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Nady [450]
3 years ago
15

-Material A has a melting point of 34°C whereas Material B has a melting point of 56°C. Both materials have been placed in an ov

en at 50°C Describe the state of each material after being being in the oven for a period of time.
I'm confused, if material B is on for the same amount of time does that mean that it has liquified and material A has melted slightly?
Chemistry
1 answer:
attashe74 [19]3 years ago
6 0

Answer:

Explanation:

Material A has a melting point of 34°C

Material B has a melting point of 56°C

Both materials, lets say a metal have been subjected to a temperature of 50°C

After a period of time, both of them would have melted to their liquid state.

The atoms of the solids would vibrates and the bonds would begin to break to form melt.

Material A would be the first to melt as it has a lower melting point. The lower the melting point, the faster and quicker it would reach its melting temperature.

Material B would need to accumulate more heat and its temperature would continue to rise for the phase change to occur. As it reaches the 50°C mark, the bonds are set free and a melt forms.

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The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

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2.275 M

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The equation of the reaction is;

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Let;

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CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

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