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Question Number 20972 Answers: 0 Comments: 0

Boyle temperature is given by (1) T_B = (a/(Rb^2 )) (2) T_B = (a/(Rb)) (3) T_B = (a/(27b^2 )) (4) T_B = (b/(aR))

$$\mathrm{Boyle}\:\mathrm{temperature}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{a}}{\mathrm{Rb}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{a}}{\mathrm{Rb}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{a}}{\mathrm{27b}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\mathrm{T}_{\mathrm{B}} \:=\:\frac{\mathrm{b}}{\mathrm{aR}} \\ $$

Question Number 20976 Answers: 0 Comments: 0

What would be the percentage composition by volume of a mixture of CO and CH_4 , whose 10.5 mL requires 9 mL oxygen for complete combustion?

$$\mathrm{What}\:\mathrm{would}\:\mathrm{be}\:\mathrm{the}\:\mathrm{percentage} \\ $$$$\mathrm{composition}\:\mathrm{by}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{mixture}\:\mathrm{of} \\ $$$$\mathrm{CO}\:\mathrm{and}\:\mathrm{CH}_{\mathrm{4}} ,\:\mathrm{whose}\:\mathrm{10}.\mathrm{5}\:\mathrm{mL}\:\mathrm{requires} \\ $$$$\mathrm{9}\:\mathrm{mL}\:\mathrm{oxygen}\:\mathrm{for}\:\mathrm{complete}\:\mathrm{combustion}? \\ $$

Question Number 18202 Answers: 1 Comments: 0

An open vessel at 27°C is heated until (3/5) parts of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated.

$$\mathrm{An}\:\mathrm{open}\:\mathrm{vessel}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}\:\mathrm{is}\:\mathrm{heated}\:\mathrm{until} \\ $$$$\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{the}\:\mathrm{air}\:\mathrm{in}\:\mathrm{it}\:\mathrm{has}\:\mathrm{been}\:\mathrm{expelled}. \\ $$$$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vessel} \\ $$$$\mathrm{remains}\:\mathrm{constant},\:\mathrm{find}\:\mathrm{the}\:\mathrm{temperature} \\ $$$$\mathrm{to}\:\mathrm{which}\:\mathrm{the}\:\mathrm{vessel}\:\mathrm{has}\:\mathrm{been}\:\mathrm{heated}. \\ $$

Question Number 18199 Answers: 0 Comments: 0

What is the equivalent weight of KH(IO_3 )_2 as an oxidant in presence of 4 (N) HCl when ICl becomes the reduced form? (K = 39, I = 127)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equivalent}\:\mathrm{weight}\:\mathrm{of} \\ $$$$\mathrm{KH}\left(\mathrm{IO}_{\mathrm{3}} \right)_{\mathrm{2}} \:\mathrm{as}\:\mathrm{an}\:\mathrm{oxidant}\:\mathrm{in}\:\mathrm{presence}\:\mathrm{of} \\ $$$$\mathrm{4}\:\left(\mathrm{N}\right)\:\mathrm{HCl}\:\mathrm{when}\:\mathrm{ICl}\:\mathrm{becomes}\:\mathrm{the} \\ $$$$\mathrm{reduced}\:\mathrm{form}?\:\left(\mathrm{K}\:=\:\mathrm{39},\:\mathrm{I}\:=\:\mathrm{127}\right) \\ $$

Question Number 18198 Answers: 1 Comments: 0

Mixture X = 0.02 mol of [Co(NH_3 )_5 SO_4 ]Br and 0.02 mol of [Co(NH_3 )_5 Br]SO_4 was prepared in 2 litre of solution 1 litre of mixture X + excess AgNO_3 → Y 1 litre of mixture X + excess BaCl_2 → Z Number of moles of Y and Z are

$$\mathrm{Mixture}\:\mathrm{X}\:=\:\mathrm{0}.\mathrm{02}\:\mathrm{mol}\:\mathrm{of} \\ $$$$\left[\mathrm{Co}\left(\mathrm{NH}_{\mathrm{3}} \right)_{\mathrm{5}} \mathrm{SO}_{\mathrm{4}} \right]\mathrm{Br}\:\mathrm{and}\:\mathrm{0}.\mathrm{02}\:\mathrm{mol}\:\mathrm{of} \\ $$$$\left[\mathrm{Co}\left(\mathrm{NH}_{\mathrm{3}} \right)_{\mathrm{5}} \mathrm{Br}\right]\mathrm{SO}_{\mathrm{4}} \:\mathrm{was}\:\mathrm{prepared}\:\mathrm{in}\:\mathrm{2} \\ $$$$\mathrm{litre}\:\mathrm{of}\:\mathrm{solution} \\ $$$$\mathrm{1}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{mixture}\:\mathrm{X}\:+\:\mathrm{excess}\:\mathrm{AgNO}_{\mathrm{3}} \:\rightarrow\:\mathrm{Y} \\ $$$$\mathrm{1}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{mixture}\:\mathrm{X}\:+\:\mathrm{excess}\:\mathrm{BaCl}_{\mathrm{2}} \:\rightarrow\:\mathrm{Z} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{moles}\:\mathrm{of}\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are} \\ $$

