a) The least number of decimal points is 0 so you should round the answer to 43.
b) The least number of significant figures is 2 so round to 7.3
c)The least number of decimal places is 1 so round to 225.7
d) The least number of significant figures is 3 so round to 92.0
e) The least number of significant figures is 3 so round to 32.4
f) The least number of decimal places is 0 so round to 104m^3
Basically, for addition and subtraction round to the least number of decimal places found in the factors, and for multiplication and division round to the least number of significant figures found in the factors.
Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).
Molarity is one of the method of expressing concentration of solution. Mathematically it is expressed as,
Molarity =
Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l
∴ 5 =
∴
number of moles = 3.75Answer: Number of moles of KOH present in solution is 3.75.
<span>When two electrical charges, of
opposite sign and equal magnitude, are separated by a distance, a dipole is
established. The size of a dipole is measured by its dipole moment (</span>μμ). Dipole moment is measured in Debye
units, which is equal to the distance between the charges multiplied by the
charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment
of a molecule can be calculated by Equation 1.11.1:
μ = qr
where
<span>
<span>μ⃗ μ→ is the dipole moment vector</span>
<span>qiqi is the magnitude of the ithith charge, and</span>
<span>r⃗ ir→i is the vector representing the position
of ithith charge.</span>
</span>
r = μ/q
<span>r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19
C) *0.124
</span>
r = 1.41 x10^-10 m
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
