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AysviL [449]
3 years ago
11

There are molecules of carbon dioxide (CO2) in 102.5 grams.

Chemistry
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

1.403x10²⁴ molecules

Explanation:

In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first<u> convert grams to moles</u> using its <em>molar mass</em>:

  • 102.5 g ÷ 44 g/mol = 2.330 mol CO₂

Now we <u>convert moles into molecules </u>using <em>Avogadro's number</em>:

  • 2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ molecules
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Which occurs when a warm fluid cools down?
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Answer:

Explination:

Given Data:

                  Trail 1          Trial 2         Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm         446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e.  435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

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3 years ago
How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
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The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

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Polair molecule, because the oxygen is slightly more negative than the two hydrogen atoms. Making water slightly negative. Due to the fact oxygen holds the to valence electrons more than the two hydrogen atoms do.
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