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Schach [20]
3 years ago
8

Both isopropyl alcohol, C3H8OH, and ethylene glycol, C2H6O2, are used as antifreeze. When equal masses of each are added to wate

r, which, if either, will be more effective at preventing freezing? A. Isopropyl alcohol B. Ethylene glycol C. They will produce the same effect D. Neither, because they do not dissolve in water.
Chemistry
2 answers:
kiruha [24]3 years ago
6 0
I believe the correct answer from the choices listed above is option B. When equal masses of each are added to water, ethylene glycol would be more effective. Ethylene glycol is the most widely used automotive cooling-systemantifreeze, although methanol, ethanol, isopropyl alcohol, and propylene glycol are also used.
Hope this answers the question. Have a nice day.
blagie [28]3 years ago
5 0
Hello there.

<span>Both isopropyl alcohol, C3H8OH, and ethylene glycol, C2H6O2, are used as antifreeze. When equal masses of each are added to water, which, if either, will be more effective at preventing freezing? 

</span><span>B. Ethylene glycol </span>
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Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
Which is an advantage of biomass energy?
mylen [45]

Answer:

A. It's a 'clean' energy. Your cooch need a cleanup

5 0
3 years ago
Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

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3 years ago
What is a wet celled battery composed of? Give an example of a wet celled battery.
svet-max [94.6K]

Answer:

A wet-cell battery is the original type of rechargeable battery.

An example of a wet cell battery is a lead-acid battery.

Explanation:

3 0
2 years ago
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A student obtained 1.57 g of product in a chemical reaction in the lab although she expected to produce 2.04 g of product. What
Veronika [31]

Answer:

D. 77.0%

Explanation:

% yield = actual yield/theoretical yield x 100%

% yield = 1.57 g / 2.04 g x 100%

% yield = 76.96 (round up)

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