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just olya [345]
3 years ago
15

For ethanol, propanol, and n-butanol the boiling points, surface tensions, and viscosities all increase. what is the reason for

this increase?
Chemistry
1 answer:
Contact [7]3 years ago
3 0
Moving from Ethanol through Propanol to Butanol the physical properties like boiling points, surface tension and viscosity increases because of the increases in intermolecular interactions between the molecules of given compounds.

Explanation:
                   Ethanol, propanol and butanol all have hydroxyl groups in common, means all have hydrogen bond intractions between their molecules. So, taking the hydrogen bonding interaction constant we are left with only the difference in the number of carbon atoms.
                    Butanol has the greatest physical properties than other two because it has four carbon atom chain. So, as we know the London Dispersion forces or Van der Waal forces increases with increase in molecular size and chain length of hydrocarbon.
                    Therefore, the strength of London forces is greater in butanol than other two while ethanol has the smallest chain comparatively hence, lowest physical properties.  
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The mass of an iron(II) sulfate crystal is 8.36 g.How many moles of FeSO4 are in the crystal?
Yuki888 [10]

The number of moles present in the FeSO4 are 0.055 mol.

<u>Explanation:</u>

  • The mass of a substance containing the same number  atoms in 12.0 g of 12C is known as mole. One mole of any substance is equal to  6.023 x 10^23. The moles of a substance can be determined by using the formula,

                 Number of moles = mass in grams / molecular mass

Given,

    mass = 8.36 g,

    molecular mass of FeSO4 = 151.908 g / mol

         number of moles = 8.36 / 151.908

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3 years ago
How is chemical and physical properties of matter like in different
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4 0
3 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
3 years ago
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