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Neporo4naja [7]
2 years ago
10

1) In molecules with the same number of electron groups but different molecular geometries, discuss what happens to the bond ang

le?
Chemistry
1 answer:
BartSMP [9]2 years ago
7 0

In molecules with the same number of electron groups but different molecular geometries, One of the chances is that the central atoms alter in the way in which their electrons are arranged.

Explanation:

  • Consider molecules with steric number =  four sets of electrons around the central atom.A tetrahedral geometry with 109.5 deg bond angles is predicted by VSEPR
  • NH3 is pyramidal with a lone pair on one end of the tetrahedron. The HNH angle is 107.3 deg, generally described by stating that the lone pair is fatter than the pairs that bond.
  • PH3 is also pyramidal, the HPH angles are close to being 90-degree angles.
  • The assumption is that the small atoms (like C and N) use hybrid sp3 orbitals to form the bonds with hydrogen in their compounds. All N and P merely make three ammonia and phosphine bonds.
  • The hydrogens remain further apart if the nitrogen uses hybrid orbitals. Phosphorus is large enough that the hydrogens don’t associate too much in phosphine. Since the P doesn’t have to exhaust the energy required to hybridize its bonding orbitals.

You might be interested in
What is the maximum radiation pressure exerted by sunlight in space ( s = 1350 w/m 2) on a flat black surface? a. 2. 25 × 10−5 p
mel-nik [20]

The maximum radiation pressure exerted by sunlight in space on a flat black surface is  4.5 × 10^{-6}  P a. So, the correct option is (b).

Radiation pressure is the name for the force electromagnetic wave particles exert on a surface. It is inversely related to the wave's speed. Given data

Solar constant  ( S )  =  1350W / m ^2

Now, the radiaton pressure is given by

P = 2 S /c

where c is the speed of the light

P = 2 × 1350 /3 × 10 ^8

P = 9 × 10^{-6}  P a

For a black surface, P = 4.5× 10^{-6}  P a

Therefore, maximum radiation pressure exerted by sunlight in space on a flat black surface is  4.5 × 10^{-6}  P a

Learn more about radiation pressure here;

brainly.com/question/23972862

#SPJ4

8 0
1 year ago
What is the mass of 8.23 x 10^23 atoms of Ag
Gnom [1K]

Answer:

\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}

Explanation:

<u>Convert Atoms to Moles</u>

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}

8.23 *10^{23}  *\frac{1 \ mol \ Ag}{6.022*10^{23} }

Condense into one fraction.

\frac{8.23 *10^{23}  }{6.022*10^{23} } \ mol \ Ag

1.366655596 \ mol \ Ag

<u>Convert Moles to Grams</u>

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

  • Ag: 107.868 g/mol

Let's make another ratio using this information.

\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

Multiply by the number of moles we calculated.

1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

The moles of silver cancel out.

1.366655596 *\frac {107.868 \ g \ Ag}{1 }

1.366655596 * {107.868 \ g \ Ag}

147.4184058 \ g\ Ag

<u>Round</u>

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

  • 147.<u>4</u>184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

147 \ g\ Ag

8.23 ×10²³ silver atoms are equal to approximately <u>147 grams.</u>

3 0
3 years ago
In order to determine if 2 atoms are cooper what must be the same for each
galina1969 [7]
-number of electrons
-atom size
-charge of the atom
-number of protons
4 0
3 years ago
In right triangle EFG, mE = 25. In right triangle HJK, mH = 25. Which similarity postulate or theorem proves that EFG and HJK ar
Rus_ich [418]
The angle- angle similarity postulate. It states that if a triangle has two equal corresponding angles the angles are similar.
6 0
3 years ago
Read 2 more answers
1. Draw a wedge/dash structure for trans-1,2-dimethylcyclohexane.
Angelina_Jolie [31]

Answer:

The structure with the ring flipped is the most stable

Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

Hope this helps

6 0
3 years ago
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