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1 g of a moist sample of KCl And KClO3 was dissolved in water and made up to 250 ml. .25 ml of the solution was treated with SO2 to reduce the chlorate to chloride and excess SO2 was removed by boiling . The total chloride wa ...
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1 g of a moist sample of KCl And KClO3 was dissolved in water and made up to 250 ml. .25 ml of the solution was treated with SO2 to reduce the chlorate to chloride and excess SO2 was removed by boiling . The total chloride wa ...
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How many litres of SO2 has to be passed through HClO3 to reduce 16.9 g of it to HCl.. Ans 13.44 ...
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anyone received it ????????????????? ...
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anyone received it ????????????????? ...
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How many litres of SO2 has to be passed through HClO3 to reduce 16.9 g of it to HCl.. Ans 13.44 ...
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anyone received it ????????????????? ...
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How newton founded his II and IIIrd laws ? ANS >>>> He gave a force " F " by kicking a goat, it made a sound " Ma " so he formulated IInd law " F = Ma " AFTER a few seconds goat kicked him back. Then he formulated ...
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*Image* ans is a) but shouldnt it be b ???? ...
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anyone received it ????????????????? ...
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The exponential function ex is strolling along the road insulting the functions he sees walking by. He scoffs at a wandering polynomial for the shortness of its Taylor series. He snickers at a passing smooth function of compa ...
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15 ml of a gaseous hydrocarbon was required for combustion in 357 ml of air ( 21% of O2 by volume ) . If the gaseous products occupied 327 ml (at NTP) what is the formula of the hydrocarbon )... Ans c3h8 ...
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15 ml of a gaseous hydrocarbon was required for combustion in 357 ml of air ( 21% of O2 by volume ) . If the gaseous products occupied 327 ml (at NTP) what is the formula of the hydrocarbon )... Ans c3h8 ...
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15 ml of a gaseous hydrocarbon was required for combustion in 357 ml of air ( 21% of O2 by volume ) . If the gaseous products occupied 327 ml (at NTP) what is the formula of the hydrocarbon )... Ans c3h8 ...
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anyone received it ????????????????? ...
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Guys do u feel BT aits are getting tougher that FIITJEE even??????????????????wat i want to know is ......i jus got the STimulator tests ( YELLOW BOOK) ...the tests in them are easier ..........bit are they of JEE LEVEL...... ...
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15 ml of a gaseous hydrocarbon was required for combustion in 357 ml of air ( 21% of O2 by volume ) . If the gaseous products occupied 327 ml (at NTP) what is the formula of the hydrocarbon )... Ans c3h8 ...
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15 ml of a gaseous hydrocarbon was required for combustion in 357 ml of air ( 21% of O2 by volume ) . If the gaseous products occupied 327 ml (at NTP) what is the formula of the hydrocarbon )... Ans c3h8 ...
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2 g of P4 reacts completely with 2 g of o2 to produce P4O6 and P4O10... find the weight of P4O6 and P4O10 produced.? ...
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HARD copy of Application Number********** has been received and under scrutiny. Last updated on : 02-03-2010 is everyone getting like this ??????????????? still under scrutiny ???? ...
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2 g of P4 reacts completely with 2 g of o2 to produce P4O6 and P4O10... find the weight of P4O6 and P4O10 produced.? ...
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2 g of P4 reacts completely with 2 g of o2 to produce P4O6 and P4O10... find the weight of P4O6 and P4O10 produced.? ...
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1....... *Image* ...
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A line intersencts hte ellipse x2/a2 +y2/b2=1 at P and Q and the parabola y2=4d(x+a) at R and S.The line segment PQ subtend right angle at the centre of ellipse.Find the locus of hte point of intersection of tangents to the p ...
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1 ) if iiiiii...∞ = a+ib prove that tan 1/2 pi a = b/a and a2+b2=e-pib... 2) if z1 and z1 are the roots of the equation αz2+βz+γ=0 prove that |z1|+|z2| = 1/|α|[ |-β+√αγ|+|-β-√αγ|] ...
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1 ) if iiiiii...∞ = a+ib prove that tan 1/2 pi a = b/a and a2+b2=e-pib... 2) if z1 and z1 are the roots of the equation αz2+βz+γ=0 prove that |z1|+|z2| = 1/|α|[ |-β+√αγ|+|-β-√αγ|] ...
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1 ) if iiiiii...∞ = a+ib prove that tan 1/2 pi a = b/a and a2+b2=e-pib... 2) if z1 and z1 are the roots of the equation αz2+βz+γ=0 prove that |z1|+|z2| = 1/|α|[ |-β+√αγ|+|-β-√αγ|] ...
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1 ) if iiiiii...∞ = a+ib prove that tan 1/2 pi a = b/a and a2+b2=e-pib... 2) if z1 and z1 are the roots of the equation αz2+βz+γ=0 prove that |z1|+|z2| = 1/|α|[ |-β+√αγ|+|-β-√αγ|] ...
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Arrenge the fllowing ions in correct order of decreasing nucleophilicity in non polar solvent OH- ,NH2- ,F- ,CH3- ...
