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erastovalidia [21]
3 years ago
11

Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. solutes formula hydroio

dic acid hi calcium hydroxide ca(oh)2 hydrofluoric acid hf methyl amine ch3nh2 sodium bromide nabr propanol c3h7oh sucrose c12h22o11
Chemistry
2 answers:
Andreyy893 years ago
6 0
The correct classification of the solutes are as follows:

<span>hydroiodic acid hi                           =   strong electrolyte
calcium hydroxide ca(oh)2             =   weak electrolyte 
hydrofluoric acid hf                         =   weak electrolyte
methyl amine ch3nh2                     =   weak electrolyte
sodium bromide nabr                     =   strong electrolyte
propanol c3h7oh                            =   non electrolyte
sucrose c12h22o11                        =   non electrolyte

Strong electrolytes are substances that completely ionizes in aqueous solution while weak electrolytes are those that partially ionizes. Non electrolytes are substances that cannot conduct electric charge since there are no ions in the solution.</span>
zhenek [66]3 years ago
4 0

hydroodic acid h = strong electrolyte

calcium hydroxide ca (oh) 2 = weak electrolytes

hydrofluoric acid hf = weak electrolytes

methyl amine ch3nh2 = weak electrolytes

sodium bromide nabr = strong electrolyte

propanol c3h7oh = non electrolyte

c12h22o11 sucrose = not an electrolyte

<h2>Further Explanation </h2>

Electrolytes are solutions that can conduct electric current.

There are three types of solutions. The solution that can make the lamp turn bright is the strong electrolyte solution, the solution that can make the lamp turn dim is the weak electrolyte solution and the solution that does not turn on the lamp is the non-electrolyte solution.

Strong electrolyte solutions are perfectly ionized compounds when dissolved in water. Strong electrolyte solutions actually come from three types of solutions, namely water-soluble salts, strong acids, and strong bases. Strong electrolyte solutions derived from salt can be exemplified with a solution of NaCl salt. This solution can dissolve in water to produce cations and anions.

A weak electrolyte solution is a partially ionized solution in water. So this type of solution produces only a few ions in the water. Weak electrolytes usually come from two types of solutions, namely weak acids, and weak bases. One example of a weak acid which is also a weak electrolyte is Acetic Acid (HC2H3O2). Acetic acid has a different character from strong acids because if dissolved in water, acetic acid will not be fully ionized, only about 1% of the molecule will dissociate into ions in aqueous solution. Examples of weak acids: Acetic acid (HC2H3O2)

Learn more

electrolyte brainly.com/question/4284571

Details

Grade:  College

Subject:  Chemistry

keywords: electrolyte

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Find the total surface area of the rectangular prism in the figure. A) 174 cm^ 2 B) 522c * m ^ 2; 348c * m ^ 2 D) 432c * m ^ 2
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Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

  • Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

                        Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄.
  • Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and                                Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
  • To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is =  5.451 X 10³ kg.

 

5 0
3 years ago
A 0.245-L flask contains 0.467 mol co2 at 159 °c. Calculate the pressure using the ideal gas law.
lubasha [3.4K]

Answer:

Pressure, P = 67.57 atm

Explanation:

<u>Given the following data;</u>

  • Volume = 0.245 L
  • Number of moles = 0.467 moles
  • Temperature = 159°C
  • Ideal gas constant, R = 0.08206 L·atm/mol·K

<u>Conversion:</u>

We would convert the value of the temperature in Celsius to Kelvin.

T = 273 + °C

T = 273 + 159

T = 432 Kelvin

To find the pressure of the gas, we would use the ideal gas law;

PV = nRT

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of substance.
  • R is the ideal gas constant.
  • T is the temperature.

Making P the subject of formula, we have;

P = \frac {nRT}{V}

Substituting into the formula, we have;

P = \frac {0.467*0.08206*432}{0.245}

P = \frac {16.5551}{0.245}

<em>Pressure, P = 67.57 atm</em>

4 0
3 years ago
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