Compound X has a molar mass of 283.89 mol^-1 and the following composition:

How can you which molecule is positive and negative in a dipole-dipole relationship?

marysya [2.9K]

Dipole interactions are observed in covalent bonds. In ionic bonding, permanent transfer of electrons occurs and due to this dipole-dipole interactions are not observed. In covalent bonding, electron cloud is shared between 2 atoms. If this electron cloud is not shared equally between them, polarities are formed in a molecule. And hence we say that the molecule is polar. For a molecule to be polar, there should be electronegativity difference between them. Atom with greater electronegative attracts electron cloud more towards itself whereas atom with lesser electronegative attracts electron cloud less. But there is no permanent transfer of electrons. Due to this electronegativity differences, atom with more electronegative gains partial negative charge and atom with lesser electronegative value gains partial positive charge. The charge is partial because there is no complete transfer of electrons.

5 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

3 years ago

Approximately 1 mL of two clear, colorless solutions-0.1 M Ba(NO3)2 and 0.1 M Na2SO4- were combined. Upon mixing, a thick, milky

murzikaleks [220]

Answer: balanced chemical equation: $Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Net ionic equation : $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation is:

$Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

$Ba^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2Na^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are sodium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

4 0

3 years ago

Help help help help help

Rina8888 [55]

Answer:

Temporary hardness is a type of water hardness caused by the presence of dissolved bicarbonate minerals (calcium bicarbonate and magnesium bicarbonate). ... However, unlike the permanent hardness caused by sulfate and chloride compounds, this "temporary" hardness can be reduced by boiling the water.

5 0

3 years ago

Why are noble metals for example Rhodium, Palladium, Platinum, and Iridium largely used in chemical manufacturing as heterogeneo

Elanso [62]

They are also called the noble gases or inert gases. They are virtually unreactive towards other elements or compounds. They are found in trace amounts in the atmosphere. Their elemental form at room temperature is colorless, odorless and monatomic gases. They also have full octet of eight valence electrons in their highest orbitals so they have a very little tendency to gain or lose electrons to form ions or share electrons with other elements in covalent bonds.

4 0

3 years ago