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sveticcg [70]
2 years ago
9

A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud

ent discovers that the density of pentane is ·0.626gcm−3. Calculate the mass of pentane the student should weigh out. Round your answer to 3 significant digits.
Chemistry
1 answer:
Masja [62]2 years ago
5 0

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

                             = 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence,   Mass = gcm-3 * cm3

              Mass = g

The mass of pentane the student should weigh out is 28.17g

Learn more about Density on

brainly.com/question/1354972

#SPJ1

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Which one is an expression of distance in Sl units?
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Answer:

500 meters

Explanation:

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The flotation process used in metallurgy involves Multiple Choice the roasting of sulfides. separation of gangue from ore. elect
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The flotation process used in metallurgy involves the separation of gangue from ore.

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<h3>What is the process of separating minerals from gangue known as?</h3>

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8 0
2 years ago
Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!
e-lub [12.9K]
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol


Find the molar mass
2Na = 23 * 2 = 46 grams
1S   = 32 * 1 = 32 grams
O4   = 16 * 4 = 64 grams
Total =            142 grams / mol

Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???

given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams. 

85.2 are in a 2 L solution that has a concentration of 0.6 mol/L


4 0
3 years ago
Read 2 more answers
How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?
slavikrds [6]

Answer:

V_{LiOH}=21.8mL

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

n_{LiOH}=n_{HBr}

That it terms of molarities and volumes we have:

M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}

Next, solving for the volume of lithium hydroxide we obtain:

V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL

Best regards.

8 0
3 years ago
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