The symbol for free energy is a. DG. b. DS. c. DT. d. DH.

Is chlorine a substance or a mixture

WARRIOR [948]

Answer:

its a substance

Explanation:

8 0

2 years ago

What other kinds of empirical evidence can be used to show common ancestry among organisms?

kotegsom [21]

Answer:There are, however, other forms of empirical evidence that support the theory of evolution. The fossil record, anatomical and embryological homologous, and genetic similarities are also used to support the theory of evolution. Many pieces of evidence support the theory that birds and dinosaurs are related.

Explanation:hope i am helpful

4 0

3 years ago

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

3 years ago

Help me on question 5 pls

Gre4nikov [31]

Ca +2
At -1
In +3
Sr +2
Ra +2
Fr +1
Ba +2
As -3

7 0

3 years ago

Why are the molecules of hydrocarbons nonpolar?

laiz [17]

This is not the case in the hydrocarbon tail. The electronegativity of hydrogen and carbon are very similar, so the electron cloud is distributed evenly over the two atoms. Carbon-hydrogen bonds are said to be non-polar because they do not have positive and negative poles within themselves. Hope this helps.

4 0

3 years ago

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