See if the carbon atoms are SP2 or Sp they the coplanirty is more and if its Sp3 hybridization it cant be in coplanar as Sp3 is having Td shape where as sp2 and sp are not :)

7 0

3 years ago

Jasper made a list of the properties of electromagnetic waves. Identify the mistake in the list.

ch4aika [34]

Answer:

B. Line 2 has a mistake

Explanation:

1. All electromagnetic waves are made of both electric and magnetic fields (hence the name electromagnetic).

2. All electromagnetic waves are transverse, and not longitudinal.

3. All electromagnetic waves can travel through space because they do not need particles to travel (like sound waves).

4. Electromagnetic waves all travel at the speed of light (299 792 458 m/s).

3 0

3 years ago

What do the parentheses mean in a chemical reaction and how they are used

galben [10]

Parantheses mean present of a subgroup or a group of polyatomic ions in a chemical reaction.

Explanation:

In a chemical reaction, sometimes, a compound may be composed of group of polyatomic ions with other ions. Some of the polyatomic ions are sulfate, carbonate, nitrate, hydroxide, bicarbonate, ammonia etc.

So these polyatomic ions are formed by combining two or more elements. Thus, if the number of polyatomic ions in a compound is more than one, then we use parathesis and write those polyatomic ions in it and write the number of polyatomic ions present in the compound as subscript of the parathesis.

For example, Fe₂(SO₄)₃

So here SO₄ is a polyatomic ion and in the compound 3 atoms of SO₄ is required to neutralize the compound and thus paranthesis is used. Thus, parantheses mean present of a subgroup or a group of polyatomic ions in a chemical reaction.

5 0

3 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

Thus, percentage abundance of 121 Sb is = 57.2 %

percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %

6 0

3 years ago