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loris
16 days ago
15

Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right

arrow .CCl3 + HCl.b. CHCl2 + Cl2 right arrow CHCl3 + Cl..c. CH2Cl + Cl2 right arrow CH2Cl2 + Cl..d. CHCl2 + Cl. Right arrow CHCl3

Chemistry
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Why are salt and sugar both able to dissolve in water, even though the solutes have different types of chemical bonding?
castortr0y [3046]

Response: Water is a polar substance, facilitating the dissolution of ionic compounds due to the principle that similar types mix.

Ionic interactions occur between salt and water

Sugar contains hydroxyl groups that can form hydrogen bonds with water molecules.

[Hydrogen bond: this refers to the attraction between a hydrogen atom bonded to a highly electronegative atom (like F, O, or N) and another highly electronegative atom (F, O, or N)]

Thus, due to the presence of hydrogen bonds, sugar dissolves in water.

Clarification: Water molecules are polar, exhibiting partial positive charges on the hydrogen atoms and a partial negative charge on the oxygen atom. This allows them to interact with ionic compounds such as salt (NaCl). These interactions occur through the partial charges on water, which attract opposite charges. When dissolved in water, NaCl dissociates into sodium and chloride ions; sodium ions are surrounded by negatively charged oxygen from water, while chloride ions are surrounded by positively charged hydrogens from water. As a result, salt dissolves in water.

Sugar, being a covalent compound, has bonds where electrons are shared unevenly, creating slight positive and negative charges. This characteristic allows sugar to interact with the polar ends of water, facilitating its dissolution. Therefore, it can be stated that sugar dissolves in water due to both substances being polar.

In summary, water is capable of dissolving most polar or ionic substances, as seen with sugar and salt.

6 0
3 months ago
Read 2 more answers
How much CO2 (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0 °C and 1.23 atm? A)
KiRa [2933]

The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:

NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)

The substances NaHCO_3 and HCl combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.

NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol.

2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol.

We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of CO_2 equal 24.977 moles.

Next, we apply the ideal gas equation PV=nRT, where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V

PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3

The final answer should be expressed in liters, 1L = 1000\ m^3, hence

49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced

6 0
3 months ago
Read 2 more answers
Fructose is a type of sugar found in fruits and honey. Analysis of a 2.00-g
Anarel [2989]
The molecular formula is C6H12O6.
3 0
2 months ago
Determine the pH of a solution that is 0.15 M HClO2 (Ka = 1.1 x 10-2) and 0.15 M HClO (Ka = 2.9 × 10-8).
Tems11 [2777]
The pH level is 1.39. To explain, we start with the given information: the concentration of HClO is 0.15 M, with an acid dissociation constant of 2.9 × 10-8. The objective is to calculate the pH of the solution. Through the process, we find that the equilibrium concentration after applying the formula yields 0.04069 M for H3O⁺, leading us to a pH of 1.39.
4 0
3 months ago
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