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3 years ago

Used to measure an exact volume of liquids

Chemistry

1 answer:

irakobra [83]3 years ago

6 0

Breaker measures a large amount and graduated cylinder measures a small amount

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Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

Usimov [2.4K]

Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ
1 mol of water weighs = 18.015g
1.55 kg = 1550/18.015 mol = 86.03 mol

Heat required to vaporize :
= 86.03 mol x 40.7 kJ

= 3501.421 kJ

5 0

3 years ago

List three conditions for separating crude oil constituents pls someone help me out

Alborosie

Answer:

All refineries have three basic steps: separation, conversion and treatment. During the separation process, the liquids and vapors separate into petroleum components called factions based on their weight and boiling point in distillation units.

Explanation:

4 0

2 years ago

Why do you use the same amount of water for each shape of ice?

Fiesta28 [93]

Answer:

Ice is a solid form of liquid. Water simply expands when it freezes.

The same amount of water for each shape of ice because the melting rate of ice depends on the surface area of each shape. So the same amount of water or volume with a large surface area will remain frozen for longer time.

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2 years ago

A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio

Pachacha [2.7K]

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ( $K_{a}$ ) of HA is represented as-

$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$

6 0

2 years ago

At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar

Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr

The partial pressure of acetic acid is 73.5 torr

4 0

3 years ago