If we analyze the situation analytically, there are situations or states. Then, we are also given with 2 values of pressure and 1 value of volume. Lastly, temperature was set as constant. Thus, this means we use the Boyle's Law.

P₁V₁ = P₂V₂
Let's find V₂.
(1 atm)(1.72 L) = (35 atm)(V₂)
Solving for V₂,
<em>V₂ = 0.049 L</em>

7 0

3 years ago

How many molecules of Na are in 3.2 moles of Na?

pav-90 [236]

Answer: 1.9 x 10²⁴ molecules Na

Explanation: To solve for the molecules of Na, we will use the Avogadro's number.

3.2 moles Na x 6.022 x10²³ molecules Na/ 1 mole Nà

= 1.9 x 10²⁴ molecules Na

7 0

2 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

3 years ago