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  1. 6

    Troubles in chem p1

    This is a disproportion reaction oxidation number of I in HIO is +1, in I2 is 0 and in HIO3 is +5 so HIO to I2 is (+1 -> 0)x2 which is -2 because there is 2 Is in I2. and HIO to HIO3 is (+1->+5) which is +4 ratio is -2:4 is 1:2 so mol ratio must be 2:1
  2. 6

    Sitting A level physics and chemistry!

    Sitting A level physics and chemistry!
  3. 6

    Troubles in chem p1

    Here is a quick equation P1V1 + P2V2 = (total pressure) / (V1 + V2) so new P = (P1V1 + P2V2)/(V1 + V2) = (5x12 + 10x6) /(5+10) = 120/15 = 8 so correct answer is A
  4. 6

    Chemistry: Post your doubts here!

    group 2 nitrate decomposition 2 M(NO3)2 -> 4NO2 + O2 + 2 MO first we find the mass of MO solid formed as NO2 and O2 are both gases m(MO) = 3-1.53 = 1.47g and we know m(M(NO3)2) = 3g so 1.47/3 = m(MO)/m(M(NO3)2) = 16+M / M + (14 + 16x3) x2 = 16 + M /M + 124 solve the equation M = 87.7 so must be...
  5. 6

    Chemistry: Post your doubts here!

    w12 qp 11 #4: first remember bond breaking is + (endothermic) and bond forming is - (exothermic) N2 + O2 → 2NO so 1x NN triple bond and 1x O=O is broken and 2 N->O dative bond is formed the overall enthalpy change is +180 so +944+496 -2(N->O) = +180 2(N->O) = +1260 N->O = around 630 answer...
  6. 6

    Chemistry: Post your doubts here!

    #15 answer is D, -8 in oxidation number 5H2SO4 + 8KI → 4K2SO4 + 4I2 + H2S + 4H2O there are 2 ways to approach this problem, the obvious one is by by finding the change in oxidation number of the element oxidised which is I 8I- -> 4I2 oxidised from -1 -> 0 and there are 8 I-...
  7. 6

    AS Physics P1 MCQs Preparation Thread.

    hm ok original v is A, new v is B Ek = A^2 4Ek = B^2 so A^2 = (B^2)/4 square root both sides A = B/2 B = 2A so velocity is doubled
  8. 6

    AS Physics P1 MCQs Preparation Thread.

    for s12 qp11 #15 L is moved further to the right, so there is a bigger clockwise moment, meaning there need to be a bigger anticlockwise moment to counter it. note the system is in equilibrium so there must be no net force and no net torque immediately eliminate A and B because there is a...
  9. 6

    Chemistry: Post your doubts here!

    Answer is B, activation too high A: irrelevant, state only determines whether reaction will occur or not, and rate C: the bond that is broken is N=N not triple bond D: exothermic reactions occur easily, unless high EA so B is correct answer
  10. 6

    AS Physics P1 MCQs Preparation Thread.

    for w11 #16: initial Ek = (mv^2)/2 since in a collision momentum must be conserved, initial momentum = mv . final = 2m x (v/2) final Ek = (m (0.5v)^2)/2 = (mv^2)/4 so loses 50% of original Ek for s11 #4 amplitude is 4x1.5= 6 period = 4 x 5ms = 20ms #14 this one is tricky, utilize the...
  11. 6

    AS Physics P1 MCQs Preparation Thread.

    F(net)= W = mg= (120 - 80) x 9.81 = 400 Fnet = ma = (120 + 80) x a = 400 a=2 kinematics equation V^2= u^2 + 2as v^2 = 0 + 2x2x9 v = 6
  12. 6

    AS Physics P1 MCQs Preparation Thread.

    you resolve the motion into horizontal and vertical components the vertical initial velocity is 0 m/s with -9.81 m/s^2 acceleration and the distance is 1.25 so we apply the equation s(displacement) = ut + (gt^2)/2 1.25 = 0 + (9.81 x t^2)/2 = 4.905 t^2 square root, so t= 0.505seconds now we...
  13. 6

    AS Physics P1 MCQs Preparation Thread.

    variance 3, so taking paper 13
  14. 6

    AS Physics P1 MCQs Preparation Thread.

    #4 there are 5 divisions between 2-4 so each division is 0.4mA the pointer passes roughly 3.5 marks so 0.4x3.5 = 1.4mA 1.4+2 = 3.4mA answer is C #18 Ep=mgh = 50x9.81x1.6=784.8 so answer is C
  15. 6

    AS Physics P1 MCQs Preparation Thread.

    hey sorry, I'm sitting the exam tomorrow.
  16. 6

    Physics: Post your doubts here!

    The question states that "X is measured with a percentage uncertainty of ±1 % of its value at all temperatures." meaning that the percentage uncertainty is unchanged, so A and B is eliminated. then you need to apply some logic here, will the actual uncertainty get bigger as you approach 100 or...
  17. 6

    AS Physics P1 MCQs Preparation Thread.

    first we find the resistance of the wires 0.005 x 800 =4 and because there are 2 wires which forms the loop so 4x2 = 8 now we find the V dissipated by the resistance so V= IR = 8x0.6 = 4.8V this means that 4.8 will be lost during transmission and the relay requires 16, so 16+4.8 = 20.8
  18. 6

    AS Physics P1 MCQs Preparation Thread.

    ok each ocsillation occupies 6 squares so 0.2msx6= 1.2ms = 1.2x10^-3 s f = 1/T = 1/(1.2x10^-3) = 830 answer is B
  19. 6

    Physics: Post your doubts here!

    ok first you find the wavelength λ which is 2x the distance between 2 antinodes X and Y. so λ= 2x 33cm = 66cm = 0.66m and then v=f λ so f= v/ λ=330/0.66=500 we know that period (T) = 1/f = 1/500 = 0.2ms for 1 oscillation so answer is B
  20. 6

    AS Physics P1 MCQs Preparation Thread.

    np, good luck on exam! I'm having mine tomorrow, hopefully I'll get 90%+ this time
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