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lbvjy [14]
2 years ago
8

List the experiment procedures. Some steps have been filled in for you. You may need to adjust the numbers to match the steps yo

u decide on. 1. Identify the variables. 2. Write out your hypothesis in an if/then format. 3. Gather your materials. 4. 5. 6. 7. 8. 9. Analyze the data. 10. Analyze the data. Data Analysis Label a data table so that the experimenter can record observations for the sand and water temperatures at various points.
Chemistry
2 answers:
sergeinik [125]2 years ago
7 0

Answer:

I sadly don’t know what 7 & 8 is, but I hope this helps!

Explanation:

Quizlet.

Whitepunk [10]2 years ago
4 0

Answer:all u need to do is list what u are doing. Do u do FLVS? cuz tht was in ... have been filled in for you. You may need to adjust the numbers to match the steps you decide on. 1. Identify the variables. 2. Write out your hypothesis in an if/then format. 3.Gather your materials. 4. 5.

Explanation:

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The product of the nuclear reaction in which 31p is subjected to neutron capture followed by alpha emission is ________.
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4 0
3 years ago
You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. This reaction occurs 2AI(S)+ 3Pb(NO
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<u>Answer:</u> The theoretical yield of solid lead comes out to be 5.408 grams.

<u>Explanation:</u>

To calculate the moles, we use the following equation:  

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

  • <u>Moles of Lead nitrate:</u>

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles

  • <u>Moles of Aluminium:</u>

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles

For the given chemical reaction, the equation follows:

2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = \frac{2}{3}\times 0.0261=0.0174moles of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = \frac{3}{3}\times 0.0261=0.0261moles of lead metal.

  • Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:

Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

0.0261mol=\frac{\text{Given mass}}{207.2g/mol}

Mass of lead = 5.408 grams

Hence, the theoretical yield of solid lead comes out to be 5.408 grams.

8 0
3 years ago
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