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BabaBlast [244]
3 years ago
5

The freezing point of a solution the freezing point of the pure solvent

Chemistry
2 answers:
nignag [31]3 years ago
6 0

Answer:

Freezing Point Depression. The freezing point of a solution is less than the freezing point of the pure solvent. This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur

babymother [125]3 years ago
5 0

Answer:

This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.

Explanation:

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One application of Hess's Law (which works for ΔH, ΔS, and ΔG) is calculating the overall energy of a reaction using standard en
VikaD [51]

Answer:

The standard change in free energy for the reaction =  - 437.5 kj/mole

Explanation:

The standard change in free energy for the reaction:

                              4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)

Given that   ΔGf(KClO3(s)) = -290.9 kJ/mol;

                    ΔGf(KClO4(s)) = -300.4 kJ/mol;

                    ΔGf(KCl(s)) = -409 kJ/mol

According to Hess's law

ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)

               ⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}

                            = - 901.2 - 409 + 872.7

                            =  - 437.5 kj/mole

3 0
3 years ago
______ use energy from their enviornment to make food .
wlad13 [49]

Answer:

autotrophs

Explanation:

3 0
3 years ago
Read 2 more answers
The International Date Line
geniusboy [140]

Answer:

The International Date Line, established in 1884, passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich, England, in 1852.

8 0
3 years ago
Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0
3 years ago
What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol)?
NemiM [27]

the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g

The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn

3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2

the formula is n= mass/M so, now substituting values

m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3

so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g

so mass of aluminum oxide obtained = 1.36g

To learn more about Mass:

brainly.com/question/19694949

#SPJ4

3 0
2 years ago
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