Answer:
No this statement is false.
Explanation:
when an atom lose or gain the electron ions are formed. If the atom lose an electron the positive ions are formed called cations while anion is formed by the gaining of electron by an atom The elements having less electrons in valance shell usually lose their electrons while the elements like halogens having 7 valance electrons gain an electron to complete the octet. while p block elements noble gases are inert, their outer most valance shell is complete so they do not form ions.
P block elements are non-metals, metals and metalloids. These are thirty five elements. The P-block elements are present on right side of periodic table. There valance electrons are present in P orbital. The p-block metals are shiny and good conductor of heat and electricity. These metal lose the electron which is accept by non metals and form ionic bond. They have high melting points.
Metalloids includes boron, silicon, germanium, arsenic, antimony and tellurium. Metalloids contain both the properties of metals and non metals, Some metalloids are toxic like arsenic.
Most of p-block elements are non metals. They are bad conductor of heat and electricity and have low boiling points. The non metals mostly accept the electron from the metals and usually from ionic bond like in case of chlorine. It form the ionic compound with sodium.
The sodium chloride which is an ionic compound, formed by the complete transfer of electron from sodium to chlorine atom and form ionic bond. In this ionic compound sodium carry positive charge and chlorine carry negative charge there is attraction between these oppositely charged atoms.
I would say mass lost by nuclear collisions. The mass defect is the mass difference between the mass of an atomic nucleus and the sum of the mass of its constituent particles. It equals the energy given off in the formation of the nucleus.
Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:

The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M
Answer:
24.6g of NaCl
Explanation:
Expression of the reaction:
2NaCl → 2Na + Cl₂
Given parameters:
Mass of Cl₂ = 15g
Unknown:
Mass of NaCl = ?
Solution:
To solve this problem, we have to use mole relationships.
Find the number of moles of the mass of the given specie;
Number of moles =
Molar mass of Cl₂ = 2(35.5) = 71g/mol
Number of moles =
= 0.21mole
Now;
From the balanced reaction equation;
1 mole of Cl₂ is produced from 2 moles of NaCl;
0.21 mole of Cl₂ will be produced from 0.21 x 2 = 0.42mole of NaCl
So,
Mass of NaCl = number of moles x molar mass
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass of NaCl = 0.42 x 58.5 = 24.6g of NaCl
The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.