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8090 [49]
2 years ago
8

39. suppose an aqueous solution contains both table sugar and table salt. can you separate either of these solutes from the wate

r by filtration? explain your reasoning.
Chemistry
1 answer:
Degger [83]2 years ago
7 0

Answer:

No, assuming that the salt/sugar is already dissolved

Explanation:

As long as the particle size is too big, it won't filter through. Therefore, if it is dissolved, it will pass through the filter.

If you were to throw rocks in there or something, and they are non-dissolvable, then yes.

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Why is cl2 a subcript​
vazorg [7]

Answer:

In Cl_2, the 2 is a subscript because it indicates there are 2 of the same elements. The Lewis structure would display it as Cl-Cl.

On the other hand, a superscript would indicate a specific charge.

All subscripts show the amount of the specific element there is.

An example would be O_2\\ or N_2, they both show that there are 2 of the same elements.

If the subscript is outside a parenthesis such as Ca_3(PO_4)_2 it indicates there are 2 PO_4 molecules.

5 0
3 years ago
A baker is decorating cupcakes using 3 cherries for every 1 cupcake. If the baker has 12 cherries and 5 cupcakes, what is the th
11111nata11111 [884]
C. 4 decorated cupcakes.

Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries
7 0
3 years ago
Read 2 more answers
Is Bromine from the noble gasses? if yes why? if no why? too​
Jet001 [13]

Answer: Bromine is Actually a Halogen!

Explanation: Bromine is in the 7th column of the Periodic Table. It needs that magic 8 valence electrons and it has 7.

Hoped that helped!

8 0
4 years ago
In which of these statements are protons, electrons, and neutrons correctly compared
Marat540 [252]
<span>Quarks are present in protons and neutrons but not in electrons.</span>
8 0
3 years ago
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
4 years ago
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