Question Number 18197 Answers: 1 Comments: 0

Rearrange the following (I to IV) in the the order of increasing masses. I. 1 molecule of oxygen II. 1 atom of nitrogen III. 10^(10) g molecular weight of oxygen IV. 10^(−18) g atomic weight of copper

$$\mathrm{Rearrange}\:\mathrm{the}\:\mathrm{following}\:\left(\mathrm{I}\:\mathrm{to}\:\mathrm{IV}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{increasing}\:\mathrm{masses}. \\ $$$$\mathrm{I}.\:\mathrm{1}\:\mathrm{molecule}\:\mathrm{of}\:\mathrm{oxygen} \\ $$$$\mathrm{II}.\:\mathrm{1}\:\mathrm{atom}\:\mathrm{of}\:\mathrm{nitrogen} \\ $$$$\mathrm{III}.\:\mathrm{10}^{\mathrm{10}} \:\mathrm{g}\:\mathrm{molecular}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{oxygen} \\ $$$$\mathrm{IV}.\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{g}\:\mathrm{atomic}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{copper} \\ $$

Question Number 18186 Answers: 2 Comments: 0

3.92 g of ferrous ammonium sulphate are dissolved in 100 ml of water. 20 ml of this solution requires 18 ml of potassium permanganate during titration for complete oxidation. The weight of KMnO_4 present in one litre of the solution is

$$\mathrm{3}.\mathrm{92}\:\mathrm{g}\:\mathrm{of}\:\mathrm{ferrous}\:\mathrm{ammonium}\:\mathrm{sulphate} \\ $$$$\mathrm{are}\:\mathrm{dissolved}\:\mathrm{in}\:\mathrm{100}\:\mathrm{ml}\:\mathrm{of}\:\mathrm{water}.\:\mathrm{20}\:\mathrm{ml} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{requires}\:\mathrm{18}\:\mathrm{ml}\:\mathrm{of} \\ $$$$\mathrm{potassium}\:\mathrm{permanganate}\:\mathrm{during} \\ $$$$\mathrm{titration}\:\mathrm{for}\:\mathrm{complete}\:\mathrm{oxidation}.\:\mathrm{The} \\ $$$$\mathrm{weight}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{present}\:\mathrm{in}\:\mathrm{one}\:\mathrm{litre} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is} \\ $$

Question Number 18261 Answers: 1 Comments: 0

What is the maximum angle to the horizontal at which a stone can be thrown and always be moving away from the thrower?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{angle}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{at}\:\mathrm{which}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{thrown}\:\mathrm{and}\:\mathrm{always}\:\mathrm{be}\:\mathrm{moving}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{thrower}? \\ $$

Question Number 18142 Answers: 1 Comments: 1

An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0, the plank starts moving along the x- direction with an acceleration 1.5 ms^(−2) . At the same instant a stone is projected from the origin with a velocity u^→ as shown. A stationary person on the ground observe the stone hitting the object during its downward motion at an angle of 45° with the horizontal. Take g = 10 m/s^2 and consider all motions in the x-y plane. 1. The time after which the stone hits the object is 2. The initial velocity (u^→ ) of the particle is

Question Number 18140 Answers: 1 Comments: 0

From a tower of height H, a particle is thrown vertically upward with speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (1) 2gH = n^2 u^2 (2) gH = (n − 2)^2 u^2 (3) 2gH = nu^2 (n − 2) (4) gH = (n − 2)u^2

$$\mathrm{From}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{of}\:\mathrm{height}\:{H},\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{speed} \\ $$$${u}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle},\:\mathrm{to} \\ $$$$\mathrm{hit}\:\mathrm{the}\:\mathrm{ground},\:\mathrm{is}\:{n}\:\mathrm{times}\:\mathrm{that}\:\mathrm{taken}\:\mathrm{by} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{highest}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}. \\ $$$$\mathrm{The}\:\mathrm{relation}\:\mathrm{between}\:{H},\:{u}\:\mathrm{and}\:{n}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{gH}\:=\:{n}^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right)^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}{gH}\:=\:{nu}^{\mathrm{2}} \left({n}\:−\:\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right){u}^{\mathrm{2}} \\ $$

Question Number 18131 Answers: 0 Comments: 3

A stone is projected from a point on the ground in such a direction so as to hit a bird on the top of a telegraph post of height h, and then attain a height 2h above the ground. If, at an instant of projection, the bird were to fly away horizontal with a uniform speed, find the ratio of the horizontal velocities of the bird and the stone, if the stone still hits the bird.

