54248.Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then
minimum molecular weight of peroxidase anhydrous enzyme is
minimum molecular weight of peroxidase anhydrous enzyme is
$1.568 x 10^4$
$1.568 x 10^3$
15.68
$2.136 x 10^4$
Explanation:
let molecular weight of enzyme be x
weight of Se in enzyme = 0.5% of x
= 0.5 x / 100
at. wt. of Se = 78.4 (which is given in the question)
so,
78.4=0.5 x / 100
78.4 * 100 = 0.5 x
7840 = 0.5 x
x = 7840 / 0.5
x = 7840 * 10 / 5
therefore , x = 15680 or 1.5680 * 10 to the power 4
let molecular weight of enzyme be x
weight of Se in enzyme = 0.5% of x
= 0.5 x / 100
at. wt. of Se = 78.4 (which is given in the question)
so,
78.4=0.5 x / 100
78.4 * 100 = 0.5 x
7840 = 0.5 x
x = 7840 / 0.5
x = 7840 * 10 / 5
therefore , x = 15680 or 1.5680 * 10 to the power 4
54249.Molarity of liquid HCl, if density of solution is 1.17 g/cc is
36.5
18.25
32.05
42.10
Explanation:
Given:Density of HCl=1.17g/cm³
Molarity=(density/molar mass)x1000 moles per liter
molar mass of HCl=[1+35.5]=36.5g/mol
Molarity=(1.17/36.5)x1000 =32.05M
∴Molarity of liquid HCl with density equal to 1.17 g/cc is 32.05
Given:Density of HCl=1.17g/cm³
Molarity=(density/molar mass)x1000 moles per liter
molar mass of HCl=[1+35.5]=36.5g/mol
Molarity=(1.17/36.5)x1000 =32.05M
∴Molarity of liquid HCl with density equal to 1.17 g/cc is 32.05
54250.Specific volume of cylindrical virus particle is $6.2 x 10^{-2}$ cc/g whose radius and length are 7
A and 10 A respectively. If $N = 6.02 x 10^{23}$ , find molecular weight of virus.
A and 10 A respectively. If $N = 6.02 x 10^{23}$ , find molecular weight of virus.
$15.4 kg/mol$
$1.54 x 10^4 kg/mol$
$3.08 x 10^4 kg/mol$
$3.08 x 10^3 kg/mol$
Explanation:
Volume of one virus = voume of cylinder = π r²h
= 22/7 × 7 × 7 × 10 A°³
=22 × 7 × 10 A°³
=$1540 × 10^-30 m³$
$=1.54 × 10^-27 m³$
so, volume of one mole of virus = 6.023 × 10²³ × volume of one virus
= $6.023 × 10²³ × 1.54 × 10^{-27} m³$
= $9.27542 × 10^-4 m³$
= 9.27542 × 10² cm³
now, molecular weight = volume of 1mole/ specific volume
= $9.27542 ×10² / 6.02 × 10^-2 g/mol$
= 1.54 × 10⁴ g/mol
Volume of one virus = voume of cylinder = π r²h
= 22/7 × 7 × 7 × 10 A°³
=22 × 7 × 10 A°³
=$1540 × 10^-30 m³$
$=1.54 × 10^-27 m³$
so, volume of one mole of virus = 6.023 × 10²³ × volume of one virus
= $6.023 × 10²³ × 1.54 × 10^{-27} m³$
= $9.27542 × 10^-4 m³$
= 9.27542 × 10² cm³
now, molecular weight = volume of 1mole/ specific volume
= $9.27542 ×10² / 6.02 × 10^-2 g/mol$
= 1.54 × 10⁴ g/mol
54251.In quantitative analysis of second group in laboratory, H2S gas is passed in acidic medium
for precipitation. When Cu2+ and Cd2+ react with KCN, then for product, true statement is
for precipitation. When Cu2+ and Cd2+ react with KCN, then for product, true statement is
K2 [Cu(CN)4 ] more soluble
K2 [Cd(CN)4 ] less stable
K3 [Cu(CN)2 ] less stable
K2 [Cd(CN)3 ] more stable.
Explanation:
Cu2+ + H2S ⟶ CuS(black) + 2Hsup>+
Cd2+ + H2S ⟶ CdS(black) + 2Hsup>+
CuS + 4KCN ⟶ K2 [Cu(CN)4 ] (I) + K2S
CdS + 4KCN ⟶ K2 [Cd(CN)4 ] (II) + K2S
K2 [Cu(CN)4 ] is more soluble than [Cd(CN)4 ].Because of Ksp of (I) is greater than (II).
