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3 years ago

What is the total number of moles of solute contained in 0.5 liter of 3.0 m

Chemistry

1 answer:

myrzilka [38]3 years ago

4 0

Based on science 4) 1) 0.25M

2) 0.66M

3) 1.5M

4) 4.0M ––> Therefore, this is the most concentrated

5) 30g NaOH(1 mol NaOH / 40gNaOH) = 0.75 mol NaOH / 0.500L = 1.5M NaOH (3)

6) 1) 1.0 mol / L(1 L) = 1 mol H2SO4

2) 1 mol / L (2 L) = 2 mol H2SO4

3) 0.50 mol / L ( 1.0L) = 0.50 mol H2SO4 ––>

7) 0.200 mol / L (1 L)(74.6g / mol) = 14.92g (2)

8) 0.25 mol / 0.250 L = 1 mol / L = 1M (1)

This is the answer : 1m

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We have water flowing in a pipe of 1.05 in diameter.

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The mean velocity can be calculated as

$u=\frac{Q}{A}=\frac{Q}{\pi*D^2/4}=\frac{35gpm }{3.14*(1.05in)^2/4}\\\\ u=\frac{35}{0.865}*\frac{gal}{min}\frac{1}{in^2}*\frac{231in^3}{1gal}*\frac{1}{60s} \\\\ u=156\,in/s=13\,ft/s$

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$Re=\frac{\rho*u*D}{\mu}$

being ρ: density, u: mean velocity of the fluid, D: internal diameter of the pipe and μ the dynamic viscosity.

To simplify the calculation, we can first make all the variables have coherent units.

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$\mu=1.2cP=\frac{1.2}{100}\frac{g}{cm*s}*\frac{1lb}{453.6g}*\frac{30.48cm}{1ft}= 0.0008\frac{lb}{ft*s}$

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Then the Reynolds number is

$Re=\frac{\rho*u*D}{\mu}\\\\Re=62.3\frac{lb}{ft^3}*13\frac{ft}{s} *0.0875ft*\frac{1}{0.0008}*\frac{ft*s}{lb}\\\\Re=88,583$

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