All categories

3 years ago

What is deposition ?

Chemistry

2 answers:

Angelina_Jolie [31]3 years ago

8 0

Answer:

the act of deposing someone especially a monarch

slavikrds [6]3 years ago

7 0

The act of depositing material, especially by a natural process; the resultant deposit.

You might be interested in

State which separation method you would use to separate potassium iodine solution

navik [9.2K]

Could be separated by distillation.

3 0

2 years ago

If 0.138g of cyclohexene (c6h10) was obtained from 0.240g of cyclohexanol (c6h120), what is the percentage yield of cyclohexene?

tino4ka555 [31]

Given in the question- 1 mole of cyclohexanol = > 1 mole of cyclohexene Molar mass 100.16 g/mol moles of cyclohexanol = .240 / 100.16= 0.002396 moles Molar mass 82.143 g/mol moles of cyclohexene formed @100 % yield = 0.002396 Molar mass 82.143 g/mol mass of cyclohexene @ 100 % = .002396 x 82.143 = 0.197g bur we have .138g so % yield = .138 / .197 = 70.0 % Ans- 70 percentage yield of cyclohexene.

4 0

3 years ago

Riley is doing an experiment working with friction. He wants to find two ways to modify an object that could increase or decreas

Shtirlitz [24]

Riley can either change the surface area of the object or can change the slipperiness of the material.

5 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]

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

Hydrogen peroxide can decompose to water and oxygen by the following reaction

mestny [16]

To begin calculating, there is one thing you need to remember : $1 mole of H2O2=34.0148 g$
Then we have $5.00g of H2O2=5/34.0148=0.146995$
As you know decomposition of 2moles now has prodused 196kj
So, q is made due $0.146995 moles of H2O2$ $=(-196/2)*0.146995=-14.40551Kj$
I'm sure it will help.

6 0

3 years ago