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Lena [83]
3 years ago
5

¿Que mide la solubilidad en una disolución?

Chemistry
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

En lugar de utilizar las ya conocidas unidades de concentración, para la solubilidad es más común emplear la siguiente: Medimos la solubilidad como la cantidad de gramos del soluto que podemos disolver en 100 gramos de disolvente.

Explanation:

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How many seconds does it take to deposit 0.94 g of Ni on a decorative drawer handle when 14.9 A is passed through a Ni(NO3)2 sol
Goshia [24]

Answer:

Time take to deposit Ni is 259.02 sec.

Explanation:

Given:

Current I = 14.9 A

Faraday constant = 96485 \frac{C}{mole}

Molar mass of Ni = 58.69 \frac{g}{mole}

Mass of Ni = 0.94 g

First find the no. moles in Ni solution,

Moles of Ni = \frac{0.94}{58.69}

                   = 0.02 mol

From the below reaction,

  Ni^{2+}  + 2e ⇆ Ni_{(s)}

Above reaction shows "1 mol of Ni^{2+} requires 2 mol of electron to form 1 mol of Ni_{(s)} "

So for finding charge flow in this reaction we write,

    = 0.02 \times \frac{2  }{1 }  \times 96485 \frac{C}{mol}

Charge flow = 3859.4 C

For finding time of reaction,

  I = \frac{q}{t}

Where q = charge flow

   t = \frac{q}{I}

   t = \frac{3859.4}{14.9}

   t = 259.02 sec

Therefore, time take to deposit Ni is 259.02 sec.

3 0
3 years ago
ANSWER QUICK I VOTE AS BRAINLIEST
77julia77 [94]
The relative humidity of the air is the ratio of the actual amount of moisture in the air to the fully saturated amount. There for the answer is 33.1% or 3.3g/kg
5 0
3 years ago
Help me with this please!
Annette [7]

The correct answer is 9.7 grams because mg are 1,000 difference to grams.

8 0
3 years ago
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
2 years ago
A chemical reaction in which compounds break up into simpler constituents is a _______ reaction.
klio [65]
C. decomposition reaction
8 0
3 years ago
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