Seniorhighub

a. ∇N = √N = √1600 = 40.0
b. The count rate is 1600 counts / 10s = 160 count/s.
c. The activity decreases by half every 200 s do the half-life is 200 s
d. Activity = ∇n = 0.693T1/2 = 0.693 × 106 / 200 S = 3465 Bq

1.   a. producers
0.08 = 4
0.02.

b. herbivorous fish
20 = 1,000
0.02

c. predatory fish
60 = 3,000
0.02

d. hawk
90 = 4,500
0.02

2. The concentration factor keeps increasing and larger and larger as you move up and as such it can be concluded that it is a persistence chemical it is not easily broken down it is stored rather than metabolising that is why it keeps getting a higher and higher concentration.

3. 4th trophic level (Top carnivore)
4. 4th trophic level

a.  suppose the speed is u and the fall time is t ; then ut = 200 m. the time to fall h = 20 m from rest is t = √2h/g = 2.00 s so u = 200.0 m / (2.00 s) = 0.100 kms^-1

b.   the vertical component of the velocity at the foot of the cliff vy = uy + gt = 10m s^-2 × 2 s = 20.0 m s^-1 (or vy = √2gh = 20.0 m s^-1) whereas the horizontal component is the firing speed. So the speed at the foot of the cliff V = √v ²/_x + v ²/_y = √(100. M s^-1)² + (20.0 m s^-1)² = 102 m s^-1

c.  The impulse on the impact has a magnitude | = mv = 0.100kg × 02 m s^-1 = 10.2N s so the average stopping force is |/t = 2.04 N

a.  Logxⁿ = nlogx = np

b.  Logx, logx², logx³,…,… are simply p, 2p, 3p, 4p, … a linear sequence with first term a=p and arithmetic difference d = p

c.  Logx + logx² + logx³ + …+ logx¹⁰ = p + 2p + 3p + 5p + . . . + 10p =p(1 + 2 + 3 + . . . + 10)
=p(10/2)(1 + 10) = 55p

Hence k = 55

With phenolphthalein, the reaction in the titration is as follows;
HCI + Na₂Ca₃ → NaHCO₃ + NaCl

Molar mass of Na₂Ca₃ = 46 +12 + 48 = 106

Moles of Na₂Ca₃ weighed = 0.106/106 = 0.001 mmole of Na₂Ca₃ weighed = 1.0

Since HCI and Na₂Ca₃ react in a 1:1 ratio, mmole of HCI used = 1.0
But mmole = moldm^-3 × V in cm³

Therefore, concentration of HCI = 1.0 mmol/20 cm³
= 0.050 moldm^-3

1.  The pot containing the small watered plant is covered up to the base of the stem with the plastic sheet to prevent evaporation.

2.  The pot is cut across the stem just above the soil level with the knife

3.  The glass tubing is attached to the cut end of the stem using the rubber tubing.

4.  The eosin solution is poured into the glass tube and the level is marked.

5.  The setup is left for several hours after which the final level of the eosin solution is measured.

6.  The difference between the first level and the final level of the eosin solution gives the column of water supported by the root pressure.

a)  The parameter a accounts for interactions between the molecules of the gas. The parameter b accounts for non-zero volume of the molecules of the gas.

b)  R is the universal gas constant (R = N_Ak_B = 8.314 J K^-1 mol^-1).

c)  i)  Substitute the given values of the parameters to obtain T = ( p + n² a/V²) (V/n - b) / R = (140000 Pa + (2 mol)² × 4.17 J m³ / (0.025 m³)²) × (0.025 m³ / 2 mol – 3.71 × 10^-5 m³ mol^-1) / 8.314 J K^-1 mol^-1 = 250 K

ii)  An ideal gas at the same volume and pressure will have a temperature of Pv / nR = 210 K.

a)  Maximum power density equals solar constant times efficiency, that is, 200 W ms^-2.

b)  Each panel has 2 m², so each panel can give 400 W. all five can give 2000W.

c)  Maximum energy the installation can deliver in 2 h is the product of power and time, which is 2000 W × 7200 s = 14400 kJ.

a.  The perimeter of a rectangle is twice the sum of its length and breadth, giving 1100m.

b.  The volume of the wall equals its perimeter times its height times its thickness, that is, 1100 m × 2 m × 0.2 m = 440 m³ ; so its mass, given by its density multiplied by its volume, is 1.1 × 10⁶ kg.

c.  Pressure equals weight of wall divided by base area: P = F/A =mg/A = 1.1 × 10⁶ kg × 10 ms² / ( 1100 m × 0.2 m) = 50 000 Nm^-2

a.  The plant lost 40g in 10 hours, so its rate of transpiration was 40/10 = 4g per hour.

b.  I)  in daylight, the mass loss due to transpiration will be reduced by a gain in mass resulting from photosynthesis.

ii)  In darkness some of the decrease in mass will be due to the loss of water and carbon dioxide produced by respiration

c.  If the plant had been short of water, this might have restricted the rate of transpiration.

d.  The plastic bag prevented evaporation taking place from the pot or the soil. Had this evaporation not been prevented, the mass loss could not have been attributed solely to transpiration.

