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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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What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?
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Answer:

\boxed {\boxed {\sf 3.3 \ liters}}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

  • molarity= 1.2 mol/L
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Substitute the values into the formula.

1.2 \ mol/L = \frac{4.0 \ mol}{x}

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}

4.0 \ mol *1=1.2 \ mol/L *x

4.0 \ mol = 1.2 \ mol/L *x

x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L

\frac{4.0 \ mol}{1.2 \ mol/L} = \frac{1.2 \ mol/L *x}{1.2 \ mol/L}

\frac{4.0 \ mol}{1.2 \ mol/L}=x

The units of moles (mol) will cancel.

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The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.

The 3 in the hundredth place tells us to leave the 3 in the tenths place.

3.3 \ L\approx x

Approximately  <u>3.3 liters of solution</u> are needed to make a 1.2 M solution with 4.0 moles of sodium chloride.

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