The transition metal with the smallest atomic mass is Scandium (Sc).
Hope this helps~
<span>100.
ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water
Molarity, M
M = number of moles of solute / volume of solution in liters
number of moles of solute = mass of CHCl3 / molar mass of CHCl3
molar mass of CHCl3 = 119.37 g/mol
number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol
using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters
M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer
Molality, m
m = number of moles of solute / kg of solvent
number of moles of solute = 0.838
kg of solvent = kg of water = 1,000,000 kg
m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer
mole fraction of solute, X solute
X solute = number of moles of solute / number of moles of solution
number of moles of solute = 0.838
number of moles of solution = number of moles of solute + number of moles of solvent
number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles
number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles
X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer
mass percent, %
% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =
% = 10 ^ - 6 % <------- answer
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Answer: (i)The object is at a distance of 120cm from the lens while the image is at a distance of 160cm from the lens. (ii)The image is Real
Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?
Explanation:
The required formulas are;
1. lens formula given by, <em>1/s' + 1/s = 1/f</em>
2. magnification = <em>h'/h = s'/s</em>
where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.
h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)
solving for s' and s in eqn (2)
4.50/3.20=s'/s
1.4=s'/s
s'=1.4s
to find the values of s' and s, we use equation (1) and substitute s' = 1.4s
1/1.4s + 1/s = 1/70
2.4/1.4s =1/70
s = 2.4*70/1.4 = 120cm
s' = 1.4*120 = 160cm
Therefore, the object is on the left, 120cm from the lens while the image is 160cm on the right of the lens
<em>Since the object is at a distance between f and 2f from the lens, the image formed is</em> real, inverted and magnified
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
