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Natalka [10]
3 years ago
15

One and a half gallons of paint are used on refrigeration units which are then baked dry in a 275F oven for 2 hours. If the pain

t contains 5% solids and uses hexane as the solvent, what is the evaporation rate in pints per minute
Chemistry
1 answer:
brilliants [131]3 years ago
4 0

Answer:

Hence the evaporation rate in pints per minute is 0.001583.

Explanation:

Now,

1.5 gallons of paint are used.

It has 5% solid, so total 95% of 1.5 gallon can evaporate(only the solvent part will evaporate).

95% of 1.5 gallon=(95/100)*1.5 = 1.425 gallons.

1gallon = 8 pints.

So 1.425 gallons=8*1.425 pints

=11.4 pints.

Evaporation requires 2 hours. That is 3600*2 seconds = 7200 seconds.

So evaporation rate= 11.4 pints/7200 seconds.

=0.001583 pints per second

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A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa?
Naddik [55]

Answer:

  • <u>0.40 L</u>

Explanation:

Boyle's law for gases states that, at constant temperature, the volume and pressure of a fixed amount of gas are inversely related.

Mathematically, that is:

  • PV = constant

  • P₁V₁ = P₂V₂

Here, you have:

  • V₁ = 2.4 L
  • P₁ = 14.1 Kpa
  • P₂ = 84.6 KPa
  • V₂ = ?

Then, you can solve for V₂:

  • V₂ = P₁V₁ / P₂

Substitute and compute:

  • V₂ = 14.1 KPa × 2.4L / 84.6 KPa = 0.40 L ← answer
3 0
3 years ago
How is copper that is suitable for making electrical wires produced?
patriot [66]
Copper is a good conductor of heat and electricity so it will not fry easily
4 0
4 years ago
A gas effuses 4.0 times faster than oxygen (o2). what is the molecular mass of the gas? 1.0 g/mol 1.0 g/mol 2.0 g/mol 2.0 g/mol
Anastaziya [24]

Answer:

32(molecular mass has no unit )

Explanation:

(16)(o2)

16×2

=32

5 0
2 years ago
At this lower concentration, about how many extra hydrogen bonds would be needed to hold a and b together tightly enough to form
Scrat [10]

2..............................................

8 0
3 years ago
A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found
pshichka [43]

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for <em>freezing point depression</em>:

  • ΔTf = Kf * m

In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

  • 350 g X ⇒ 350/1000 = 0.35 kg X
  • 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea

Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to <u>solve for Kf</u>:

  • ΔTf = Kf * m
  • 3.3 °C = Kf * 1.51 m
  • Kf = 2.2 °C/m
5 0
3 years ago
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