Which is not a disadvantage to wind energy?

Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360

Aleks [24]

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

3 0

3 years ago

Read 2 more answers

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction i

Natasha2012 [34]

Answer: Oxidation reaction: $Al(s)\rightarrow Al^{3+}(aq)+3e^-$

Reduction reaction: $Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$

Overall redox reaction : $2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)$

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$

Overall redox reaction : $2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)$

5 0

3 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

7 0

4 years ago

A container of what volume would be required to store 60.0 moles of a gas at a pressure of 20.0 atm and a temperature of 50.0 Ce

belka [17]

Answer:

The answer to your question is: 79.6 L

Explanation:

Data:

n = 60 moles

Pressure = P = 20.0 atm

Temperature = T = 50°C

Volume = V = ?

Convert temperature to Kelvin

°K = 273 + 50

°K = 323

Formula (ideal gas law)

PV = nRT solve for V

V = nRT / P

V = (60)(0.08206)(323) / 20 Substitution

V = 1590.32 /20

V = 79.6 L

7 0

4 years ago

I know this is a very broad question but can someone explain ions, I am so confused.

Vinil7 [7]

There are negative ions and positive ions known as anions and cations.

Atoms of elements (besides the noble gases) are not stable.

These atoms lose or gain electrons to gain the stable electronic configuration of a noble gas which has either 2 electrons or 8 electrons in the valence shell.

Since an atom is neutral, once it loses or gain electrons, it gains charge and is known as an ion

4 0

3 years ago