Cu2+ + H2S ⟶ CuS(black) + 2Hsup>+
Cd2+ + H2S ⟶ CdS(black) + 2Hsup>+
CuS + 4KCN ⟶ K2 [Cu(CN)4 ] (I) + K2S
CdS + 4KCN ⟶ K2 [Cd(CN)4 ] (II) + K2S
K2 [Cu(CN)4 ] is more soluble than [Cd(CN)4 ].Because of Ksp of (I) is greater than (II).
54252.Volume of CO2 obtained by the complete decomposition of 9.85 g of BaCO3 is
2.24 L
1.12 L
0.84 L
0.56 L
Explanation:
BaCO3→BaO+Co↑
Molecular weight of BaCO3=137+12+3×16=197
∵197g produces 22.4L at S.T.P.
∴9.85g produces (22.4/197)×9.85 = 1.12L at S.T.P.
BaCO3→BaO+Co↑
Molecular weight of BaCO3=137+12+3×16=197
∵197g produces 22.4L at S.T.P.
∴9.85g produces (22.4/197)×9.85 = 1.12L at S.T.P.
54253.Oxidation numbers of A, B, C are +2, +5 and -2 respectively. Possible formula of compound is
A2 (BC2)2
A3 (BC4)2
A2 (BC3)2
A3 (B2C)2
Explanation:
In A3 (BC4)2 , (+2) * 3 + 2[+5 +4(-2)]
=> +6+10-16=0
Hence in the compound A3 (BC4)2 , the oxidation no. of A, B and C are +2,+5 and -2 respectively.
In A3 (BC4)2 , (+2) * 3 + 2[+5 +4(-2)]
=> +6+10-16=0
Hence in the compound A3 (BC4)2 , the oxidation no. of A, B and C are +2,+5 and -2 respectively.
54254.The number of atoms in 4.25 g of NH3 is approximately
4 x 1023
2 x 2023
1 x 1023
6 x 1023
Explanation:
Molar mass : Mass of the one mole of the substance is called molar mass
Molar mass is numerically equal in g(gram) to atomic mass in u(unified mass).
Molar mass of the NH3
= 14g + (1×3)g
= 17g/mol
Mass = number of moles × molar mass
4.25 = number of moles × 17
4.25/17 = number of moles
0.25 = number of moles
One mole of any substance contains avogadro number (6.022×10^23) of molecules.
Molecules = number of moles×6.022×10^23
Molecules = 0.25 × 6.022×10^23
Molecules = 1.5055×10^23 “ NH3 ” molecules.
Then,
Atoms = 4 × 1.5055 × 10^23 (because there are 4 atoms i.e 3 atoms of Hydrogen and 1 atom of oxygen )
Atoms = 6.022 × 10^23 “ NH3 atoms ”
Molar mass : Mass of the one mole of the substance is called molar mass
Molar mass is numerically equal in g(gram) to atomic mass in u(unified mass).
Molar mass of the NH3
= 14g + (1×3)g
= 17g/mol
Mass = number of moles × molar mass
4.25 = number of moles × 17
4.25/17 = number of moles
0.25 = number of moles
One mole of any substance contains avogadro number (6.022×10^23) of molecules.
Molecules = number of moles×6.022×10^23
Molecules = 0.25 × 6.022×10^23
Molecules = 1.5055×10^23 “ NH3 ” molecules.
Then,
Atoms = 4 × 1.5055 × 10^23 (because there are 4 atoms i.e 3 atoms of Hydrogen and 1 atom of oxygen )
Atoms = 6.022 × 10^23 “ NH3 atoms ”
54255.Given the numbers: 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for
the three numbers is
the three numbers is
3, 3 and 4 respectively
3, 4 and 4 respectively
3, 4 and 5 respectively
3, 3 and 3 respectively.
Explanation:
We know that all non-zero digits are significant and the zeros at the beginning of a number are not significant. Therfore number 161 cm, 0.161 cm and 0.0161 cm have 3,3 and 3 significant figures respectively
We know that all non-zero digits are significant and the zeros at the beginning of a number are not significant. Therfore number 161 cm, 0.161 cm and 0.0161 cm have 3,3 and 3 significant figures respectively
54256.Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is
approximately 67200. The number of iron atoms (Atomic weight of Fe is 56) present in one
molecule of haemoglobin is
approximately 67200. The number of iron atoms (Atomic weight of Fe is 56) present in one
molecule of haemoglobin is
4
6
3
2
Explanation:
The molecular weight of haemoglobin=67200
It contains 0.334% of iron by weight.
