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kherson [118]
3 years ago
7

List at least one reason for not touching the magnesium metal with bare hands. 3. list two reasons for using crucible tongs to h

andle the crucible and lid after their initial firing in the experimental procedure.
Chemistry
2 answers:
Usimov [2.4K]3 years ago
5 0
The magnesium metal is highly reactive and can ignite in contact with moisture or even air. The skin is usually covering in moisture, so the contact with it may cause a reaction that can produce irritation, lesions and even burns.

Crucible tongs are mandatory to handle hot crucibles to avoid skin burns and accidents during experimental procedures. Crucible tongs work with the crucible; their shape was designed to firmly hold it to avoid spills.

lutik1710 [3]3 years ago
3 0

Touching the magnesium metal can actually contaminate it and bring with it impurities which may not be removed by heating. So this leads to error in weighing.

 

There are two reasons not to touch the crucible tong with bare hands:

1. The crucible tong is used during firing so it is extremely hot and may burn your hands

2. Any impurities in your hand may contaminate the tong hence leading to error just like the case for the magnesium metal

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If an element has 3 valence electrons, what charge will likely form on its ion? Hint: It will lose those electrons. What happens
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Explanation:

If an atom lose 3 electrons its charge will be positive, it will be +3

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What is the volume of 500g of CO2?
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What happen when you add heat to solids and liquids?​
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andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

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Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

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Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

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