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2 years ago

What is the freezing point of radiator fluid that is 50% antifreeze by mass?

Chemistry

1 answer:

crimeas [40]2 years ago

7 0

Answer:

Freezing point = -30.0 °C

Boiling point = 108.25 °C

Explanation:

Step 1: Data given

Kf for water is 1.86 ∘C/m.

fluid is 50% antifreeze by mass

Antifreeze = ethylene glycols (C2H6O2)

Step 2: Determine mass of antifreeze

Let's suppose a total mass of 1000 grams

Since 50 % is antifreeze (ethylene glycol) it has a mass of 500 grams

The other 50 % is water

Step 3: Calculate moles antifreeze

500 grams / 62.07 g/mol = 8.055 moles C2H6O2

Step 4: Calculate molality

The molality = 8.055moles / 0.5 kg = 16.11 moles / kg = 16.11 molal

Step 5: Calculate freeing point

ΔT = Kf * molality

ΔT = 1.86 °C/m * 16.11 molal = 30.0 °C

This means the freezing point is 30 °C below the freezing point of water (0°C)

0°C - 30.0°C = -30.0°C

The freezing point is -30.0 °C

What is the boiling point of radiator fluid that is 50% antifreeze by mass?

Kb for water is 0.512 ∘C/m.

Step 1: Data given

Kb for water is 0.512 ∘C/m

The fluid is 50% antifreeze by mass

Step 2: Determine mass of antifreeze

Let's suppose a total mass of 1000 grams

Since 50 % is antifreeze (ethylene glycol) it has a mass of 500 grams

The other 50 % is water

Step 3: Calculate moles antifreeze

500 grams / 62.07 g/mol = 8.055 moles C2H6O2

Step 4: Calculate molality

The molality = 8.055moles / 0.5 kg = 16.11 moles / kg = 16.11 molal

Step 5: Calculate boiling point

ΔT = Kf * molality

ΔT =0.512 °C/m * 16.11 molal = 8.25 °C

This means the boiling point is 8.25 °C higher than the boiling point of water (100°C)

Boiling point = 100 °C + 8.25 °C = 108.25 °C

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2 years ago

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Label each statement with its corresponding type of scientific knowledge. law theory fact hypothesis The melting point of ice is

EleoNora [17]

Answer:

Follows are the solution to this question:

Explanation:

Fact: In this, the ice is melted at 0 ° C.
law: It is used to repeated experiments consistently showed which objects marked to both the contrary attract each law.
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Theory: Protons, nuclei, or ions are made of Atoms. As these molecules could not be seen explicitly, such legal structure several experimental observations.

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2 years ago

Consider the following hypothetical reaction: 2 P + Q â†’ 2 R + S The following mechanism is proposed for this reaction: P + P Q

aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T (fast)

Q + T → R + U (slow)

U → R + S (fast)

we would be applying the Rate law to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.

The mechanism steps for the reactions in the question are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

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2 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5C w/ 1,150 g of water at 25.0C . if the final temperature of the system is

vichka [17]

Remember: heat lost = heat gained

When calculating heat loss or gain, remember

mass*(spec heat cap)*(change in T)

The unknown loses heat- we don't know the spec heat cap, so we'll call it x.

The water gains. I've omitted the units, but always use when solving problems on your own.

75*x*(96.5-37.1) = 1150*4.184*(37.1-25)
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Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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2 years ago

"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O

o-na [289]

Answer:

Yes, the chemist can determine which compound is in the sample.

Explanation:

In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.

In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.

If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.

If the ratio is 0.830, the compound is pure K₂O.
If the ratio is 0.710, the compound is pure K₂O₂.
If the ratio is not 0.830 or 0.710, the sample is a mixture.

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2 years ago