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3 years ago

Which compound has the the highest melting point?

Chemistry

2 answers:

Olin [163]3 years ago

8 0

Al₂(CO₃)₃ would have the highest melting point.

If it existed, it would consist of Al³⁺ and CO₃²⁻ ions, and the strong attraction between them would give the ionic compound a high melting point.

C₁₂H₂₂O₁₁ and C₈H₁₆ are organic compounds (molecular). Molecular compounds typically have low melting points.

Water, of course, freezes at 0 °C.

Kobotan [32]3 years ago

5 0

The answer is
Al2(co3)3

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Give three examples,from the lab where potential energy was converted to kinetic energy?

creativ13 [48]

Answer:

*a car geared neutral that is at a stop. two men start pushing it

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Explanation:

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3 years ago

At the normal melting point of NaCl, 801 degrees C, its enthalpy of fusion is 28.8 kJ / mol. The density of the solid is 2.165 g

melisa1 [442]

Answer: The increase in pressure is 0.003 atm

Explanation:

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm

$P_2$ = final pressure = ?

$\Delta H$ = Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $801^oC=[801+273]K=1074K$

$T_2$ = final temperature = $(801+1.00)^oC=802.00=[802+273]K=1075K$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm$

Change in pressure = $P_2-P_1=1.003-1.00=0.003atm$

Hence, the increase in pressure is 0.003 atm

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when you receive fluid in the hospital, it is often 0.9% saline solution. what would happen if the nurse used distilled water in

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3 years ago

A piston contains 0.39L of air at 1.0 atm. If the pressure on the piston is increased to 5.6 atm, what is the new volume?

bazaltina [42]

The new volume : 0.07 L

<h3>Further explanation</h3>

Boyle's Law

At a constant temperature, the gas volume is inversely proportional to the pressure applied

$\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}$

P₁=1 atm

V₁=0.39 L

P₂=5.6 atm

$\tt V_2=\dfrac{P_1.V_1}{P_2}\\\\V_2=\dfrac{1\times 0.39}{5.6}\\\\v_2=\boxed{\bold{0.07~L}}$

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