Graham's law; rate is proportional to 1/√(molar mass)

Therefore R₂/R₁ = √(M₁ /M₂); and

R₂ = R₁ √(M₁ /M₂)

Where R₁ and M₁ are the rate and molar mass of gas-1 respectively and R₃ and M₂ for gas-2

(1) For hydrogen gas, M = 2 RH2 = X√(20 /2) = X√10

(2) For Helium; RHe = X√(20 /4) = X√5

(3) For argon; RAr = X√(20 /40) = X/√2

(4) For bromine; RBr = X√(20 /160) = X/2√2

(5) For ammonia; RNH3 = X√(20 /17) = X√(20/17)

i.   1.  LL – Long nose

2.  Dd – Dark body

3.  Mm – Muscular body

4.  RR – Round eye

5.  Rr – Round eye

6.  II - Stubby nose

7.  MM - Muscular body

8.  DD - Dark body


ii. Write the genotypes for each of the following

k.  Yellow body – dd

l.  Skinny body – mm

m.  Oval eyes – rr

n.  Long nose – LL and LI

o.  Stubby nose – II

p.  Round eyes – RR and Rr

q.  Muscular body – MM and Mm

r.  Dark body - DD and Dd

s.  Heterozygous round eyes - Rr

t.  Homozygous long nose - LL

(1 + kx)⁶ = 1 + 6(kx) + 15(kx)² + 20(kx)³ + ….
(Pascal's triangle)
= 1 + 12x + px2 + qx3 + …

Comparing coefficients i) For x, 6k = 12, hence k =2

ii) For x², 15k² = p, hence p = 15(4) = 60

iii) For x³, 20k³ = q, hence q = 20(8) = 180

iv) The term in x⁴ is 15(kx)⁴ = 15(2)⁴x⁴

The coefficient of x⁴ = 15(16) = 240

The total mass is M = m₁ + m₂ + m₃ and the acceleration of each of them is a=F/M. The only forces acting on m₃ is the tension T₁ in the string attaching it to m₂ so, by Newton's second law,
m₃a = T₁ = m₃F / (m₁ + m₂ + m₃).

The forces acting on m₂ are the tension T₁ in the string connecting it to m₁ and the tension T₁ in the string connecting it to m₃. By Newton's second law, T₂ – T₁ = m₂a + T₁ = (m₂ + m₃)a = (m₂ + m₃) F / (m₁ + m₂ + m₃)

a)  Y = x³ – 27x (stationary points) dy/dx = 3x² – 27 = 0 X2 – 9 = 0, x = ±3

For x = 3, y = 33- 27(3) = 27 – 81 = -54 i.e. (3, -54) (stationary point)

for x = -3, y = -27 + 81 = 54, i.e.(-3, 54) (stationary point) d²y/dx² = 6x,

for x = 3, d²y/dx² = 3(6) = 18 > 0, hence (3, -54) is a minimum point.

For x = -3, d²y/dx² = 6(-3) = -18 < 0. Hence (-3, 54) is a maximum point. b.  point of inflexion

d²y/dx² = 6x = 0, hence x = 0 for x = 0, y = 0, hence point of inflexion is (0, 0)

C₆H₆ + 7½O₂ ⟶ 6 CO₂ + 3 H₂O
Δ H combustion = Σ ΔHf(products) − Σ ΔHf(reactants)

= (−6 × 394− 3 × 242) − 49 = (− 2364 − 726) − 49
= −3139 or −3.14×103 kJmol^-1

With phenolphthalein, the reaction in the titration is as follows;

HCI + Na₂Ca₃ ⟶ NaHCO₃ + NaCl Molar mass of Na₂Ca₃ = 46 + 12 + 48 = 106

Moles of Na₂Ca₃ weighed = 0.106/106 = 0.001 mmole of Na₂Ca₃ weighed = 1.0
Since HCI and Na₂Ca₃ react in a 1:1 ratio, mmole of HCI used = 1.0

But mmole = moldm^-3 × V in cm³
Therefore, concentration of HCI = 1.0 mmol/20 cm³ = 0.050 moldm^-3

1. a.  Clouds  b. Plants/Vegetation  c.Sea
2. Humans are likely to interrupt the water cycle
   • Between rivers and the sea
   • Between soil and rivers
   • Beween plants and clouds
3. • By irrigation and domestic and industrial use
   • By drainage schemes, afforestation or deforestation
   • By deforestation