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3 years ago

Frozen CO2 or dry ice is different from CO2 gas in the atmosphere because it has

2 answers:

8 0

A fixed shape. gasses do not have a definite volume (which stems from lack of definite shape) which is why their volume is the same as the container they are being held in.

Schach [20]3 years ago

6 0

Answer: Option (b) is the correct answer.

Explanation:

Dry ice is the solid form of carbon dioxide whereas carbon dioxide gas present in the atmosphere is present in gaseous phase.

Also it is known that in solids, particles are held together tightly due to strong intermolecular forces of attraction. As a result, solid substances have fixed shape and volume.

On the other hand, in gases particles are held by weak intermolecular forces due to which they are far apart from each other. As a result, they do not hace any fixed shape.

Hence, we can conclude that frozen $CO_2$ or dry ice is different from $CO_2$ gas in the atmosphere because it has a fixed shape.

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The graph shows how milk production from a herd of cows has changed.

sammy [17]

Answer:

drink it

Explanation:

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3 years ago

En un matraz, disponemos de 100 g de gas oxígeno que se encuentran a 1 at de presión y 273 K de temperatura. Calcular : a) el nú

Misha Larkins [42]

Answer:

Explanation:

Dado que:

masa de oxígeno gaseoso = 100 g

presión = 1 atm

temperatura = 273 K

(a)

número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno

= 100 g / 16 gmol⁻¹

= 6.25 moles

(b) El número de moléculas de oxígeno es el siguiente:

Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.

Entonces, 6.25 moles contendrán:

= (6.25 × 6.023 * 10²³) moléculas de oxígeno.

≅ 3.764 × 10²³ moléculas de oxígeno.

= 2 × 3.764 × 10²³

= 7.528 × 10²³ átomos de oxígeno

(d) Usando la ecuación de gas ideal

PV = nRT

El volumen ocupado por el oxígeno = $\dfrac{nRT}{P}$

Volumen ocupado por oxígeno = $\dfrac{ 6.25 * 8.314 *273}{1}$

Volumen ocupado por oxígeno= 14185.76 m³

3 0

3 years ago

LA MATERIA SE PUEDE PRESENTAR COMO

olya-2409 [2.1K]

Answer:

La materia está constituida por átomos, que a su vez forman moléculas. Las moléculas constituyen la mínima parte en la que se puede fragmentar una sustancia para que conserve sus propiedades. Todo lo anterior permite que la materia se pueda presentar en estado sólido, líquido y gaseoso.

Explanation:

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3 years ago

Nitration of aniline?

OleMash [197]

In nitration u add two concentration acids HNO3 and H2SO4 both are strong acids. always the acid base neutralization reactions are fastest. even though NH2 group in aniline is a weak base, its neutralization with a very strong acid is still faster than nitration. hence if u carry the acid base reaction u would end up with an NH3+ group. hence the meta substitution can't be rule out here and u would get all products ortho,meta and para...

8 0

3 years ago

Read 2 more answers

Mass of metal=0.0291. Volume of gas collected over water=25.67mL. Temperature=24.1C. Atmospheric pressure=754.6mmHg. Note: Metal

NISA [10]

Answers:

1) 732.1 mmHg; 2) 0.001 014 mol; 3) 0.001 014 mol; 4) 28.7 g/mol; 5) 187 ppt.

Explanation:

1) Partial pressure of hydrogen

You are collecting the gas over water, so

$p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}$

$p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}$

$p_{\text{atm}} = \text{754.6 mmHg}$

At 24.1 °C, $p_{\text{H}_{2}\text{O}} = \text{22.5 mmHg}$

$p_{\text{H}_{2}} = \text{754.6 mmHg} - \text{22.5 mmHg} = \textbf{732.1 mmHg}$

===============

2) Moles of H₂

We can use the Ideal Gas Law.

pV = nRT Divide both sides by RT and switch

n = (pV)/(RT)

p = 732.1 mmHg Convert to atmospheres

p = 732.1/760 Do the division

p = 0.9633 atm

V = 25.67 mL Convert to litres

V = 0.025 67 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 24.1 °C Convert to kelvins

T = (24.1 + 273.15 ) K = 297.25 K Insert the values

n = (0.9633 × 0.025 67)/(0.082 06 × 297.25) Do the multiplications

n = 0.02473/24.39 Do the division

n = 0.001 014 mol

===============

3) Moles of metal

The partial chemical equation is

M + … ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.001 014 × 1/1 Do the operations

Moles of M = 0.001 014 mol M

===============

4) Atomic mass of M

Atomic mass = mass of M/moles of M Insert the values

Atomic mass = 0.0291/0.001 014 Do the division

Atomic mass = 28.7 g/mol

===============

5) Relative deviation in ppt

Your metal must be in Group 2 because of the 1:1 molar ratio of M:H₂.

The metal with the closest atomic mass is Mg (24.305 g/mol).

Relative deviation in ppt = |Experimental value – Theoretical value|/Theoretical value × 1000

Relative deviation = |28.7 – 23.405|/23.405 × 1000 Do the subtraction

Relative deviation = |4.39|/23.405 × 1000 Do the operations

Relative deviation = 187 ppt

3 0

3 years ago