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3 years ago

How could you separate a mixture of sulfur, gold and ethanol?

Chemistry

2 answers:

ahrayia [7]3 years ago

5 0

Answer:

Explanation: There can be may ways to separate a mixture of sulfur, gold and ethanol nut one of the simplest ways are written below-

As we know that the sulfur as well gold both are soluble in ethanol but the solubility of gold in ethanol is more than that of solubility of sulfur.

Gold dissolved in ethanol can be easily collected by the means of steam condensation. and in this way sulfur can be separated.

Now the steams can be condensed off in order to get the gold so that the two of them could get separated.

stira [4]3 years ago

3 0

Answer: Dissolve it

Explanation:

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11. What is another name for the transition metals? Group B elements noble gases Group A elements Group C elements

finlep [7]

Answer:

Group B elements

Explanation:

Group B metals are referred to as transition metals. They are located in the middle of the periodic table between Group IIA and Group IIIA. They provide transition between the "base formers" on the left and "acid formers" on the right.

3 0

3 years ago

What is the molarity of a solution containing 8.0 moles of solute in 120mL of solution?

Nataliya [291]

Answer:
200/3 M which is approximately equal to 66.6667 M

Explanation:
Molarity is defined as the number of moles of solute per liter of solution.

It can be calculated as follows:
$molarity = \frac{number of moles of solute}{liters of solution}$

We are given that:
number of moles of solute = 8 moles
volume of solution = 120 ml = 0.12 liters

Substitute with the givens in the above equation to get the molarity as follows:
molarity = $\frac{8}{0.12} = \frac{200}{3} = 66.6667 M$

Hope this helps :)

3 0

3 years ago

Carbon 14 has a half-life of 5730 years. A geologist has dated a fossil sample, at roughly 28650 years. How much carbon 14 remai

IgorC [24]

<h3>Answer </h3>

After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left. after three half lives or 17190 years, 8.75 g of C-14 will be

Explanation:

hope this help

7 0

3 years ago

Help me with this question so I can move on from it.

fomenos

Just look it up on google kodvjkngkrefsnjkvjfrnefsjkj

7 0

3 years ago

The Goodyear blimp contains 5.7 x 10^6 L of helium at 25 degrees Celsius and 1 atm. What is the mass in grams of the helium insi

anygoal [31]

Answer:

1.72x10⁻⁵ g

Explanation:

To solve this problem we use the PV=nRT equation, where:

P = 1 atm

V = 5.7x10⁶ L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 25 °C ⇒ (25+273.16) = 298.16 K

And we <u>solve for n</u>:

1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

n = 4.29x10⁻⁶ mol

Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:

4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g

8 0

3 years ago

How could you separate a mixture of sulfur, gold and ethanol? ​

How could you separate a mixture of sulfur, gold and ethanol?