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muminat
3 years ago
15

Brainliest if correct LINKS WILL BE REPORTED

Chemistry
2 answers:
AlladinOne [14]3 years ago
6 0

Answer:

1.transparent

2.able to pass through

Explanation:

nordsb [41]3 years ago
3 0
1. Transparent
2.able to pass through
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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
7. A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?​
USPshnik [31]

Answer:

The volume is

<h2>180 mL</h2>

Explanation:

In order to solve for the volume we use the formula for Boyle's law which is

<h3>P _{1}  V _{1} = P _{2}V _{2}</h3>

where

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Since we are finding the final volume we are finding V2

Making V2 the subject we have

<h3>V _{2}  = \frac{P _{1}  V _{1}}{P _{2}  }</h3>

From the question

P1 = 300 mmHg

V1 = 300 mL

P2 = 500 mmHg

Substitute the values into the above formula and solve for the final volume obtained

That's

<h3>V _{2} =  \frac{300 \times 300}{500}  \\  =  \frac{90000}{500}  \\  =  \frac{900}{5}</h3>

We have the final answer as

<h3>180 mL</h3>

Hope this helps you

7 0
4 years ago
Which group in the periodic table is known as the as the noble gases?
tekilochka [14]
ANSWER:

Group 18:
Elements are
-helium(He)
-neon (Ne)
-argon (Ar)
-krypton (Kr)
-xenon (Xe)
-radon (Rn)
-oganesson (Og)
6 0
3 years ago
Decide which element probably has a boiling point most and least similar to the boiling point of cesium.
natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.

<em>Hope that helps!</em>

5 0
3 years ago
What is the name of the compound NO₄?
Kryger [21]

Answer:

hello

Explanation:

answer is oxido nitrate please follow me

3 0
3 years ago
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