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Viktor [21]
3 years ago
6

Suppose a thermometer has marks at every one degree increment and the mercury level on the thermometer is exactly between the 25

and 26 degree Celsius marks. We should properly report the temperature measurement as:
Chemistry
1 answer:
alekssr [168]3 years ago
5 0

Answer: 25.5°C

Explanation: take the average of the reading i.e (25 + 26)/2= 25.5

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According to Newton's third law of motion, for every action force, there is
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A. an equal and opposite reaction force

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This means that there is a natural reaction of force

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How is burning wood a good illustration of the law of conservation of mass?
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The wood turns into ash and smoke so mass is nor destroyed or created.
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1. Which groups of atoms become MORE stable when they lose an electron?
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Which alcohol will undergo acid-catalyzed dehydration under the mildest conditions?
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<span>Tertiary alcohols are the type of alcohols that will undergo acid-catalyzed dehydration under the mildest conditions. Types of tertiary alcohols are 2-methylpropan-2-ol and 2-methylbutan-2-ol. Other types of alcohols are referred to as primary alcohols and secondary alcohols.</span>
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Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de car
jarptica [38.1K]

Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>

Usando la ley general de los gases, las moles de dioxido de carbono son:

PV = nRT.

<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>

<h3>67.8%</h3>
7 0
3 years ago
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