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vodka [1.7K]
3 years ago
12

You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that

contains 5% naocl by weight. assuming that the density of the stock solution is 1.0 g/ml, how many milliliters of bleach should you add to each liter of test solution?
Chemistry
1 answer:
Alecsey [184]3 years ago
4 0

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

                                     = 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

                        = 0.005 g

B = \frac{0.005}{0.05}

  = 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

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A nonvolatile solute is dissolved in benzene so that it has a mole fraction of 0.139. What is the vapor pressure of the solution
lapo4ka [179]

Answer:

The vapor pressure of the solution is 3.69 torr

Explanation:

Step 1: Data given

Mole fraction of benzene in the solution = 0.139

P° of benzene is 26.5 torr

Step 2: Calculate the vapor pressure of the solution

Psolution = Xbenzene * P°benzene

⇒with Psolution = the vapor pressure of the solution

⇒with Xbenzene = the mole fraction of benzene = 0.139

⇒with P°benzene = the vapor pressure of pure benzene = 26.5 torr

Psolution = 0.139 * 26.5 torr

Psolution = 3.69 torr

The vapor pressure of the solution is 3.69 torr

7 0
3 years ago
I need help ASAP please
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Answer:

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Explanation:

6 0
2 years ago
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BASED ON THE INFORMATION BELOW. CALCULATE THE MASS FOR ONE OF THESE ATOMS:
stepladder [879]

C = 12 g

O = 16 g

H = 1 g

<h3>Further explanation </h3>

Conservation of mass stated that  

<em>In a closed system, the masses before and after the reaction are the same </em>

we can calculate the mass of each atom in the compound :

O in O₂ :

mass O₂ = 32

mass O = 32 : 2 = 16 g

H in H₂O

mass H₂O = 18

mass 2.H + mass O = 18

mass 2.H + 16 = 18

mass 2.H=2

mass H = 1 g

C in CH₄

mass CH₄ = 16

mass C + mass 4.H = 16

mass C + 4.1=16

mass C = 12 g

or we can use formula :

Mass of a single C :

\tt =\dfrac{Ar~C}{MW~CO_2}\times mass~CO_2\\\\=\dfrac{12}{44}\times 44=12

7 0
3 years ago
What is tins atomic symbol
qaws [65]
Tin
Chemical Element
Tin is a chemical element with the symbol Sn and atomic number 50. It is a main group metal in group 14 of the periodic table. Wikipedia
Symbol: Sn
Electron configuration: [Kr] 4d105s25p2
Atomic number: 50
Melting point: 449.5°F (231.9°C)
Atomic mass: 118.71 u
Boiling point: 4,717°F (2,603°C)
Electrons per shell: 2, 8, 18, 18, 4
4 0
3 years ago
Read 2 more answers
The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th
Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at T_2

K_1 = rate of reaction at T_1

Ea = activation energy of the reaction

R = gas constant = 8.314 J/K mol

E_a=300 kJ/mol=300,000 J/mol

T_2=625^oC=898.15 K,T_1=525^oC=798.15 K

\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]

\log (\frac{K_2}{K_1})=2.185666

K_2=153.344\times K_1

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0
3 years ago
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