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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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Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th
Tpy6a [65]

pH = 13.5

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}

The mixture would contain

  • 0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 0.5 = 0.05 \; \text{mol} of \text{Ac}^{-}

if \text{Ac}^{-} undergoes no hydrolysis; the solution is of volume 0.1 + 0.4 = 0.5 \; \text{L} after the mixing. The two species would thus be of concentration 0.30 \; \text{mol} \cdot \text{L}^{-1} and 0.10 \; \text{mol} \cdot \text{L}^{-1}, respectively.

Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

... where the water self-ionization constant pK_w \approx 14 under standard conditions. Thus pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3. By the definition of pK_b:

[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}

x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}

[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}

pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5

6 0
3 years ago
You have a mass of 54 kg and are sitting on a stool. The normal force the stool applies on you is ...
Karo-lina-s [1.5K]

Answer:

529.2 N

Explanation:

As we have studied the first law of motion, which states that every action has some reaction, equal in magnitude but having an opposite direction.

The force that is acting on the student will be due to gravitational force, that is equal to his weight.

                                              F=mg: 54kg x 9.8m/s^2 =529.2 N

So the weight of student is exerting downwards towards the stool and land. The stool will also exert a force on the student that will be equal in magnitude but opposite in direction, then it will be 529.2 N.

This is because the student is sitting in a constant state and all the weight is exerted on the stool.

Note: This answer is very generic supposing that all the weight of the student is on stool. But, if we suppose that student's legs are on floor so it means the force of gravity acting on the stool has become less because student's mass on stool is less. So the answer would be a force somehow less than 529.2 N.  However, since the question asked normal force, it would be weight of student in general terms.

Hope it helps!

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cestrela7 [59]
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Answer:

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Well, the definition of climate is the weather conditions prevailing in an area in general or over a long period so I don't really know

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