We have a 500 litre tank containing a liquid product contaminated by small particles that must be removed before the product can be sold. the particle concentration is, initially, 10' particles/m and the product may only be sold if the particle concentration is reduced to 10 particles/m. we have a filter that will effectively remove all particles from the liquid product. (a) at what rate must the product be pumped through the filter to make the product acceptable after 2 days; you may assume that the tank is well- mixed and the filtered product is continuously returned to tank. draw a sketch of this process (b) at the pumping rate determined in part (a) how long will it take to filter the product if a second vessel is used to hold the filtrate. what will the particle concentration in the filtered product be in this case? (c) repeat part (a) assuming the filter only removes 90% of the particles on each pass.

the answer is the first option

explanation:

just googled it m8. not that hard

lol

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

i would say b

explanation:

but i think i would need more information than that