Question

Table 1. Seawater major constituent concentrations. (Concentrations for salinity = 34.7‰.) Table 1 information: Constituent: Chlorine...

Table 1. Seawater major constituent concentrations. (Concentrations for salinity = 34.7‰.)

Table 1 information: Constituent: Chlorine , % of total salinity: 55.04, parts per million: 18980

1. Table 1 indicates that the concentration of Cl-  in a sample of sea water with a salinity of 34.7 is 18,980 ppm. This means that there are 18,980g of Cl- every ____ of water.

a. 1,000 g
b. 10,000 g
c. 100,000 g
d. 1,000,000 g
e. 1,000,000 kg

2. Therefore, the concentration of Cl- in the same sample can also be expressed as ____ (remember that 1,000g = 1 kg)

a. 18,980 g/1,000 g = 18,980 g/kg
b. 18,980 g/10,000 g = 1,898 g/kg
c. 18,980 g/100,000 g = 189.8 g/kg
d. 18,980 g/1,000,000 g = 18.98 g/kg
e. 18,980 g/1,000,000 kg = 0.019 g/kg

3. Using the Principle of Constant Proportions we know that the ratio of the major constituents always remains the same in seawater, which also means that the ratio ion concentration/seawater salinity is constant. Therefore we can write _____, where x is the unknown concentration of Cl-.

a. (34.7/0.019)=(32.8/x)
b. (18.98/34.7)=(x/32.8)
c. (189.8/34.7)=(x/32.8)
d. (34.7/1,898)=(32.8/x)
e. (18,980/34.7)=(x/32.8)

4. Hence, the concentration of Cl- in seawater with a salinity of 32.8 is __ g/kg.

a. 0.018
b. 17.94
c. 179.4
d. 1,794
e. 17940.7

Homework Answers

Answer #1

1) Assume the density of sea water is 1 g/mL.

The Cl- concentration is 18980 ppm. We know that 1 ppm = 1 mg/L; therefore, 18980 ppm = 18980 mg/L.

Again, the mass of 1 L of water = (1 L)*(1000 mL/1 L)*(1 g/1 mL) = 1000 g.

Therefore, we can write 18980 ppm = 18980 mg/L = 18980 mg/1000 g; thus, we have (a) 1000 g of water.

2) We know that 1000 g = 1 kg; therefore, we can write the Cl- concentration as 18980 mg/1000 g = 18980 mg/kg. Therefore, (a) is the correct answer.

3) The concentration for salinity is 18980 ppm while the salinity is 34.7%; therefore, we have 18980/34.7 = x/32.8 where 32.8% is the salinity and x is the unknown Cl- concentration. (e) is the correct answer.

4) Solve the above equation and obtain

x = 18980*32.8/34.7 = 17940 g/kg (ans).

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The concentrations of Mg2+ and Cl- in seawater are 52.8 mmol kg-1 and 545.9 mmol kg-1...
The concentrations of Mg2+ and Cl- in seawater are 52.8 mmol kg-1 and 545.9 mmol kg-1 , respectively. Globally, the average concentration of Mg2+ in river water is 0.128 mmol kg-1 , while that of Cl- is 0.220 mmol kg-1 . Given a global ocean mass of 1.38 x 1021 kg and an annual river flow rate of 3.5 x 1016 kg y-1 , how much longer is the residence time of Cl- in the ocean to that of Mg2+?
1.) You will work with 0.10 M acetic acid and 17 M acetic acid in this...
1.) You will work with 0.10 M acetic acid and 17 M acetic acid in this experiment. What is the relationship between concentration and ionization? Explain the reason for this relationship 2.) Explain hydrolysis, i.e, what types of molecules undergo hydrolysis (be specific) and show equations for reactions of acid, base, and salt hydrolysis not used as examples in the introduction to this experiment 3.) In Part C: Hydrolysis of Salts, you will calibrate the pH probe prior to testing...
1) Describe an example of each of the following that may be found of your kitchen:...
1) Describe an example of each of the following that may be found of your kitchen: Explain how your choice falls into this category, and if there is a chemical name or symbol for it, provide that as well. Provide a photo of your example with your ID card in it. a) a compound b) a heterogeneous mixture c) an element (symbol) Moving to the Caves… Lechuguilla Caves specifically. Check out this picture of crystals of gypsum left behind in...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT