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inessss [21]
3 years ago
7

a bottle of rubbing alcohol is labeled 70%(v/v) how many milliliters of isopropyl alcohol are in a 400ml bottle of this solution

?
Chemistry
1 answer:
xz_007 [3.2K]3 years ago
3 0
 The number of   milliliters  of  isopropyl alcohol that are  in   a 400 ml bottle  of this solution  is 280 Ml

   calculation

  % V/V =  volume of  solute/ volume  of  solution  x 100
 Let the volume  of solute  be =  y  ml
 the volume  of  solution =  400 ml
convert 70%  to fraction = 70/100

 by  substituting  the value  in the equation 

 = y  ml/  400 ml =  70/100

by cross  multiplication

100y =400 x70
100y=  28000
y= 280 ml
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<h3>1200 J</h3>

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8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c
dangina [55]

Answer:

Partial pressure of O_{2} in the gas was 733 torr and mass of KClO_{3} in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of O_{2})

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of O_{2}= (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that O_{2} behaves ideally. Hence-

                                            PV=nRT

where P is pressure of O_{2}, V is volume of O_{2} , n is number of moles of O_{2} , R is gas constant and T is temperature in kelvin

here P = 733 torr = (733\times 0.001316)atm = 0.9646 atm

        V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

   So, n=\frac{PV}{RT}

                   = \frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}

                   = 0.0259 moles

As 3 moles of O_{2} are produced from 2 moles of KClO_{3} therefore 0.0259 moles of O_{2} are produced from (\frac{2\times 0.0259}{3}) moles or 0.0173 moles of KClO_{3}.

Molar mass of KClO_{3}= 122.55 g

So mass of KClO_{3} in sample = (0.0173\times 122.55)g

                                                                    = 2.12 g

7 0
3 years ago
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svetoff [14.1K]

The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 \times 10^-5 V.

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where p is a permanent electric dipole,

           ∈ο is permittivity,

            r is the radius from the axis of a dipole,

            V is the electric potential.

        V = 1 / (4 \times 3.14 \times 8.85 \times 10^-12)  \times (4.90 \times 10^-30 \times 1) / (55.3 \times 10^-9)^2

        V  = 1.44 \times 10^-5 V.

8 0
3 years ago
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