Answer:
The amount of NaF produced is doubled.
(d) is correct option.
Explanation:
Given that,
A 2 mole sample of F₂ reacts with excess NaOH according to the equation.
The balance equation is
If the reaction is repeated with excess NaOH but with 1 mole of F₂
The balance equation is
Hence, The amount of NaF produced is doubled.
(d) is correct option.
Answer:
A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.
.
Explanation:
There are two types of beta decay modes: beta-minus and beta-plus.
In both decay modes, the mass number of the nucleus stays the same.
However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.
Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.
Look up the atomic number and relative atomic mass for the element scandium.
This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.
Each element (e.g, ) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.
The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.
The atomic number of the nucleus will increase by 1. . That corresponds to titanium. The mass number stays the same at
. Hence the daughter nucleus would be titanium-48. Note that two other particles: one electron and one electron
and one antineutrino
(note the bar.) The neutrino helps balance the lepton number of this reaction.
Answer:
The of the given reaction is -129.6 kJ
Explanation:
The given chemical reaction is as follows.
Enthalpy of each reactant and products are as follows.
In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.
Therefore, The of the given reaction is -129.6 kJ
Answer :
(1) The hybridization of central atom beryllium in is,
(2) The hybridization of central atom nitrogen in is,
(3) The hybridization of central atom carbon in is,
(4) The hybridization of central atom xenon in is,
Explanation :
Formula used :
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
Now we have to determine the hybridization of the following molecules.
(1) The given molecule is,
The number of electron pair are 2 that means the hybridization will be and the electronic geometry of the molecule will be linear.
(2) The given molecule is,
If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, .
In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, .
(3) The given molecule is,
The number of electron pair are 4 that means the hybridization will be and the electronic geometry of the molecule will be tetrahedral.
(4) The given molecule is,
Bond pair electrons = 4
Lone pair electrons = 6 - 4 = 2
The number of electrons are 6 that means the hybridization will be and the electronic geometry of the molecule will be octahedral.
But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.