Explanation:
Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.
The azide ion that is 
, is a symmetrical ion, all of whose contributing structures have formal charges.
Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.
Contributing structures of azide ion are drawn in the image attached.
The value 6.0 x 10^3- 2.3 × 10^3 in scientific notation is 3.7 × 10^3.
<h3> What is scientific notation?</h3>
Scientific notation is a way to write very large or very small numbers so that they are easier to read and work with.
You express a number as the product of a number greater than or equal to 1 but less than 10 and an integral power of 10 .
<h3>Why it is used? </h3>
There are two reasons to use scientific notation.
- The first is to reveal honest uncertainty in experimental measurements.
- The second is to express very large or very small numbers so they are easier to read.
Given,
= 6.0 x 10^3- 2.3 × 10^3
= (6.0 - 2.3) × 10^3
= 3.7 × 10^3
Thus, we find that the value 6.0 x 10^3- 2.3 × 10^3 in scientific notation is 3.7 × 10^3.
learn more about scientific notation :
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The law of conservation of mass states that matter cannot be created nor destroyed the answer would be D. It remains the same
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