$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{hit}\:\mathrm{a} \\ $$$$\mathrm{bird}\:\mathrm{on}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{telegraph}\:\mathrm{post}\:\mathrm{of} \\ $$$$\mathrm{height}\:{h},\:\mathrm{and}\:\mathrm{then}\:\mathrm{attain}\:\mathrm{a}\:\mathrm{height}\:\mathrm{2}{h} \\ $$$$\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If},\:\mathrm{at}\:\mathrm{an}\:\mathrm{instant}\:\mathrm{of} \\ $$$$\mathrm{projection},\:\mathrm{the}\:\mathrm{bird}\:\mathrm{were}\:\mathrm{to}\:\mathrm{fly}\:\mathrm{away} \\ $$$$\mathrm{horizontal}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed},\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{velocities}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{bird}\:\mathrm{and}\:\mathrm{the}\:\mathrm{stone},\:\mathrm{if}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{still} \\ $$$$\mathrm{hits}\:\mathrm{the}\:\mathrm{bird}. \\ $$

Question Number 18107 Answers: 0 Comments: 0

The first and second ionization potentials of helium atoms are 24.58 eV and 54.4 eV per mole respectively. Calculate the energy in kJ required to produce 1 mole of He^(2+) ions.

$$\mathrm{The}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{ionization} \\ $$$$\mathrm{potentials}\:\mathrm{of}\:\mathrm{helium}\:\mathrm{atoms}\:\mathrm{are}\:\mathrm{24}.\mathrm{58}\:\mathrm{eV} \\ $$$$\mathrm{and}\:\mathrm{54}.\mathrm{4}\:\mathrm{eV}\:\mathrm{per}\:\mathrm{mole}\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{kJ}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{produce}\:\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{He}^{\mathrm{2}+} \:\mathrm{ions}. \\ $$

Question Number 18106 Answers: 0 Comments: 0

The ionization potential of hydrogen is 13.60 eV/mole. Calculate the energy in kJ required to produce 0.1 mole of H^+ ions. Given, 1 eV = 96.49 kJ mol^(−1) )

$$\mathrm{The}\:\mathrm{ionization}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{hydrogen}\:\mathrm{is} \\ $$$$\mathrm{13}.\mathrm{60}\:\mathrm{eV}/\mathrm{mole}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in} \\ $$$$\mathrm{kJ}\:\mathrm{required}\:\mathrm{to}\:\mathrm{produce}\:\mathrm{0}.\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{H}^{+} \\ $$$$\left.\mathrm{ions}.\:\mathrm{Given},\:\mathrm{1}\:\mathrm{eV}\:=\:\mathrm{96}.\mathrm{49}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \right) \\ $$

Question Number 18095 Answers: 1 Comments: 0

A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.

$$\mathrm{A}\:\mathrm{boy}\:\mathrm{travelling}\:\mathrm{in}\:\mathrm{an}\:\mathrm{open}\:\mathrm{car}\:\mathrm{moving} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{levelled}\:\mathrm{road}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed} \\ $$$$\mathrm{tosses}\:\mathrm{a}\:\mathrm{ball}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{in}\:\mathrm{the}\:\mathrm{air}\:\mathrm{and} \\ $$$$\mathrm{catches}\:\mathrm{it}\:\mathrm{back}.\:\mathrm{Sketch}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{ball}\:\mathrm{as}\:\mathrm{observed}\:\mathrm{by}\:\mathrm{a}\:\mathrm{boy}\:\mathrm{standing} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{footpath}.\:\mathrm{Give}\:\mathrm{explanation}\:\mathrm{to} \\ $$$$\mathrm{support}\:\mathrm{your}\:\mathrm{diagram}. \\ $$

Question Number 18031 Answers: 1 Comments: 0

What is the solution set of ((x + 2)/(x + 1)) = 1

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\:\:\frac{\mathrm{x}\:+\:\mathrm{2}}{\mathrm{x}\:+\:\mathrm{1}}\:=\:\mathrm{1} \\ $$

Question Number 23722 Answers: 0 Comments: 4

A block of mass m is pulled on the smooth horizontal floor using two methods I and II. The ratio of acceleration (a_I /a_(II) ) is

$$\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{pulled}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{floor}\:\mathrm{using}\:\mathrm{two} \\ $$$$\mathrm{methods}\:\mathrm{I}\:\mathrm{and}\:\mathrm{II}.\:\mathrm{The}\:\mathrm{ratio}\:\mathrm{of} \\ $$$$\mathrm{acceleration}\:\frac{{a}_{{I}} }{{a}_{{II}} }\:\mathrm{is} \\ $$

