bonding elements that are the same, our molecule is NON-POLAR. *Note: If the molecule has a central atom surrounded by three bonding elements that are not the same, as in the case of COCl2, the molecule is polar.
What are the piano notes for Halloween theme?
Right Hand:1:C#, F#, F#, C#, F#, F#, C#, F#, D, F#2: C, F, F, C, F, F, C, F, C#, F3: B, E, E, B, E, E, B, E, C, E4: Bb, Eb, Eb, Bb, Eb, Eb, Bb, Eb, B, Eb5: F#, B, B, F#, B, B, F#, B, G, BLeft Hand1: F#, A, Bb, F#, A, Bb2: E, G, Ab, E, G, Ab3: B, D, E, F#, B, D, E, F#
What are the piano notes for Halloween theme song?
Right Hand:1:C#, F#, F#, C#, F#, F#, C#, F#, D, F#2: C, F, F, C, F, F, C, F, C#, F3: B, E, E, B, E, E, B, E, C, E4: Bb, Eb, Eb, Bb, Eb, Eb, Bb, Eb, B, Eb5: F#, B, B, F#, B, B, F#, B, G, BLeft Hand1: F#, A, Bb, F#, A, Bb2: E, G, Ab, E, G, Ab3: B, D, E, F#, B, D, E, F#
Prove that the line which divides the nonparallel sides of a trapezium proportionally is parallel to the third side.?
To prove that the line which divides the nonparallel sides of a trapezium proportionally is parallel to the third side, we can use the property of similar triangles. Let the trapezium ABCD have sides AB and CD as the nonparallel sides, and side BC as the third side. Let the line dividing AB and CD be denoted as EF, with E on AB and F on CD. By the property of similar triangles, we can show that triangles AEF and BCF are similar, and hence their corresponding angles are congruent. This proves that EF is parallel to BC.
