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fenix001 [56]
2 years ago
10

Write E in the blank if the material is heterogeneous or O if it is homogeneous.

Chemistry
1 answer:
Advocard [28]2 years ago
8 0

Answer/Explanation:

Wood: E

Freshly brewed black coffee: O

Water: O

Lucky Charms: E

Salt: O

Dirt: E

Sausage and mushroom pizza: E

Air: O

Milk: O

Gold: O

Any mixture that looks the same through the substance is homogenous. The different parts are unable to be visibly seen. (Any periodic element is homogenous)

Any mixture that can be easily determined a mix and cannot be evenly separated or dispersed is heterogenous.

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Agata [3.3K]
The quality of being easily dissolved in liquid.

Hope this help and Happy holidays
3 0
3 years ago
Read 2 more answers
Choose two of the following scientists: Anton Lavoisier, John Dalton, JJ Thomson, Robert Millikan, Ernest Rutherford, James Chad
OverLord2011 [107]

Ernest Rutherford

J. J Thomson

Explanation:

<u>Ernest Rutherford</u>

In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.

 Experiment

 In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.

 Discovery and reflection on the atomic theory

To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.

<u>J. J Thomson</u>

Experiment

In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.

Discovery and reflection on the atomic theory

From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:

  • they move in a straight line
  • they possess kinetic energy
  • they attract positive charges and repels negative charges

Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.

learn more:

Rutherford brainly.com/question/1859083

#learnwithBrainly

6 0
3 years ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of
pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

<em>pH = 10.05</em>

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

<em />

I hope it helps!

3 0
2 years ago
Help which one will it be
salantis [7]

Answer:

D

Explanation:

6 0
2 years ago
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