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Murljashka [212]
3 years ago
3

Which term best describes the substances that cause oil and water molecules in salad Dressing to interact and mix?

Chemistry
2 answers:
hodyreva [135]3 years ago
6 0
<h2>Answer : Option A) Surfactants</h2><h3>Explanation :</h3>

The substances that causes the oil and water molecules in salad dressing to interact and mix are surfactants.

Surfactants are those compounds which has hydrophobic and hydrophilic parts in it which attaches according to the availability of the substrate. Oil and water does not mixes usually but in presence of a surfactant it readily mixes.


alukav5142 [94]3 years ago
3 0

the answer is A because I got it right on Edgenuity 2020

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marin [14]

Answer:

A

D

E

and maybe C I'm not sure on that one.

Explanation:

8 0
3 years ago
Atomic emission spectra are unique to each element. true false
11111nata11111 [884]
True, bc each element emits lights that have an unique wavelength and frequency
8 0
3 years ago
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Does the number of electrons reaching the detector increase or decrease as you change the frequency?
velikii [3]
If you increase the frequency, the photon will have more energy and can therefore transfer more electrons to the other side. As such, you will have an increased current.
3 0
3 years ago
In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
4 years ago
100 points please help me will also mark brainliest!
REY [17]

The maximum amount of H2O that can form is 0.07125 mole

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