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barxatty [35]
3 years ago
12

One difference between mixtures and pure substances is that A) mixtures can be physically separated. B) mixtures are made of one

type of atom. C) pure substances have no chemical bonds. D) pure substances can be physically separated.
Chemistry
2 answers:
Elanso [62]3 years ago
8 0

Answer:

A

Explanation:

Pure substances are substances made of only one compound, and mixtures, substances made of two or more compounds. To separate the compounds of a mixture, it can be used chemical or physical separation. For example, to separate sand and water it may be used as a filter, which is a physical separation. But if there are small particles in the water, maybe it would be necessary coagulation with some chemical element before a filtration, which would be a chemical separation.

To separate the elements of a pure substance, only chemical separations can be used: would be necessary a chemical reaction to form H2 and O2 from H20, for example. So, letter A is correct.

Letter B is incorrect because can be differents atoms in the mixture, depends on its components.

Letter C is incorrect because all molecules, compounds, and substances have chemical bonds, it's the force that let the elements be together.

Letter D is incorrect because, as explained, pure substances only can be separated in their elements by a chemical reaction.

Crank3 years ago
4 0
If I'm not wrong I would say C
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the reaction AB2C (g) <---> B2 (g) + AC (g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.0840 mol of
Verizon [17]

Answer:

k = 4,92x10⁻³

Explanation:

For the reaction:

AB₂C (g) ⇄ B₂(g) + AC(g)

The equilibrium constant, k is defined as:

k = \frac{[B_{2}][AC]}{[AB_2C]} <em>(1)</em>

Molar concentration of the species are:

[AB₂C]: 0,0840mol / 5L = <em>0,0168M</em>

[B₂]: 0,0350mol / 5L = <em>0,0070M</em>

[AC]: 0,0590mol / 5L = <em>0,0118M</em>

Replacing this values in (1):

k = \frac{[0,0070][0,0118]}{[0,0168]}

<em>k = 4,92x10⁻³</em>

I hope it helps!

7 0
3 years ago
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate th
Andreas93 [3]

Complete question:

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.

Answer:

the heat of combustion of the methanol is 402.31 kJ/mol

Explanation:

Given;

mass of water, m_w = 100 g

initial temperature of water, t₁ = 28 ⁰C

final temperature of water, t₂ = 58 ⁰C

specific heat capacity of water = 4.184 J/g⁰C

reacting mass of the methanol, m = 1.00 g

molecular mass of methanol = 32.04 g/mol

number of moles = 1 / 32.04

                             = 0.0312 mol

Apply the principle of conservation of energy;

n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} =  mc\Delta t\\\\n\Delta H_{methanol} =  100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} =  12,552 \ J\\\\n\Delta H_{methanol} =  12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol

                       

5 0
3 years ago
N yogurt de 125 g muestra en su etiqueta el contenido de algunas de sus sustancias nutricionales:¿Cuál es la composición en tant
poizon [28]

Answer:

Sustancia  Masa (g)  Composición (% m/m)

<em>Proteínas</em>      5,5                  4,4

<em>Azúcares</em>      5,0                  4,0

<em>Sodio            </em>0,06               0,048

<em>Potasio         </em>0,15                 0,12

                         

Explanation:

Para calcular la composición porcentual en masa de las sustancias debemos usar la siguiente ecuación:

\%_{m/m} = \frac{m_{s}}{m_{sol}} \times 100

En donde:

m_{s} es la masa de la sustancia

m_{sol} es la masa del yogurt = 125 g

Para las proteínas:

\%_{m/m} = \frac{m_{s}}{m_{sol}} \times 100 = \frac{5,5 g}{125 g} \times 100 = 4,4 \%

Para los azúcares:

\%_{m/m} = \frac{5,0 g}{125 g} \times 100 = 4,0 \%

Para el sodio:

\%_{m/m} = \frac{0,06 g}{125 g} \times 100 = 0,048 \%

Para el potasio:

\%_{m/m} = \frac{0,15 g}{125 g} \times 100 = 0,12 \%

Espero que te sea de utilidad!    

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