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Phantasy [73]
2 years ago
5

How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt? Give your answer in kg/day.

Chemistry
1 answer:
ser-zykov [4K]2 years ago
7 0

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0m^{3}/s

1m^{3}/s = 1000L

Multiply by 1000 to convert  to L/s

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

                     = 2,592,000,000L/day

if the river contains 50mg of salt  in 1L of solution

lets find how many mg of salt (X) is contained in 2,592,000,000L/day

X= \frac{ 2,592,000,000*50}{1}

X= 129,600,000,000 mg/day

convert this value to kg/day by multiply by 10^{-6}

X= 129,600kg/day

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A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb
Flura [38]

Answer:

C3H6O2

Explanation:

To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.

H = 4.10/1.36 = 3

O = 1.36/1.36 = 1

C = 2.05/1.36 = 1.5

We then multiply through by 2 to yield the compound with the empirical formula C3H6O2

7 0
2 years ago
If you weigh 100 kg, how much would you weigh if all the water were removed from your body? A65 kg B45 kg C50 kg D35 kg
Mekhanik [1.2K]

Answer:

B. 45k

The human body is about 60 to 70% water.

(:

5 0
2 years ago
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Anna11 [10]
Molar mass  CaCl₂ =  110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 =  0.7686 / V

V = 0.7686 / 0.788

V = 0.975 L

hope this helps!
5 0
2 years ago
Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.
Travka [436]

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means  1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × \frac{254 g}{1 mole}= 508 g of Phenobarbital sodium.

1 L ×\frac{1000 mL}{ 1 L} = 1000 mL of solution

Thus, % w/v is:

\frac{508 g}{1000 mL} × 100 = 50,8 % w/v

And radio strengths:

\frac{1000 mL}{508 g}  = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

5 0
2 years ago
You have a mixture that contains 0.380 moles of Ne(g), 0.250 moles of He(g), and 0.500 moles CH4(g) at 400 K and 7.25 atm. What
boyakko [2]
I think it is the pacific ocean
5 0
2 years ago
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