questions 1 and 2 A home owner has a pond and uses a mixture that contains 11.3%

Question

questions 1 and 2

A home owner has a pond and uses a mixture that contains 11.3% Bromine by mass to kill an algae bloom. An ideal Bromine level for the pond is 3 parts per million (3ppm). (Think of 1 ppm as being 1 g Bromine per million grams of water) If you assume densities of 1.17 g/mL for the Bromine solution and 1.00 g/mL for the pond water, what volume of the bromine solution, in liters is required to produce a bromine level of 3 ppm in a 50,000-gallon pond? In seawater, Beryllium has a concentration of 2.44 g Beryllium per kilogram of seawater. What volume of Seawater, in cubic meter, would have be processed to produce 3.25 times 10^4 tons of Beryllium (1 ton = 2000 Ib)? Assume a density of 1.045 g/mL for seawater. If a piece of copper wire is 1.34-m in length, how many atoms are present? The wire has a diameter of 0.02496 in, and the density of copper is 8.92 g/cm^3. TreeAzin ^Tm, an insecticide treatment for Emerald Ash Borer contains 5% by mass of Azadirachtin, C_35H_44O_16. How many mg of Carbon (from chemical) does a larvae ingest by eating 0.0134 oz of cambium tree tissue?

Explanation / Answer

Question 2:

Concentration Be = 2.44 g Be/kg seawater

this means that:

2.44 g Be -------------> 1 kg seawater

Let's convert the tons to gram:

3.25x104 tons * 2000 lb/ton * 454 g/lb = 2.951x1010 g

2.44 g Be ----------------> 1 kg seawater

2.951x1010 g Be ------------> X

X = 2.951x1010 / 2.44 = 1.21x1010 kg seawater

1.21x1010 kg * 1000 g/kg = 1.21x1013 g seawater

V = 1.21x1013 g /1.045 g/cm3 = 1.16x1013 cm3

V = 1.16x1013 cm3 * 1 m3/1x106 cm3 = 1.16x107 m3 of seawater.

I have a little doubt with a conversion for question 1, so instead of posting something unsure, it's better for you to post that question in another question thread.

Hope this helps you

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