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Arturiano [62]
1 year ago
14

Why is it important to test the crucible for cracks before heating it?

Chemistry
1 answer:
solmaris [256]1 year ago
7 0

It important to test the crucible for cracks before heating it because it can break while performing the chemical reaction.

The crucible and the lid should be inspected to check for the before heating it to prevent the breaking while performing the reactions. if there are some breaks in the crucible then if we heat the crucible it can lead to breaking of the whole crucible and the material  from the crucible to come out and it will come in contact with the lid.

The crucible plays a important role while preforming the reactions, so. it important to check the cracks of the crucible before the heating of the crucible.

To learn more about crucible here

brainly.com/question/10237849

#SPJ4

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Which of the following statement(s) are true about the atoms of any element?
Sunny_sXe [5.5K]

Answer: The answers are:

The number of protons in an atom of an element is unique to each element.

A proton in an atom of one element is identical to a proton in an atom of another element.

Explanation: The number of protons in an atom of an element is unique to each element. The number of protons in an atom of an element is the atomic number of that element. Atomic number determines the chemical properties and reactivity of the atom.

All atoms are made up of three particles: protons, electrons and neutrons. These particles are identical in all the elements, what distinguishes one element from another is the number of each of these particles it contains. Therefore, a proton in an atom of one element is identical to a proton in an atom of another element likewise the electrons and neutrons.

3 0
3 years ago
A 3.00 L flexible container holds a sample of hydrogen gas at 153 kPa. If the pressure increases to 203 kPa and the temperature
dybincka [34]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

 

P1V1 =P2V2

V2 = P1 V1 / P2

V2 = 153 x 3.00 / 203

<span>V2 = 2.26 L</span>

3 0
2 years ago
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
2 years ago
Bacteria consume other organisms for food, making them heterotrophs. True or False?​
11111nata11111 [884]

Answer:

I would say true!

Explanation:

5 0
2 years ago
Which of the following is a physical change?
Anit [1.1K]
I’m pretty sure cooking an egg could be it
8 0
2 years ago
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