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babunello [35]
3 years ago
5

Examining the equations or equilibrium constants related to a base, salt, or an acid is an indirect way to determine strength of

an electrolyte. The strength of an electrolyte can be examined directly by placing a solution into a circuit so that the voltage or amount of current can be measured. Although conductivity refers to the flow of charged species, we usually examine conductivity with respect to resistance. As the name implies, a solution that does not conduct electricity very well also has a very high resistance. If solutions containing various acids, bases, and salts were prepared and connected to a circuit that powers a light bulb, the strength of the electrolyte could be estimated by examining the intensity of the light bulb. Complete the following sentences regarding the experimental determination of the electrolyte strength for various molecules dissolved in solution.
a. strong electrolyte
b. weak electrolyte
c. non-electrolyte
1. A 20 mL solution containing 2 mmol of Ca3(PO4)2(s) was integrated into to a circuit that powers a light but, When the power supply was turned on, the light bulb produced a glow. Ca3(PO4)2(s) is__.
2. A 20 mL, solution containing 2 mmol of C12H22O11(aq) was integrated into a circuit that powers a light bulb. When the power supply was turned on the light bulb remained off. C12H22O11(aq) is___.
3. A 20 mL, solution containing 2 mmol of HF(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb faintly flickered. HF(g) is an____.
4. A 20 mL solution containing 2 mmol of LiOH(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb produced a bright glow LiOH(s) is a____.
Chemistry
1 answer:
Gwar [14]3 years ago
7 0

Answer:

1 - Weak electrolyte

2- Non electrolyte

3- Weak electrolyte

4- Strong electrolyte

Explanation:

A strong electrolyte refers to an electrolyte that decomposes completely in solution. This means that there are more charge carriers in solution when a strong electrolyte is dissolved in water. A strong electrolyte produces a strong glow. LiOH is a strong electrolyte.

A weak electrolyte is not completely dissociated in water. Only a small amount dissociates in water. HF is a weak electrolyte. A weak electrolyte does not produce a bright light.

A non-electrolyte does not dissociate in solution at all hence it does not power a bulb E.g  C12H22O11.

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According to the molar gas volume, one mole of a gas occupies a volume of 22400 cm³ or 22.4 liters. The standard temperatures and pressures conditions are temperature of 0 degrees Celsius or 273 kelvin and an absolute pressure of 103.325 kPa or 1 atm. In this case, the density is 0.902 g/l and the volume is 22400 cm³,
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A sealed container can hold
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0.216 moles of gas can the container hold if a sealed container can hold 0.325 L of gas at 1.00 atm and 293 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

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Putting value in the given equation:

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n= \frac{1.05 atm\; X \;5 L}{ 0.08206 L.atm / mol K  X 296 K}

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2 years ago
Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
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Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

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2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

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Now that you have k, m and n you can use the formula of the rate with the concentrations given

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