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matrenka [14]
3 years ago
11

After Earth formed, the emerging biosphere had several effects on Earth’s lithosphere. For example, rocks and minerals showed ev

idence of
A.cobustion B.oxidation C.fossilization D.respiration caused by the presence of oxygen in air. In a second example, A. minerals B.organisms C.carbon dioxide D.sunlight actually created new rock after being dissolved in water or buried underground.
Chemistry
1 answer:
kogti [31]3 years ago
3 0

The answer is; B

Initially, the earth’s atmosphere was mainly filled with carbon dioxide and methane. However, with the evolution of photosynthetic organisms, the composition of the atmosphere changed to be mainly oxygen. This geological time is referred to as the great oxygenation event. The high oxygen levels resulted in oxidation of most the upper lithosphere.  

The answer is; B

During this time, the organism on earth began to evolve and live on land. Therefore these geologic events are marked by fossilized of dead primitive organisms such as archaebacteria within the early rocks

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The state of matter with a specific volume but no definite shape is the <u>Liquid</u> state.

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There are three states of matter according to chemical science: Solid, Liquid and Gas. There is not much space in among the molecules. The particles in a liquid are free to float, so while a liquid has a certain volume it has no definite form. Liquids are composed of atoms or molecules bound by inter molecular bonds.

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What is the total mass of 25.04 gram
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Answer:

What is the absolute temperature of this gas sample when the pressure is ... The total mass of the water vapor is 0.252 g ... A silver spoon with a mass of 25.04 g at a temperature of 100.00 ... A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH and heated to drive off the NH, a.

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Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.
attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

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5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>
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