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3 years ago

How did life begin? need asap

Chemistry

2 answers:

3241004551 [841]3 years ago

5 0

From KimbeKimbe21 in a comment section found on the page:

"The earth began to be formed over 4.5 billion years ago,but for millions of years nothing could live here.Gradually,the earth's crust and the atmosphere formed.The simplest fo…rm of life, bacteria and algae,probably began to grow less than four billion years ago.Human beings did not appear until about two million years ago."

stealth61 [152]3 years ago

4 0

Birds and bees. Adam and Eve

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Acids or bases can be tested by chemical_________

DedPeter [7]

Answer:

acids or bases can be tested

by chemical indicators

7 0

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]

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

Explanation:

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

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

For Iron:

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

2 years ago

a = Proton transfer d = SN2 Nucleophilic substitution g = Nucleophilic subs at carbonyl(acyl Xfer) b = Lewis acid/base e= Electr

marissa [1.9K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution to this question is shown on the second uploaded image

Explanation:

4 0

3 years ago

Formula semidesarrollada de los hidrocarburos aromáticos de 1 a 10 átomos de carbono

Tju [1.3M]

Answer:

El principal componente del gas natural es también el hidrocarburo más simple: el metano. Este compuesto está formado por un átomo de carbono y cuatro átomos de hidrógeno y se representa de dos formas:

El hidrocarburo que le sigue en simplicidad es aquel que está constituido por dos átomos de carbono. Su fórmula condensada es C2H6 y se le conoce como etano.

Si se continúan colocando átomos de carbono con enlaces sencillos entre ellos e hidrógenos en los enlaces libres, se crean largas cadenas de compuestos. Al etano le sigue el propano (C2H8) y a éste, el butano (C4H10). Todos estos compuestos forman parte de la familia de los alcanos, y sus nombres terminan con el sufijo –ano para indicar que pertenecen a la misma familia.

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2 years ago

Is the maximum population that a given area can support

Anastaziya [24]

Answer:

Carry capacity

Explanation:

I took the test

Hope this helps

pls mark brainliest :3

8 0

3 years ago

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