Weight of iron= $ /dfrac{0.334}{100} $ × 67200 = 224.448
No. of atoms= $ /dfrac{Weight of iron in haemoglobin}{Atomic weight} $
=$ /dfrac{224.448}{56} $
=4.008
No. of iron atoms present in one molecule of haemoglobin is 4.
The molecular weight of haemoglobin=67200
It contains 0.334% of iron by weight.
Weight of iron= $ /dfrac{0.334}{100} $ × 67200 = 224.448
No. of atoms= $ /dfrac{Weight of iron in haemoglobin}{Atomic weight} $
=$ /dfrac{224.448}{56} $
=4.008
No. of iron atoms present in one molecule of haemoglobin is 4.
54257.In the reaction,4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O when 1 mole of ammonia and 1 mole of O2 are made to react to completion,then
All the oxygen will be consumed.
1.0 mole of NO will be produced.
1.0 mole of HO2O is produced.
All the ammonia will be consumed.
Explanation:
4NH3(g) (4 mole) + 5O2(g) (5 mole) —> 4NO(g) (4 mole)+ 6H2O (6 mole)
=> 1 mole of NH3(g) requires =(5/4)= 1.25 mole of oxygen
while 1 mole of O2(g) requires =(4/5)= 0.8 mole of NH3(g)
As there 1 mole of ammonia and 1 mole of O2 , so all the oxygen will be consumed.
4NH3(g) (4 mole) + 5O2(g) (5 mole) —> 4NO(g) (4 mole)+ 6H2O (6 mole)
=> 1 mole of NH3(g) requires =(5/4)= 1.25 mole of oxygen
while 1 mole of O2(g) requires =(4/5)= 0.8 mole of NH3(g)
As there 1 mole of ammonia and 1 mole of O2 , so all the oxygen will be consumed.
54258.The pair of quantities having same dimensions is
Impulse and Surface Tension
Angular momentum and Work
Work and Torque
Youngs modulus and Energy
54259.The dimensions of $(\mu_{0}\xi_{0})^{-1/2}$ are
$[L^{1/2}T^{-1/2}]$
$[L^{-1}T] $
$[LT^{-1}$
$[L^{1/2}T^{1/2}]$
54260.Which pair has the same dimensions
Work and power
Density and relative density
Momentum and impulse
Stress and strain
54261.The speed of light in vacuum is $3 × 10^8 m / s$. How many nanosecond does it take to travel one metre in a vacuum?
8 ns
10/3 ns
3.34 ns
none of these
54262.The density of a material in CGS system of units is 4 g $cm^{-3}$
In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be
In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be
0.04
0.4
40
400
54263.1 revolution is equivalent to 360. The value of 1 revolution per minute is
2πra /s
0.104rad/ s
3.14rad /s
None of these
54264.If in a system the force of attraction between two point masses of 1kg each situated 1km apart is taken as a unit force and is called notwen (newton written in reverse order) If
$ G = 6.67 × 10− 11^N $
$−m^2kg^{− 2}$
in SI units, the relation of newton and nowton is
$ G = 6.67 × 10− 11^N $
$−m^2kg^{− 2}$
in SI units, the relation of newton and nowton is
$1notwen = 6.67 × 10 ^ {− 11}newton$
$1newton = 6.67 × 10 ^ {− 17}notwen$
$1notwen = 6.67 × 10 ^ {− 17}newton$
$1newton = 6.67 × 10 ^ {− 12}notwen$
54265.A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numberical value. If the result is expressed in various units then
n ∝ sizeofu
n ∝ u2
n ∝ √u
n ∝ u1 .
54266.The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are
$kg m s^{-1}$
$kg m s^{-2} $
$kg s^{-1}$
$kg s $
54267.The frequency f of vibrations of a mass m suspended from a spring
of spring constant k is given by $f = Cm^xk^y$ , where C is a
dimensionnless constant. The values of x and y are, respectively,
of spring constant k is given by $f = Cm^xk^y$ , where C is a
dimensionnless constant. The values of x and y are, respectively,
$\dfrac{1}{2} , \dfrac{1}{2}$
$- \dfrac{1}{2} , - \dfrac{1}{2}$
$\dfrac{1}{2} , - \dfrac{1}{2}$
$- \dfrac{1}{2} , \dfrac{1}{2}$