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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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A compound was decomposed and determined to be 51.27 % C, 7.75 % H and 40.98 % O by mass. What is the subscript for H in the emp
Vlad [161]

The subscript for H in the empirical formula for this compound is 3.

There are 3 steps involved in the construction of empirical formula.

Calculation of empirical formula is as under as ...

Step 1: Divide the % of each atoms which their atomic weights.

 C = 51.27 / 12 = 4.27

 H = 7.75 / 1 = 7.75

 O = 40.98 / 16 =2.56

Step 2 : Divide all the answers with the smallest answer to get the subscripts for empirical formula.

C = 4.27/ 2.56 ≈ 2

H = 7.75/ 2.56 ≈ 3

O = 2.56 /2.56 = 1

Step 3: Construction of empirical formula by putting subscripts calculated in step 2.

Empirical formula form given data = C_{2} H_{3} O_{1}

Thus , subscript for H in the empirical formula for this compound is 3.

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2 years ago
Select all of the factors that influence GRAVITATIONAL potential energy (Lesson 3.10):
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Answer:

C) Mass

E) Distance from ground

Explanation:

The factors that influence the gravitational potential energy are mass and distance from the ground or height.

Gravitational potential energy is potential energy of a body in relation to that of another under the influence of gravity.

  Mathematically;

      Gravitational potential energy  = mgh

where m is the mass

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nataly862011 [7]

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5 0
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The theoretical yield of NaBr from
Lapatulllka [165]

Taking into account definition of percent yield, the percent yield for the reaction is 100%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 FeBr₃ + 3 Na₂S → Fе₂S₃ + 6 NaBr

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • FeBr₃: 2 moles
  • Na₂S; 3 moles
  • Fе₂S₃: 1 mole
  • NaBr: 6 moles

<h3>Moles of NaBr formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 2 moles of FeBr₃ form 6 moles of NaBr, 2.36 moles of FeBr₃ form how many moles of NaBr?

moles of NaBr=\frac{2.36 moles of FeBr_{3}x6 moles of NaBr }{2 moles of FeBr_{3}}

moles of NaBr= 7.08 moles

Then, 7.08 moles of NaBr can be produced from 2.36 moles of FeBr₃.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield}x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, being the molar mass of NaBr 102.9 g/mole, you know:

  • actual yield= 7.08 moles× 102.9 g/mole= 728.532 grams
  • theorical yield= 7.08 moles× 102.9 g/mole= 728.532 grams

Replacing in the definition of percent yields:

percent yield=\frac{728.532 grams}{728.532 grams}x100

Solving:

<u><em>percent yield= 100%</em></u>

Finally, the percent yield for the reaction is 100%.

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Energy resources are required for which of the following ?
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