All categories

3 years ago

1. What is the name of the metal which exists in liquid state at room temperature?

Chemistry

1 answer:

Bad White [126]3 years ago

5 0

Answer:

1. mercury.

2. suspensions.

3.iodine is the solute and alcohol is the solvent.

4. A element.

You might be interested in

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:

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

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>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

Which form of oxygen (solid,liquid or gas) would have the highest temperature

ElenaW [278]

Answer:

drip drip water splash (gas)

Explanation:

8 0

3 years ago

What only sure evidence for a chemical reaction is?

Vera_Pavlovna [14]

Heat because when substances react they produce heat.

3 0

3 years ago

Why is Azulene aromatic ?

evablogger [386]

Add 2 that’s why and I said if

3 0

3 years ago

A scientist has two solutions, which she has labeled solution a and solution

Lelu [443]

To determine the amount of each solution that she should use, we do mass balance for the whole process. We let x be the amount the 65% salt and y the amount of the 90% salt. We do as follows:

The total mass should be equal to 150 oz.

x + y = 150

Doing a salt balance, we obtain the second equation:
.65x + .90y = 150(.85)

Solving for x and y, we obtain:

x = 30 oz
y = 120 oz

6 0

3 years ago