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Eva8 [605]
3 years ago
8

You are given aqueous solutions of six different substances and asked to determine whether they are strong, weak, or nonelectrol

ytes. Describe how you could test the solutions. Include definitions of strong electrolyte, weak electrolyte, and nonelectrolyte in your answer and explain how your test would allow you to differentiate between the solution
Physics
1 answer:
kogti [31]3 years ago
3 0

Answer:

Answer is explained below;

Explanation:

Electrolytes are any substances that dissociate into charged particles called ions when dissolved in water. The positively charged ions called cations and the negatively charged ions called anions move toward the negative and positive terminals (cathode and anode) of an electric circuit.

When a substance dissolved in water completely dissociates into ions, it is called a strong electrolyte. The aqueous solutions containing strong electrolytes conduct electricity very well and the examples include strong acids and soluble ionic compounds such as barium chloride, sodium hydroxide, etc.  

When a substance dissolved in water does not completely dissociate into ions, it is called a weak electrolyte. Since the aqueous solutions containing weak electrolytes have relatively few ions, their electrical conductivity is very low compared to the solutions containing strong electrolytes. Examples of weak electrolytes include weak acids and bases like acetic acid, ammonia, etc.

When a substance does not dissociate into ions when dissolved in water, it is called a nonelectrolyte. Since the aqueous solutions containing nonelectrolytes do not contain any ions, such solutions do not conduct electricity. Examples of nonelectrolytes are ethanol, aldehydes, glucose, ketones, etc.

If a solution contains dissolved ions, it conducts electricity and as the ion concentration increases, the conductivity also increases. To determine whether the aqueous solutions of six different substances are strong, weak, or nonelectrolytes, we can test them by applying a voltage to electrodes immersed in the solutions and a light bulb. By observing the brightness of the light bulb or by measuring the flow of electrical current, we can find out which solution contains a strong electrolyte or weak electrolyte, or nonelectrolyte.

If the solution contains a nonelectrolyte, the current flow is nil and the light bulb does not glow. If the solution contains a strong electrolyte, the current flow is very strong and so the brightness of the light bulb is very high. If the solution contains a weak electrolyte, the current flow is much low compared to the strong electrolyte and the light bulb glows, but the brightness is very low.

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B. Ask a question - Hypothesize - Experiment - Analyze - Draw  a conclusion

Explanation:

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3 years ago
Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe
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Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

Work is said to be done if the force applied to an object cause the object to move through a distance

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Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

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Work done = Fdsin theta

Work done = 77(5.2)sin41

Work done = 400.4sin41

Work done = 262.69Joules

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An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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