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2 years ago

What are the uses of evaporative salts?

2 answers:

6 0

Answer : Evaporative salts are majorly used as common salts or halites, which are highly and widely used to preserve foods, dye fabric, and de-ice roads.

Explanation : Evaporative salts are produced by evaporation of the sea water hence it is named as evaporative salts. These are mainly extracted through evaporation from seawater. The salts from shallow ponds where the seawater gets collected in a land, which is later harvested and then purified.

abruzzese [7]2 years ago

5 0

<span>A solid can evaporate by melting into a liquid, which then evaporates; or by changing instantly into a vapor, or subliming. The rate of evaporation of a substance depends on its surface temperature, the pressure, and the humidity. Evaporation is the procedure by which a liquid or a solid changes into a vapor. A substance may evaporate by changing into a vapor at the surface, as when water evaporates from an uncovered dish.</span>

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To form an oxygen molecule o2 two oxygen atoms share two pairs of electrons what kind of bond is shown below by the electron dot

Alexus [3.1K]

They have a Covelant bond

7 0

2 years ago

Read 2 more answers

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr

lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Using the equation,

Δ $T_{f}$ = i $K_{f}$ m

where:

Δ $T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ $T_{f}$ = i $K_{f}$ m we get,

Δ $T_{f}$ = 2 × 1.86 × 0.1219

Δ $T_{f}$ = 0.4536°C

So, Δ $T_{f}$ of solution is,

Δ $T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ $T_{f_{solution} }$ = - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Learn more about freezing point here;

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7 0

1 year ago

You are given 3.0 grams of solid sodium to react to pure water which has a molarity of 55.6 M. How many milligrams of H2 can be

serious [3.7K]

Answer:

= 15.51 mL

Explanation:

Here's is the reaction:

2HgO(s) ⇒ 2 Hg(s)+O₂(g)

In this reaction 2mol HgO = 1mol O₂

The molecular weight of HgO = 216.59g

so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO

= 0.0138511 molHgO

The amount of Oxygen follows:

0.0138511 molHgOx1/2= 0.00692555 mol O₂

Now, volume of 1 any gas = 22400mL

so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂

= 15.513232mL O₂

4 0

3 years ago

Sodium hydroxide +HCI ➙ sodium chloride +H2O<br><br> Is this a:

Alexus [3.1K]

Neutralization reaction??

4 0

2 years ago

The kinetic energy of a 10.3 g golf ball traveling at 48.0 m/s is

Svetach [21]

To work out the kinetic energy of an object, you use the formula:

E = 0.5 x (mass) x (velocity)^2

One important thing, though. The units MUST be consistent. Mass needs to be in kilograms, and velocity in metres per second.

To convert the mass form grams to kilograms, we need to divide it by 1000, getting 0.0103 kg. Since the velocity is already in the units we need, we can just plug the numbers into the equation to get:

E = 0.5 x (0.0103 kg) x (48.0)^2 = 11.8656 J = 11.9 J, to 3 significant figures

Hope I helped! xx

3 0

3 years ago