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3 years ago

The diatomic molecule h2 has an extremely low boiling point because

1 answer:

5 0

The diatomic molecule H2 has an extremely low boiling point because only weak Vander waals forces are present between the atoms of hydrogen molecule. This weak inter-molecular forces are easily broken when heat is apply resulting in low boiling point.

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

In our simulation what did the σ (sigma) stand for?

jenyasd209 [6]

Answer:

The atomic radius corresponds to Sigma

Explanation:

It is the Van Der Waals radius ;)

3 0

3 years ago

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0

3 years ago

The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2).

Rufina [12.5K]

Answer:

Mass = 182.4 g

Explanation:

Given data:

Number of moles of Al₂O₃ = 3.80 mol

Mass of oxygen required = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Now we will compare the moles of aluminum oxide and oxygen.

Al₂O₃ : O₂

2 : 3

3.80 : 3/2×3.80 = 5.7

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 5.7 mol × 32 g/mol

Mass = 182.4 g

4 0

2 years ago

Can the same fertilizer be applied to all types of soil ? Explain

Arte-miy333 [17]

You cant use the same fertilizer for all soils bc differnt soils have different nutrients like lets say ur growing tomatoes with cabbage it wouldnt work bc they need different nutrients for them to grow. This is used with trees, veggies, plants and fruits. it is true somw can use the same fertilizer if they can be grown with it. if you give it the wrong nutrients it can die or not grow

4 0

3 years ago