the mass in grams of 1.00 x 10^23 molecules of N2 is 4.648 g
<u><em>calculation</em></u>
Step 1: use the Avogadro's constant to determine the moles of N2
According to Avogadro's law 1 mole = 6.02 x 10 ^23 molecules
? moles = 1.00 x 10 ^23 molecules
<em>by cross multiplication</em>
= (1 mole x 1.00 x10^23 molecules) / ( 6.02 x 10^23 molecules)
=0.166 moles of N2
Step 2 : find mass of N2
Mass = moles x molar mass
From periodic table the molar mass of N2 = 14 x2 = 28 g/mol
mass = 0.166 moles x 28 g/mol = 4.648 g
Of the same element with different mass numbers are called isotopes 3) Isotopes have the same number of protons but different numbers of neutrons in their nuclei. ... The 63Cu isotope has 63 - 29 = 34 neutrons. The 65Cu isotope has 65 - 29 = 36 neutrons.
Answer:
σ*2pₓ, also called 
Explanation:
I have drawn the MO diagram for fluorine below.
Each F atom contributes seven valence electrons, so we fill the MOs of fluorine with 14 electrons.
We have filled the 
and 
MOs.
They are the highest occupied molecular orbitals (HOMOs).
The next unfilled level (the LUMO) is the σ*2pₓ orbital. If you use the symmetry notation, it is called the orbital.
This is the orbital that fluorine uses when it acts as an electron acceptor.
The arrow that represents the transition in which dew is formed would be arrow 4.
<h3>How is dew formed?</h3>
Dew is formed by condensation. Condensation, in itself, involves a transition from a gas phase to a liquid phase.
From the diagram, the arrow that represents a transition from gas to liquid is arrow 4.
Thus, the arrow that represents the transition that leads to the formation of dew would be arrow 4.
More on dew formation can be found here: brainly.com/question/23169635
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Answer:
7,94 minutes
Explanation:
If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>
For the zero-order reactions, concentration-time equation can be written as follows:
[A] = - Kt + [Ao]
where:
- [A]: concentration of the reactant A at the <em>t </em>time,
- [A]o: initial concentration of the reactant A,
- K: rate constant,
- t: elapsed time of the reaction
<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>
Data:
K = 4.2 ×10−3atm/s,
[A]o=[HBr]o= 2 atm,
[A]=[HBr]=0 atm (all HBr(g) is gone)
<em>We clear the incognita :</em>
[A] = - Kt + [Ao]............. Kt = [Ao] - [A]
t = ([Ao] - [A])/K
<em>We replace the numerical values:</em>
t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes
So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).
