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Yuliya22 [10]
3 years ago
7

2. Explain what type of chemical reaction is taking place, and what evidence you have to support your conclusion.

Chemistry
1 answer:
Sonja [21]3 years ago
6 0
I cant see the chemical reaction taking place, and what is the evidence to support it
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Complete these equations for the ionization of an Arrhenius acid or base in water. Include the states of the products.
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So this is an Arrhenius acid because it releases H⁺.

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LiOH(s) → Li⁺ + OH⁻
So this is an Arrhenius base because it releases OH⁻
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Trans-2-butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm−1); however, ir spectroscopy is
Elza [17]
However <em>trans</em>-2-Butene does not give a characteristic peak in 1620-1680 cm⁻¹ region but still the presence of carbon double bond carbon can be detected by detecting following peaks in IR Spectrum.

1)  3010-3100 cm⁻¹:
                               As in trans-2-Butene a hydrogen atoms ate attached to sp² hybridized carbon, therefore the stretching of =C-H (C-H) bond will give a peak of medium intensity in the range of 3010-3100 cm⁻¹.

2)  675-1000 cm⁻¹:
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3 0
3 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
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