Ask question

Ask question

All categories

3 years ago

6

BRAINLIST!!

2 answers:

Natali5045456 [20]3 years ago

6 0

B erosion is the answer

Montano1993 [528]3 years ago

5 0

Answer is B erosion!!!!

You might be interested in

__________ often cause overactive eye movement. A. Stimulants B. Depressants C. Barbiturates D. None of the above

jeka57 [31]

<u>Stimulants </u> often cause overactive eye movement

Ans (A)

Stimulants are a class of psychoactive drugs that increase the brain activity. They work by acting on the central nervous system and increase the level of mental alertness and impart heightened cognitive functions. Stimulants like nicotine and cocaine cause overactive eye movement.

5 0

3 years ago

Read 2 more answers

Which statement is true for all minerals

nekit [7.7K]

What are the answer choices

8 0

3 years ago

levacccp [35]

B aaaaaaaaaaassaaaaaaaaaaaaaaaaaaaaaaaaaaaaAaaaaaaaaaaaa

4 0

3 years ago

Read 2 more answers

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

Scientist use their knowledge of the half-life of carbon and uranium isotopes to measure the age of the rock or fossil .

marusya05 [52]

They can use them to find the age of fossils.

8 0

3 years ago