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2 years ago

Upon heating a copper sample will expand, leading to a lower density? A) TrueB) False

Chemistry

1 answer:

Leona [35]2 years ago

8 0

Answer;

The above statement is true

upon heating a copper sample will expand, leading to a lower density

Explanation;

-The density of solids decreased with increase in temperature and vice versa. The increase in temperature causes the volume of the solid to increase which as a result decreases the density as Density=Mass/Volume. The temperature of a body is the average kinetic energy of the molecules present in it.

In other words; The temperature of a body is the average kinetic energy of the molecules present in it. Therefore; when heat is supplied ( or temperature is increased) the average kinetic energy increases which increases the volume and thus density decreases.

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What the 4 main parts to a flower<br> science i cant find it in the supject

kondaur [170]

stem, roots, leaves and flower

I hope it helps

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2 years ago

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

3 years ago

A scientist experiments ima unknown solution. She notices that when a base, sodium hydroxide, is added to the solution, it forms

Lelechka [254]

The unknown solution is acidic

8 0

2 years ago

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Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be

kakasveta [241]

Explanation:

Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).

The equation is show below as:

[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃ ⇒ [(CH₃)₂CH]₂N⁻Li⁺ + CH₃CH₂CH₂CH₃

N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence, LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)

Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)

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2 years ago

A student has a sample of isopropanol (C3H,OH) that has a mass of 78.6 g. The molar mass of isopropanol is 60.1 g/mol. How

Sauron [17]

I think it goes like this.

mass = 78.6 g
molar mass = 60.1 g/mol
amount (in mols) = mass/mol
78.6 g / 60.1 g/mol = 1.30782 mol

^^rounded to 3 sig figs, final answer = 1.31mol

4 0

2 years ago

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