I need help cause this teacher won’t stop cursin

Convert 73.7L to KL *

Mariulka [41]

0.737 KL would be the answer, pls give brainliest.

7 0

3 years ago

Por qué razón se fomenta la inmigración Europea?

Morgarella [4.7K]

Hola aquí va la respuesta!

Se fomentó la inmigración europea porque la economía no era buena y vinieron a Paraguay para mejorar su situación.

8 0

3 years ago

Read 2 more answers

Below are some of the steps completed in the titration process.

serg [7]

The steps are as followed in occurrence: (II),(III),(IV),(I)

8 0

4 years ago

Read 2 more answers

How many moles of nitrogen, N, are in 76.0 g of nitrous oxide, N2O?

guapka [62]

If the balanced equation reads:

2N2 + O2 ———> 2N2O

You should have 1.72 moles of N2.

8 0

3 years ago

Read 2 more answers

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