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2 years ago

Given the two compounds propane and pentanone, the one with the higher boiling point is

1 answer:

6 0

Pentanone, because it will boil at around 100 to 102 degree C, about the same as water, while propane boils at about -42 degree C, a temperature way below both pentanone and water's freezing point.

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How is star stuff connected to the sun?

skad [1K]

The sun is a type of big star.

7 0

2 years ago

A rigid tank contains 1.40 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to l

sergey [27]

Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

$PV=nRT$

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between pressure and number of moles of gas will be:

$\frac{P_1}{P_2}=\frac{n_1}{n_2}$

where,

$P_1$ = initial pressure of gas = 24.5 atm

$P_2$ = final pressure of gas = 5.30 atm

$n_1$ = initial number of moles of gas = 1.40 moles

$n_2$ = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

$\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}$

$n_2=0.301mol$

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

8 0

2 years ago

Predict the products of La(s) + O2(aq) ->

mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

Hello there!

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

I hope this helps you to understand better. Enjoy your studies

3 0

2 years ago

Use the image above and describe the

Lina20 [59]

Answer:

If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.

Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.

4 0

3 years ago

Which is the limiting reagent in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2

NemiM [27]

Answer:

CuSO4

Explanation:

Na2S + CuSO4 → Na2SO4 + CuS

The reaction is balanced (same number of elements in each side)

To determine limiting reagent you need to know the moles you have of each.

Molar mass Na2S = 23 * 2 + 32 = 78

Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5

Na2S mole = 15.5 / 78 = 0.2

CuSO4 mole = 12.1/159.5 = 0.076

*Remember mole = mass / MM

With that information now you have to divide each moles by its respective stoichiometric coefficient

Na2S stoichiometric coefficient : 1

Na2S : 0.2 / 1 = 0.2

CuSO4 stoichiometric coefficient: 1

CuSO4: 0.076 / 1 = 0.076

The smaller number between them its the limiting reagent, CuSO4

8 0

3 years ago