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Viefleur [7K]
3 years ago
9

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% soluti

on. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
Chemistry
2 answers:
Akimi4 [234]3 years ago
6 0
The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution. 
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
<span>.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
</span><span>.8x = 14
</span><span>x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

</span>
Snowcat [4.5K]3 years ago
5 0

Answer:

17.5 litres removed and 17.5 litres of pure antifreeze added

Explanation:

Let k equal the amount of the solution to be removed

amount of antifreeze to be added

0.2(70 - x) + x = 0.4(70)

14 - 0.2x +x = 28

0.8x = 28 -14

x = 14/0.8 = 17.5

x = 17.5 litres removed and 17.5 litres of pure antifreeze added

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Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

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3 years ago
Consider the following types of electromagnetic radiation:
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Explanation:

Electromagnetic wave              Wavelength

(1) Microwave  =   1 m to 1 mm = 10^9 nm to 10^6 nm

(2) Ultraviolet  =    10 nm to 400 nm

(3) Radio waves  =   1 mm to 100 km = 10^6 nm to 10^{14}nm

(4) Infrared  =    700 nm to 1 mm

(5) X-ray  =   0.01 nm to 10 nm

(6) Visible =   400 nm t0 700 nm

a) In order of increasing wavelength:

: 5 < 2 < 6 < 4 < 1 < 3

b) Frequency of the electromagnetic wave given as:

\nu=\frac{c}{\lambda }

\nu = frequency

\lambda = Wavelength

c = speed of light

\nu \propto \frac{1}{\lambda }

So, the increasing order of frequency:

: 3 < 1 < 4 < 6 < 2 < 5

c) Energy(E) of the electromagnetic wave is given by Planck's equation :

E=h\nu

E\propto \nu

So, the increasing order of energy:

: 3 < 1 < 4 < 6 < 2 < 5

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3 years ago
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3 years ago
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0
3 years ago
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