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ioda
2 years ago
10

49. How much precipitate would result if you have 50 grams of NaCl at 10 degrees Celsius and its cooled to saturation

Chemistry
1 answer:
Gennadij [26K]2 years ago
7 0

Option D is the correct answer , 13.5gm of precipitate would result if 50 grams of NaCl at 10 degrees Celsius and its cooled to saturation.

<h3>What is Solubility ?</h3>

the maximum quantity of a substance that will dissolve in a certain quantity of water at a specified temperature.

From the Solubility Graph of NaCl , we can see that the solubility of NaCl at 10 C is around 36.5gm/L.

If it is cooled down to saturation we will see the precipitate forming and that will be equal to the difference of solute present and the solubility.

As it is given that 50gm of solute is present

The precipitate will be 50- 36.5 gm, = 13.5 gm.

Therefore Option D is the correct answer , 13.5gm of precipitate would result if 50 grams of NaCl at 10 degrees Celsius and its cooled to saturation.

To know more about Solubility

brainly.com/question/8591226

#SPJ1

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A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
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<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

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mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

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