what is the expected freezing point of a 0.50 m so...
what is the expected freezing point of a 0.50 m solution of li 2so4 in water? kf for water is 1.86°c/m.

a) -0.93°c

b) -1.9°c

c) -2.8°c

d) -6.5°c
-2.8oc
heat lost by the vessel and water= [150 x 4.2 x (30 - 5) j + 100 x 0.4 x (30-5)] j = (150 x 4.2 x 25 + 100 x 0.4 x 25) j = (15,750 + 1000) j= 16,750 j heat gained by ice to change into water at 0 °c = m x specific latent heat of fusion of ice = {m x 336} heat gained by water to come to final temperature (5 °c) = m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m if there is no loss of heat to the surroundings, according to the principle of calorimetry: heat gained by ice to melt and come to final temperature = heat lost by vessel and water or 336m + 21m= 16,750 or 357m = 16,750 or m = 16750/357 = 46.92 (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2 the formulas for converting between degree celsius and degree fahrenheit are: °f = (°c * 9/5) + 32 °c = (°f - 32) * 5/9 to find the temperature when both are equal, we use an old algebra trick and just set of = oc and solve one of the equations. °c = (°c * 9/5) + 32 °c - (°c * 9/5) = 32 - 4/5 * °c = 32 °c = - 32 * 5/4 °c = - 40 °f = (°f * 9/5) + 32 °f - (°f * 9/5) = 32 - 4/5 * °f = 32 °f = - 32 * 5/4 °f = - 40 so the temperature when both the celsius and fahrenheit scales are the same is - 40 degrees. the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. the north pole sits on a floating ice sheet which is called the arctic ice sheet. there is no land underneath it, but the sea depth at the north pole has been measured at 4,261 metres (13,980 ft). because of this, the north pole is incapable of supporting any plant or animal life on it. v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> Correct answer is 26 mm. Mass of powder is constant. mi=mf  ⇒ ρiVi = ρfVf ρihi = ρf hf ⇒ hi = (ρf hf)/ρi On substitution, (1755×12)/ 810= 26 mm  


Correct answer is 26 mm.


Mass of powder is constant.


mi=mf ⇒ ρiVi= ρfVf


ρihi = ρf hf⇒ hi = (ρf hf)/ρi


On substitution, (1755×12)/ 810= 26 mm


 


[See less](#) use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4 Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm