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Igoryamba
3 years ago
7

Material A has a small latent heat of fusion. Material B has a large heat of fusion. Which of the following statements is true?

(assume equal masses of both materials) material A will take less energy to phase change to a liquid material A will take more energy to phase change to a liquid material A will take less energy to phase change to a gas material A will take more energy to phase change to a gas
Chemistry
2 answers:
neonofarm [45]3 years ago
8 0

Answer: Material A will take less energy to phase change to a liquid material.

Explanation:

The energy provided to change the enthalpy of a substance is known as latent heat of fusion. This energy actually helps in changing the state of a substance that is from solid to liquid or vice versa.

Hence, a substance with small latent heat of fusion will require less energy to change its phase. Whereas a substance with large heat of fusion will require large energy to change its phase.

There, it can be concluded that material A will take less energy to phase change to a liquid material

Olin [163]3 years ago
7 0
The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
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Trava [24]

Answer: 2.52 M

Explanation:

The product of molarity (moles/litre) and volume in litres yields moles, and the numbers of moles in two solutions means dilute and concentrated are equal, which is expressed by the following equation: 

M_1V_1=M_2V_2

$$Given that\\M $1=12.0 \mathrm{M}$ or mole $/ \mathrm{L}$\\$\mathrm{V} 1=420 \mathrm{ml}$\\$\mathrm{M} 2=$ ?\\$\mathrm{V} 2=2.0 \mathrm{~L}$ or $2000 \mathrm{ml}$\\\\$\mathrm{M} 1 \mathrm{~V} 1=\mathrm{M} 2 \mathrm{~V} 2$\\$\mathrm{M} 2=\mathrm{M} 1 \mathrm{~V} 1 / \mathrm{V} 2$\\$=12.0 * 420 / 2000$\\$=2.52 \ \mathrm{M}$

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2 years ago
What is solubility?
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2) Solubility is how much solvent will dissolve in solute.

4 0
3 years ago
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Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl
anzhelika [568]

Answer:  ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]

\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]

\Delta H=-277.4kJ

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0
3 years ago
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b: mole ratio 2:3

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True [87]

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