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3 years ago

What is the name of the state when electrons absorb energy and move to a higher energy level?

Chemistry

1 answer:

murzikaleks [220]3 years ago

3 0

Answer:

Excited state is the name of the state when electrons absorb energy and move to a higher energy level.

Explanation:

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Which of the following makes a solute (like salt) dissolve faster in a solvent (like water)?

dmitriy555 [2]

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ The correct answer would be A. Agitation (i.e. stirring)

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3 years ago

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The displacement, d, in millimeters of a tuning fork as a function of time, t, in seconds can be modeled with the equation d = 0

QveST [7]

Answer:

1760 HZ

Explanation:

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2 years ago

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SKILL: MAKING VALID GENERALIZATIONS Use the paragraph below to answer the following questions. The Chinese people also call the

miss Akunina [59]

Answer:

Hope this helps

Explanation:

A generalization I can make about the Huang River is that the Huang River is the most important river in China. Also called the Yellow River, it is where historians believe ancient Chinese people settle in to develop its civilization on the banks of the river for the many benefits it had for life and agriculture. However, early Chinese people had to learn how to control the floods of the river in order to survive ad prosper.

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2 years ago

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago

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:

<em>Equivalentes KOH:</em>

0.100L * (1eq / L) = 0.100eq = 0.100moles

<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

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

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

8 0

3 years ago