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3 years ago

Which of the following substances is the most soluble in water?

Chemistry

2 answers:

GuDViN [60]3 years ago

5 0

A. Sodium Chloride.

brilliants [131]3 years ago

5 0

Answer: A. Sodium chloride

Explanation: Sodium chloride ( $NaCl)$ is an ionic compound formed by transfer of electrons from sodium to chlorine.

Electronic configuration of sodium:

$[Na]=1s^22s^22p^63s^1$

Sodium atom will loose one electron to gain noble gas configuration and form sodium cation with +1 charge.

$[Na^+]=1s^22s^22p^63s^0$

Electronic configuration of chlorine:

$[Cl]=1s^22s^22p^63s^23p^5$

Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.

$[Cl^-]=1s^22s^22p^63s^23p^6$

These ions can be separated in polar solvent water and thus are soluble in water.

Whereas methane ( $CH_4)$ is a covalent compound formed by sharing of electrons and is soluble in non polar solvents.

Bromine $(Br_2)$ is a covalent compound formed by sharing of electrons and is soluble in non polar solvents.

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A hurricane is MOST LIKELY to occur in an area

Flura [38]

Answer:

Western Pacific, such as the Philippines, Guam, southeast Asia (including China and Taiwan) and Japan.

Explanation:

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2 years ago

What substance is reduced in the reaction 2Fe 2+ + CI2 = 2Fe 3+ + 2CI- ?

atroni [7]

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

Hence, $Fe^{2+}$ is oxidized and $Cl_{2}$ is reduced

5 0

3 years ago

What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?

sp2606 [1]

First step in answering the question is to establish a balanced chemical reaction equation. More specifically, a combustion chemical equation.

CH4 + 2O2 ---> CO2 + 2H20

Then using dimension analysis:

$4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) = 0.0184 g O_{2}$

7 0

3 years ago

Electrolysis of molten lead bromide is carried out.

bearhunter [10]

Answer:

Read Below

Explanation:

Electrolysis is not possible with solid lead (II) bromide. This is because the ions are held in a three-dimensional lattice, unable to move freely to the electrodes. Melting enables the ions to become mobile and to travel to the respective electrodes.

The bulb won't glow when the electrodes are embedded in solid lead bromide. The bulb will glow when the material surrounding the electrodes is molten lead bromide. When an ionic compound is in the molten (liquid) form the positive and negative ions are free to move around.

Hopes this Helps :D

Brainiest Please

5 0

2 years ago

Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

tangare [24]

Answer:

$\boxed{\text{23.4 mmHg}}$

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

$K_{\text{p}} = p_{\text{H2O}}$

$\text{The relationship between $\Delta G^{\circ}$ and $K_{\text{ p}}$ is}\\\Delta G^{\circ} = -RT \ln K_{\text{p}}$

Data:

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

$\begin{array}{rcl}8600 & = & -8.314 \times 298.15 \ln K \\8600 & = & -2478.8 \ln K\\-3.47 & = & \ln K\\K&=&e^{-3.47}\\& = & 0.0311\end{array}$

Standard pressure is 1 bar.

$p_{\text{H2O}} = \text{0.0311 bar} \times \dfrac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\\\\\text{The vapour pressure of water at $25 ^{\circ}\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}$

4 0

2 years ago