Given 700 ml of oxygen at 7 ºC and 80.0 cm Hg pressure, what volume does it take at 27 ºC and 50.0 cm Hg pressure?

Compare the solubility of silver chromate in each of the following aqueous solutions: Clear All 0.10 M AgCH3COO 0.10 M Na2CrO4 0

slavikrds [6]

Solution :

Comparing the solubility of silver chromate for the solutions :

$0.10 \ M \ AgCH_3COO$ ----- Less soluble than in pure water.

$0.10 \ M \ Na_2CrO_4$ ----- Less soluble than in pure water.

$0.10 \ M \ NH_4NO_3$ ----- Similar solubility as in the pure water

$0.10 \ M \ KCH_3COO$ ----- Similar solubility as in the pure water

The silver chromate dissociates to form :

$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$

When 0.1 M of $AgCH_3COO^-$ is added, the equilibrium shifts towards the reverse direction due to the common ion effect of $Ag^+$ , so the solubility of $Ag_2CrO_4$ decreases.

Both $AgCH_3COO$ and $KCH_3COO$ are neutral mediums, so they do not affect the solubility.

4 0

3 years ago

3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,

VARVARA [1.3K]

Answer:

$Cp_{liquid}=2.54\frac{J}{g\°C}$

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

$Q_{Ag}=-Q_{liquid}$

That in terms of the heat capacities, masses and temperature changes turns out:

$m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})$

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

$Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}$

Best regards.

6 0

3 years ago

In which direction does the reaction proceed after heating to 2000 °c?

e-lub [12.9K]

What do I possibly answer here?
ask a full question pls!

6 0

3 years ago

Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all

Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

$n = 4$ .
$l = 1$ .
$m_l = 0$ .
Either $m_s = 1/2$ .

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the $4s$ orbital (fourth $s$ orbital.)

The quantum number $n$ (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, $n = 4$ for this electron.

The quantum number $l$ (the angular momentum quantum number) specifies the shape ( $s$ , $p$ , $d$ , etc.) of an electron. $l = 1$ for $s\!$ orbitals (such as the one that contains this electron.

Quantum numbers $n$ and $l$ specify the shape of an orbital. On the other hand, the magnetic quantum number $m_l$ specifies the orientation of these orbitals in space.

However, $s$ orbitals are spherical. Regardless of the value of $n$ , the only possible $m_l$ value for electrons in $s\!$ orbitals is $m_l = 0$ .

The spin quantum number $m_s$ distinguishes between the two electrons in an orbital. The two possible values of $m_s \!$ are $(+1/2)$ and $(-1/2)$ . Typically, the first electron in an orbital is assigned an upward ( $\uparrow$ ) spin, which corresponds to $m_s = (+1/2)$ .