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4 years ago

Which of these phase changes does NOT involve changing a liquid into a gas?

Chemistry

1 answer:

zzz [600]4 years ago

8 0

The answer is sublimation.

Sublimation is the chemical process whereby a solid turns into a gas without first becoming a liquid.

The word sublimation comes from the Latin word sublimare which means "raised to a higher status."

Examples of sublimation include surface layer of snow and ice turning into steam or fog without melting and dry ice which is the solid form of carbon dioxide which also undergoes sublimation.

The low temperature and sublimation makes dry ice an effective coolant or cooling agent . One of its main advantages is that it has a lower temperature than water ice and also does not leave any residue.

You might be interested in

In which type of chemical reaction do two or more substances combine to form one substance?

Stella [2.4K]

A) synthesis reaction

7 0

4 years ago

Read 2 more answers

Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.

weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0

3 years ago

In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera

vredina [299]

Answer : The enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1.55J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 107.70 g

$\Delta T$ = change in temperature = $T_2-T_1=(22.80-20.34)=2.46^oC$

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

$q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]$

$q=1112.3J=1.1123kJ$

Now we have to calculate the enthalpy change of dissolution of $KClO_4$

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of $KClO_4$ = 2.84 g

Molar mass of $KClO_4$ = 138.55 g/mol

$\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole$

$\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole$

Therefore, the enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

5 0

4 years ago

How many atoms of lead are contained in 0.532mol?

Colt1911 [192]

Answer:

3.2043 x 10²³

Explanation:

No. of Mole of lead (Pb) = 0.532 mol

No. of atoms of lead = ?

Solution:

Formula Used to calculate

no. of moles = numbers of particles (ions, molecules, atoms) /Avogadro's number

Avogadro's no. = 6.023 x10²³

So,

The formula could be written as

no. of atoms of lead Pb = no. of moles x 6.023 x10²³

Put the values in above formula

no. of atoms of lead Pb = 0.532 mol x 6.023 x10²³

no. of atoms of lead Pb = 3.2043 x 10²³

so 3.2043 x 10²³ atoms of lead are contained in 0.532 mole.

6 0

3 years ago

What is the purpose of balancing chemical reactions?

Karo-lina-s [1.5K]

Answer: I believe its 1

Hope this helps!

4 0

2 years ago