If the sum of atomic mass of 3 elements forming Dobereiner traid is 69, then atomic mass of second element
A. 7
B. 46
C. 39
D. 23
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If `v = alpha sin (betat)` where v is meter per second and t is in second then. Find sum of dimesions with respect to the time of `alpha` and `beta`
Correct Answer - D `v = alpha sin (beta t)` By dimensional homogenity `alpha` has same dimension of `v = [LT^(-1)] sin (betat)` and `sin (beta t)` no dimension as it...
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An element A has three isotopes `A^(20), A^(21), A^(22)`. The % abundance of `A^(20)` is 90% by mole and average atomic mass of the element A is 20.18
Correct Answer - C `{:(,A^(20),A^(21),A^(22),M_(avg) = 20.18,),(,darr,darr,darr,,),(% "by moles",90%,(10 - x)%,x%,,):}` `M_(avg) = sum %` of isotope `xx` atomic mass `20.18 = (90 xx 20 + (10 - x) 21 +...
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If the same element is forming oxides in different oxidation states then : a)that oxide will be neutral in nature in which element will be in its high
Correct Answer - B Oxidation state `prop` electronegativity Electronegativity increases with increase in oxidation state, so the difference in electronegativity decreases (between element and oxygen) and acidic character increases.
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The electronic configuration of an element is `1s^(2)2s^(2)2p^(6),3s^(2)3p^(5)`. The atomic number of element present just below the above element in
Correct Answer - A The element which present just below the given element will have outermost electromic configuration as `4s^(2) 4p^(3),` so its full electronic configuration is `1s^(2),2s^(2),2p^(6),3s^(2)3p^(6),4s^(2), 3d^(10), 4p^(3)` and...
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The atomic mass of an element is 50 (i) Calculate the mass of one atom, in amu (ii) Calculate the mass of `6.022xx10^(23)` atoms, in gm (iii) Calculat
(i) 50 amu `" "` (ii) 50 gm (iii) `because` 50 gm of element contains `6.022xx10^(23)` atoms `therefore ` 10 gm of element will contain `(6.022xx10^(23))/(50)xx10=1.2044xx10^(22)` atoms (iv) `because 6.022xx10^(23)`...
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If 8.4 g of X element contains `9.03xx10^(22)` atoms and 0.1 moles of `X_(2)Y_(3)` compound is of 16 g. Find relative atomic mass of element?
Correct Answer - D n of x = `(9.03xx10^(22))/(6.02xx10^(23))=0.15` Molar mass of x = `(8.4)/(0.15)= 56 g` Molar mass of compound = `(16)/(0.1) = 160 g` `56 xx 2 +M_(y) xx...
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Mass of 10 atoms of an element is `24xx10^(-23)` gm, then atomic mass of that element is:
M ass of `N_(A)` atoms `=(24xx10^(-23))/(10)xx6.022xx10^(23)=14.4 gm` Atomic mass =14.4
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Atomic mass of `._(3)^(7)Li` is 7.01600u and that of `._(4)^(7)Be` is 7.01693u. Mass of one electron in atomic mass unit is 0.00055 u. if it is observ
Correct Answer - Electron capture
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Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` is
Correct Answer - B `(b)` Let `A=[{:(a,b,c),(p,q,r),(x,y,z):}]` Given `{:(a+b+c=1),(p+q+r=1),(x+y+z=1):}` `impliesA^(2)=[{:(a,b,c),(p,q,r),(x,y,z):}][{:(a,b,c),(p,q,r),(x,y,z):}]` `=[{:(a^(2)+bp+cx,,ab+bq+cy,,ac+br+cz),(pa+qp+rx,,qb+q^(2)+ry,,pc+qr+rz),(xa+yp+zx,,xb+yq+zy,,xc+yr+z^(2)):}]` Sum of elements of `R_(1)=a^(2)+bp+cx+ab+bq+cy+ac+br+cz` `=a(a+b+c)+b(p+q+r)+c(x+y+z)` `=a+b+c=0` Similarly sum of elements of `R_(2)=p(a+b+c)+q(p+q+r)+r(x+y+z)` `=p+q+r=1` `R_(3)=x(a+b+c)+y(p+q+r)+z(x+y+z)` `=x+y+z=1` `:.` sum of element...
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In a reactor, an element X decays to a radioactive element Y, at a constant rate r atoms per second. Each decay reaction releases energy `E_(1)`. Half
Correct Answer - `[(i)" "eta r[E_(1)+E_(2)(1-e^((t log 2)/T))], (ii)" "eta r(E_(1)+E_(2))]`
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