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3 years ago

What is thermal expansion?

1 answer:

7 0

Thermal expansion<span> is the tendency of matter to change in shape, area, and volume in response to a change in temperature, through heat transfer. Temperature is a monotonic function of the average molecular kinetic energy of a substance. When a substance is heated, the kinetic energy of its molecules increases.</span>

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A solution is prepared by mixing 50.0 mL toluene (C6H5CH3 d=0.867 g/mL) with 125 mL Benzene (C6H6 d=0.874 g/mL). Assuming that t

Tatiana [17]

Answer:

mass % = 28.4%

mole fraction = 0.252

molality = 3.08 molal

molarity = 2.69 M

Explanation:

Step 1: Data given

Volume of toluene = 50.0 mL = 0.05 L

Density toluene = 0.867 g/mL

Molar mass toluene = 92.14 g/mol

Volume of benzene = 125 mL = 0.125 L

Density benzene = 0.874 g/mL

Molar mass benzene = 78.11 g/mol

Step 2: Calculate masses

Mass = density * volume

Mass toluene = 50.0 mL * 0.867 g/mL = 43.35 g

Mass benzene = 125 mL * 0.874 g/mL = 109.25 g

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles toluene = 43.35 grams /92.14 g/mol = 0.470 5 moles

Moles benzene = 109.25 grams / 78.11 g/mol = 1.399 moles

Step 4: Calculate molarity of toluene

Molarity = moles / volume

Molarity toluene = 0.4705 moles / 0.175 L = 2.69 M

Step 5: Calculate mass % of toluene

Mass % = (43.35 grams / (43.35 + 109.25) )*100 % = 28.4 %

Step 6: Calculate mole fraction of toluene

Mole fraction toluene = Moles toluene / total number of moles

Mole fraction toluene = 0.4705 / (0.4705 + 1.399) = 0.252

Step 7: Molality of toluene

Molality = number of moles / mass

Molality of toluene = 0.4705 moles / (0.04335 + 0.10925)

Molality of toluene = 3.08 molal

4 0

3 years ago

1362205.2 in scientific notation

SVETLANKA909090 [29]

1.3622052x10^6 you move the . 6 places to the left making it a positive 10^6

8 0

3 years ago

This table has information about the heat of fusion and the heat of vaporization of different substances. A 3-column table with

blagie [28]

H₂S

<h3>Further explanation</h3>

Given

ΔH fusion and ΔH vaporization of different substances

Required

The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes

Solution

We can use the formula :

$\tt \Delta H=\dfrac{Q}{n}$

Q=heat/energy absorbed

n = moles

The heat absorbed : 58.16 kJ

moles = 3.11

so ΔH vaporization :

$\tt \Delta H_{vap}=\dfrac{58.16~kJ}{3.11~mol}\\\\\Delta H_{vap}=18.7~kJ/mol$

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S

(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)

3 0

3 years ago

Read 2 more answers

How does the structure of covalent bonds affects their structure.

Black_prince [1.1K]

Covalent bond is a type of chemical bond which is formed as a result of sharing of electron pairs among the elements that are involved. The structure of the covalent bond is affected by the electronegativity of the elements involved. The molecules joined by covalent bond range in size from very small to very large polymers. There are different types of structures for covalent substances, these include: macromolecular substances, molecular substances and giant covalent structures. Strong bonds hold individual molecules together but there are negligible forces of attraction among them.

6 0

3 years ago

Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa

Lapatulllka [165]

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason:

If you look at the combining masses of X and Y in each of the two compounds,

The first compound contains 0.25g of X combined with 0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3

The second compound contains 0.33 g of X combined with 0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!)

The first compound will then contain 4g of X and 12g of Y
The second compound will then contain 6g of X and 12g of Y

<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

8 0

3 years ago