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mario62 [17]
3 years ago
6

TIMED ASSIGNMENT

Chemistry
1 answer:
galina1969 [7]3 years ago
3 0

Answer:

27 grams

Explanation:

mass = density × volume

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Explanation:

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When a substance is entering a phase change, the gain or loss of heat is a result of?
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When a substance is entering a phase change, the gain or loss of heat is a result of energy gained or lost in forming or breaking intermolecular interaction.

The constant temperatures occur when a substance is undergoing a phase transition.  If heat is removed from a substance , such as in freezing and condensation , then the process is exothermic . In this instance , heat is decreasing the speed of the molecules causing then move slower.

In chemistry and thermodynamics, phase transitions (or phase changes) are the physical processes of transition between a state of a medium, identified by some parameters, and another one, with different values of the parameters.

Commonly the term is used to refer to changes among the basic states of matter: solid, liquid, and gas, as well as plasma in rare cases.

For example, a phase of a thermodynamic system and the states of matter have uniform physical properties. During a phase transition of a given medium, certain properties of the medium change, often discontinuously, as a result of the change of external conditions, such as temperature, pressure, or others.

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8 0
2 years ago
how many moles of a nonvolatile, nonelectrolyte solute are required to lower the freezing point of 1000 grams of water by 5.58
Ivan

Answer:

Explanation:

Using freezing point depression formula,

ΔTemp.f = Kf * b * i

Where,

ΔTemp.f = temp.f(pure solvent) - temp.f(solution)

b = molality

i = van't Hoff factor

Kf = cryoscopic constant

= 1.86°C/m for water

= (0 - (-5.58))/1.86

= 3.00 mol/kg

Assume 1 kg of water(solvent)

= (3.00 x 1)

= 3.00 mol.

5 0
3 years ago
Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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