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kherson [118]
4 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]4 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]4 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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1     2

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4 0
3 years ago
A 0.533 g sample of a carbonate salt is added to a flask. The only thing you know is that per formula unit, there is only one ca
dimulka [17.4K]

Answer:

\mathbf{CrCO_3}

Explanation:

The total mole of H_2SO_4 added = 15 × 1 × 10⁻³

= 15 × 10⁻³ mole

= 15 mmoles

the number of moles of NaOH added in order to neutralize the excess acid = 0.5905 ×  29.393 = 17.36 mmoles

the equation for the reaction is:

2NaOH + H_2SO_4 --------> Na_2SO_4 +2H_2O

i.e

2 moles of NaOH react with H_2SO_4

1 moles of  NaOH will react with 1/2  H_2SO_4

17.36 mmoles of NaOH = 1/2 × 17.36 mmoles of H_2SO_4

                                       = 8.68 mmoles

Number of moles of H_2SO_4 that react with MCO₃ = Total moles of H_2SO_4 added - moles of H_2SO_4 reacted with NaOH

= (15 - 8.68) mmoles

= 6.32 mmoles

H_2SO_4 + MCO_3  ------>   M_2SO_4   +    H_2O   +   CO_2

1 mole of H_2SO_4 react with 1 mole of MCO_3

6.32 mmoles of H_2SO_4 = 6.32 mmoles of MCO_3

number of moles of MCO_3 = 6.32 10⁻³ moles

mass of  MCO_3 (carbonate salt) = 0.533 g

molar mass of MCO_3  = (M+60)g/mol

We all know that ;

number of moles = mass/molar mass

Then:

6.32 10⁻³ = 0.533 / (M+60)

(M+60) = 0.533/ 6.32 10⁻³

M + 60 = 84.34

M = 84.34 - 60

M = 24.34

Thus the element with the atomic mass of 24.34 is Chromium

The chemical formula for the compound is :  \mathbf{CrCO_3}

6 0
3 years ago
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