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BaLLatris [955]
3 years ago
15

What form of waste cannot be recycled to make new products?

Chemistry
2 answers:
skelet666 [1.2K]3 years ago
4 0
~Hello there! ^_^

Your question: What form of waste cannot be recycled to make new products..?

Your answer: Food is a form of waste that cannot be recycled to make new products.

Happy Studying! =)


aivan3 [116]3 years ago
3 0

Answer: metal waste

Explanation:

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How many moles of oxygen are in 1 mole of fe(no3)3?
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One mole of Fe(NO3)3, or iron(III) nitrate, has three moles of nitrate molecules, which have three moles of oxygen atoms each. We can show this mathematically:
1 mole Fe(NO3)3 * (3 moles NO3)/(1 mole Fe(NO3)3) = 3 moles NO3
3 moles NO3 * (3 moles Oxygen)/(1 mole NO3) = 9 moles Oxygen
9 moles of Oxygen in one mole Fe(NO3)3
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Polonium -210 decays once before it becomes a stable atom. What is that stable atom?
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A. 206 Pb

82

Explanation:

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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
2 years ago
What is the width of this penny?
sergey [27]

Answer:

D. 1.2 cm

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1st, determine the units. We see that our ruler is in cm.

This rules out A and C.

2nd, determine answer.

We see that the penny extends past 1 and ends at 0.2.

So, the total is 1.2 cm.

3 0
3 years ago
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