Question Number 17999 Answers: 1 Comments: 0

A large number of bullets are fired in all direction with same speed u. The maximum area on the ground covered by these bullets will be (1) π.(u^2 /g) (2) π.(u^4 /g^2 ) (3) (π/4).(u^4 /g^2 ) (4) (π/2).(u^4 /g^2 )

$$\mathrm{A}\:\mathrm{large}\:\mathrm{number}\:\mathrm{of}\:\mathrm{bullets}\:\mathrm{are}\:\mathrm{fired}\:\mathrm{in} \\ $$$$\mathrm{all}\:\mathrm{direction}\:\mathrm{with}\:\mathrm{same}\:\mathrm{speed}\:{u}.\:\mathrm{The} \\ $$$$\mathrm{maximum}\:\mathrm{area}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{covered} \\ $$$$\mathrm{by}\:\mathrm{these}\:\mathrm{bullets}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\pi.\frac{{u}^{\mathrm{2}} }{{g}} \\ $$$$\left(\mathrm{2}\right)\:\pi.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\pi}{\mathrm{4}}.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\pi}{\mathrm{2}}.\frac{{u}^{\mathrm{4}} }{{g}^{\mathrm{2}} } \\ $$

Question Number 17995 Answers: 0 Comments: 0

A pendulum bob of mass 2kg is attached to a string 2m long and made to revolve in horizontal circle of radius 0.8 , find the tension in the string.

$$\mathrm{A}\:\mathrm{pendulum}\:\mathrm{bob}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a}\:\mathrm{string}\:\:\mathrm{2m}\:\mathrm{long}\:\mathrm{and}\:\mathrm{made}\:\mathrm{to} \\ $$$$\mathrm{revolve}\:\mathrm{in}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{0}.\mathrm{8}\:,\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}. \\ $$

Question Number 17976 Answers: 1 Comments: 0

A driver is 5.2 below the surface of water of density 1000kg/m. if the atmospheric pressure is 1.02 × 10^5 Pa. calculate the pressure on the driver.

$$\mathrm{A}\:\mathrm{driver}\:\mathrm{is}\:\mathrm{5}.\mathrm{2}\:\mathrm{below}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{water}\:\mathrm{of}\:\mathrm{density}\:\mathrm{1000kg}/\mathrm{m}.\:\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{atmospheric}\:\mathrm{pressure}\:\mathrm{is}\:\mathrm{1}.\mathrm{02}\:×\:\mathrm{10}^{\mathrm{5}} \:\mathrm{Pa}.\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{pressure}\:\mathrm{on}\:\mathrm{the}\:\mathrm{driver}. \\ $$

Question Number 17970 Answers: 1 Comments: 0

Question Number 17969 Answers: 0 Comments: 0

Question Number 17961 Answers: 1 Comments: 0

Question Number 17938 Answers: 2 Comments: 2

If only downward motion along lines is allowed, what is the total number of paths from point P to point Q in the figure below?

$$\mathrm{If}\:\mathrm{only}\:\mathrm{downward}\:\mathrm{motion}\:\mathrm{along}\:\mathrm{lines}\:\mathrm{is} \\ $$$$\mathrm{allowed},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{paths}\:\mathrm{from}\:\mathrm{point}\:\mathrm{P}\:\mathrm{to}\:\mathrm{point}\:\mathrm{Q}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{figure}\:\mathrm{below}? \\ $$

Question Number 17919 Answers: 0 Comments: 0

How do the electronic configurations of the elements with Z = 107 − 109 differ from one another?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{the}\:\mathrm{electronic}\:\mathrm{configurations}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{elements}\:\mathrm{with}\:\mathrm{Z}\:=\:\mathrm{107}\:−\:\mathrm{109}\:\mathrm{differ} \\ $$$$\mathrm{from}\:\mathrm{one}\:\mathrm{another}? \\ $$

Question Number 17918 Answers: 0 Comments: 0

Write the electronic configuration and the block to which an element with Z = 90 belongs.

$$\mathrm{Write}\:\mathrm{the}\:\mathrm{electronic}\:\mathrm{configuration}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{to}\:\mathrm{which}\:\mathrm{an}\:\mathrm{element}\:\mathrm{with} \\ $$$$\mathrm{Z}\:=\:\mathrm{90}\:\mathrm{belongs}. \\ $$

Question Number 17879 Answers: 0 Comments: 0

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