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ELEN [110]
3 years ago
10

Which of the following changes to a pipe would increase the conductance by a factor of 12?

Chemistry
1 answer:
ivanzaharov [21]3 years ago
6 0

C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.

<u>Explanation:</u>

As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.

So it is represented as,

R∝ l /A [ area is radius square]

So k is the proportionality constant used.

R = kl/A

Conductance is the inverse of resistance, so it is given as,

C= 1/R.

R₁ = kl₁ / A₁

R₂ = kl₂/A₂

R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]

        = l₂A₁ / l₁A₂

If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12

So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.

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Answer:

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Explanation:

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8 0
3 years ago
A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.
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Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

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