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3 years ago

How are elements in the last column of the periodic table similar?

Chemistry

1 answer:

maks197457 [2]3 years ago

7 0

D could be the answer

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Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

As Mass = density × volume

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

PV = nRt

V = nRT/P

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0

3 years ago

In a single replacement reaction between 2.57 moles Aluminum and 3.59 moles of Hydrochloric acid, how many moles of Hydrogen can

Alenkasestr [34]

Answer:

1.795 mole of H2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Al + 6HCl —> 2AlCl3 + 3H2

Step 2:

Determination of the limiting reactant.

From the balanced equation above,

2 moles of Al reacted with 6 moles

Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.

From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.

Step 3:

Determination of the number of mole H2 produced from the reaction.

Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.

The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:

From the balanced equation above,

6 moles of HCl reacted to produce 3 moles of H2.

Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.

From the calculations made above, 1.795 mole of H2 is produced from the reaction.

5 0

3 years ago

Give the ground-state electron configuration for silicon (SiSi) using noble-gas shorthand. Express your answer in condensed form

nydimaria [60]

Answer:

[Ne] 3s² 3p²

Explanation:

Silicon atoms have 14 electrons. The ground state electron configuration of ground state gaseous neutral silicon is 1s²2s²2p⁶3s²3p².

Using noble gas shorthand, the electronic configuration is reduced to;

[Ne] 3s² 3p². Ne s the nearest noble gas to silicon, Ne contains 8 electrons, this means there's still 4 more electrons to fill. The s orrbital can only hold 2, hence the reaing two is transferred to the p orbital.

5 0

3 years ago

Plants like the ones shown in the picture, that use seeds, flowers, or fruit for reproduction are A) flowering. B) invertebrates

tensa zangetsu [6.8K]

Plants like the ones shown in the picture, that use seeds, flowers, or fruit for reproduction are A) flowering.

The picture shows flowering plants that are using seeds, flowers or fruits and they are called flowering plant .

So the answer here is A) flowering.

Plants like the ones shown in the picture, that use seeds, flowers, or fruit for reproduction are A) flowering.

7 0

3 years ago

Read 2 more answers