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2 years ago

Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure

Chemistry

1 answer:

o-na [289]2 years ago

6 0

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

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The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi

BARSIC [14]

The answer is: D) centi

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2 years ago

En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu

lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

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2 years ago

If a large marshmallow has a volume of 2.50 i n 3 and density of 0.242 g/c m 3 , how much would it weigh in grams? 1 i n 3 =16.3

Illusion [34]

Density of a substance is defined as the mass of the substance divided by the volume.

Density of the substance= 0.242 g cm⁻³

volume of the substance= 2.50 in³

As, 1 in³= 16.39 cm³

So, 2.50 in³= 16.39× 2.50 cm³=40.97 cm³

As ,

$Density=\frac{mass}{volume}$

Mass=volume ×Density

Mass=40.97 × 0.242

Mass=9.916 g.

4 0

2 years ago

Classify the following as qualitative or quantitative statements, giving your reasons. (a) The sun is approximately 93 million m

omeli [17]

Answer:

a) quantitative

b) qualitative

c) quantitative

d) qualitative

e) quantitative

Explanation:

a) it is a quantitative statement, since it uses the concepts of limits and aproxximation in the calculation to give its estimate.

b) it's a qualitative statement, beacause its an appreciation of someone who thinks that one's work ist better than another's.

c) it is a quantitative statement, since it is an expression that is mathematically demonstrated,

d) it's a qualitative, beacuse it'a a simple opinion of someone.

e) it is a quantitative statement, since this expression is demonstrated in real life and helps to save

8 0

2 years ago

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction: C(s)+2H2O(g)→2H2

madam [21]

The question is incomplete , complete question is:

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.

Answer:

The ΔH of the reaction is -626 kJ/mol.

Explanation:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

We are given with:

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{H-H}=432 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})$

$=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol$

$\Delta H=-626 kJ/mol$

The ΔH of the reaction is -626 kJ/mol.

5 0

2 years ago