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d1i1m1o1n [39]
3 years ago
7

Trigonal planars have a 180 degree bond angle but do bent trigonal planars also have a 180 degree bond angle? Thank you ^^

Chemistry
2 answers:
Yanka [14]3 years ago
8 0

Answer:


The ideal bond angle is 120°, but the real bond angle of bent molecules is usually a few degrees less.


Explanation:


The bonds of a trigonal planar AX₃ molecule point toward the corners of an equilateral triangle.


They trisect a circle into three angles of 120° each.


An example is BF₃ (Figure 1). It is perfectly symmetrical, so the F-B-F bond angles are exactly 120°.


===============


Now, consider ozone (Figure 2).


The two O-O bonds are equivalent because of resonance.


There are two bonding groups and a lone pair, for a total of three electron groups.


The electron geometry is trigonal planar, and the molecular shape is bent.


However, the lone pair occupies more space than the bonding electrons.


It repels the bolding electrons, so the O-O-O bond angle is 116°, slightly less than the ideal angle of 120°.


earnstyle [38]3 years ago
8 0

trigonal planar means the molecule is planar n has a triangular shape. bond angle = 360/2 = 120deg.

bent trigonal planar means the molecular shape has been bent by some forces n its bond angle is no longer 120deg.


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vova2212 [387]
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A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the
yarga [219]

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X  1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of  H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

5 0
2 years ago
How long does it take me to travel 500 m E at a velocity of 50 m/s E?
konstantin123 [22]

Answer:15 s E

Explanation:

7 0
3 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
3 years ago
If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w
irinina [24]

Answer:

4.13 g

Explanation:

Data Given:

Amount of CO₂ Produced = 15 g

Amount of Oxygen = 11 g

Amount of Carbon used = ?

Solution:

Suppose Carbon dioxide (CO₂) is formed by the reaction of carbon and oxygen then the reaction will be as below

                            C   +   O₂    -------------> CO₂

                          1 mol    1 mol                  1 mol

we come to know from the above reaction that

1 mole of carbon react with 1 mole of oxygen to produce 1 mol of carbon dioxide.

molar mass of C = 12 g/mol

molar mass of O₂ = 32 g/mol

molar mass of CO₂ = 12 + 2(16) = 44 g/mol

if we represent mole in grams then

           C               +                        O₂                     ------------->        CO                 1 mol (12 g/mol)                      1 mol (32 g/mol)                      1 mol (44 g/mol)

                   

              C   +   O₂    -------------> CO₂

            12 g       32 g                   44 g

So,

we come to know that 32 g of Oxygen combine with 12 g  of oxygen produce 44 g CO₂

So now how much of Carbon will be combine with 11 g of oxygen

apply unity formula

                32 g of  O₂ ≅ 12 g of  C

                  11 g of O₂  ≅  g of  C

by doing cross multiplication

           g of C = 12 g x 11 g / 32 g

           g of C = 132 g / 32 g

           g of C = 4.13 g

So,

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

to check this answer

we use the above information

                     12 g of  C ≅ 44 g of CO₂

                     4.13 g of C ≅  g of  CO₂

by doing cross multiplication

                    g of  CO₂ = 44 g x 4.13 g / 12 g

                    g of CO₂ = 15g

So it is confirmed that

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

4 0
3 years ago
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