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4 years ago

Ethylene glycol (antifreeze) has a density of 1.11 g/cm^3. What is the volume in liters of 3.46 kg of ethylene glycol?

Chemistry

2 answers:

xenn [34]4 years ago

7 0

3.11 i'm not sure about measurements maybe like 3.11kg/cm^3

liberstina [14]4 years ago

3 0

It is 3.12 you have to round

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Describe what happens as ice changes to water.

morpeh [17]

Answer:

The energy of the particles increase and the molecules move more quickly.

Explanation:

The molecules are moving from a solid (barely moves, molecules close together) to a liquid (molecules slide past each other and take any shape), so molecules are moving more and have more energy

8 0

3 years ago

An ice cube and a scoop of table salt are left outside on a warm sunny day. explain why the ice cube melts and the salt does not

Anna11 [10]

The melting point of ice is 0 degrees Celcius, which means it exists as a liquid for any temperatures above 0 degrees. The melting point of salt is approximately 800 degrees Celcius, which is way greater than the melting point of ice. This means that for temperatures below 800 degrees, salt exists as a solid.

The temperature of the area where they were placed we can assume was somewhere between 0 and 800 degrees, greater than the melting point of ice but less than the melting point of salt. This why the ice melted but the salt did not.

I hope this helps!

3 0

3 years ago

What is the mass of 0.80 moles of Mg? (1 point)

Aneli [31]

Answer:

m = 19 grams

Explanation:

Given that,

No. of moles, n = 0.8

The molar mass of magnesium, M = 24.305 u

We need to find the mass of 0.80 moles of Mg. We know that given mass to the molar mass is equal to no of moles. Let the mass is m. So,

$n=\dfrac{m}{M}\\\\m=n\times M\\\\m=0.8\times 24.305 \\\\m=19.444\ \text{grams}$

m = 19 grams

So, the mass of 0.8 moles of Mg is 19 grams.

6 0

3 years ago

In a laboratory experiment the reaction of 3.0 mol of H2 with 2.0 mol of I2 produced 1.0 mole of HI. Determine theoratical yield

Kobotan [32]

Answer:

Theoretical yield of HI is 512 g.

The percent yield for this reaction is 25%.

Explanation:

$H_2+I_2\rightarrow 2HI$

Moles of hydrogen gas = 3.0 moles

Moles of iodine gas = 2.0 moles

According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.

Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.

According to reaction , 1 mol of iodine gas gives 2 moles of HI.

Then 2 moles of iodine gas will give:

$\frac{2}{1}\times 2 mol=2 mol$ of HI

Theoretically we will get 4 moles of HI.

Theoretical yield of HI = 4 mol × 128 g/mol= 512 g

Experimental yield of HI = 1.0 mol

= 1 mol × 128 g/mol= 128 g

$\%yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}} \times 100$

$\%yield=\frac{128 g}{512 g}\times 100=25\%$

The percent yield for this reaction is 25%.

6 0

3 years ago

A given sample of gas has a volume of 5.20 L at 60.°C and 1.00 atm pressure. Calculate its pressure if the volume is changed to

BigorU [14]

Answer:

0.78 atm

Explanation:

Applying general gas equation

PV/T= P'V'/T'................ Equation 1

Where P = initial pressure, T = Initial temperature, V = Initial Volume, P' = Final pressure, V' = Final Volume, T' = Final Temperature.

make P' the subject of the equation

P' = PVT'/TV'.............. Equation 2

From the question,

Given: P = 1.00 atm, V = 5.20 L, T = 60°C = (273+60) = 333K, V' = 6.00 L, T' = 27°C = (27+273)K = 300 K

Substitute these values into equation 2

P' = (1×5.2×300)/(333×6)

P' = 1560/1998

P' = 0.78 atm.

5 0

3 years ago