<span>There
are a number of ways to express concentration of a solution. This includes
molarity and molality. Molarity is expressed as the number of moles of solute per volume of
the solution. MOlality is expressed as moles per kg solution.
5.25 mol H2SO4 / kg solution ( 1 kg / 1000 g ) ( 1.266 g / mL ) ( 1000 mL / 1L ) = 6.6 M H2SO4</span>
C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
The matter is going to have to be in a <em>plasma </em>state! =)
Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2
From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams
Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams
From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid
To get the amount of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
<span>The equation that describes the problem is Fe(NO3)3(aq) + 3NaOH(aq) ---> Fe(OH)3(s) + 3 NaNO3(aq)
The Net ionic equation is written as follows:
Fe^3(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq) ---> Fe(OH)3(s) + 3Na+(aq) + 3NO^3-(aq)</span>
