I think it's just water right..?
Explanation :
As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.
He placed the elements with similar nature in the same group.
According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.
However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.
As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127 and tellurium isotopes are tellurium-128 and tellurium-130.
Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.
M{(NH₄)₂Cr₂O₇}=2A(N)+8A(H)+2A(Cr)+7A(O)
M{(NH₄)₂Cr₂O₇}=252.065 g/mol
M(Cr)=51.996 g/mol
m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}=m(Cr)/2M(Cr)
m(Cr)=2M(Cr)m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}
m(Cr)=2*51.996*35.8/252.065=14.770 g
m(Cr)=14.770 g
Answer: sodium amide undergoes an acid -base reaction
Explanation:
sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.
This leads to formation of ammonia and sodium methoxide.
Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.
Hence the 3rd statement is a corrects statement.
So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.
The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.
The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.
The following reaction occurs:
NaNH₂+CH₃OH→NH₃+CH₃ONa
Answer:
q=mAHf
Explanation:
When water change form from solid(ice) to liquid, it will need heat that depends on the heat of fusion(Hfus). If the form changed from liquid to gas it will use the heat of vaporization(Hvap) instead. The heat of fusion usually written just AT
q=MCAT is used to calculate heat when temperature change(AT = delta T= difference in temperature)
q=mAHvap is wrong because you need if the form change from liquid to gas
q=mc^2 is wrong because it describes the relation of mass and energy, not the form.
The answer should be q=mAHf
