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Olegator [25]
3 years ago
6

The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what

do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 751 torr ?
Chemistry
1 answer:
valina [46]3 years ago
3 0

As the atmospheric pressure is, P = dgh

Here d is the density of the mercury,

g is gravitation = 9.8 m/s²

h is height of the column, P = 751 torr = (751 torr × 1 atm / 760 torr) (101325 Pa) (1 N/m² / 1 Pa) = 100125 N/m²

Where, 1 N = 1 Kg / ms²

Thus, P = 100125 Kg / m³. s²

Therefore, height of the mercury column, when the atmospheric pressure is 751 torr,

h = P / gd

= (100125 kg / m³. s²) / (9.8 m/s²) (13.6 × 10³ kg / m³) = 0.751 m

As, d₁h₁ = d₂h₂

Here, d₁ is the density of the non-volatile liquid = 1.20 g/ml

d₂ is the density of the mercury = 13.6 g/ml

h₂ = 0.751 m

Thus, putting the values we get,

h₁ = d₂h₂ /d₁ = 13.6 g/ml × 0.751 m / 1.20 g/ml

= 8.5 m


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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
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A solution is usualy a diluted liquid that cleans for example bleach solution.
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A substance is a liquid at room temperature and dissolves well in water. It also conducts electricity. What type of bonds does i
kolbaska11 [484]
Ionic, covalent, molecular
7 0
3 years ago
onvert the value of Kc to a value of Kp for the following reaction: N2(g)+3H2(g)⇌2NH3(g) Kc = 0.50 at 400 °C.
ale4655 [162]

Answer:

K_p= 0.00016

Explanation:

The relation between Kp and Kc is given below:

K_p= K_c\times (RT)^{\Delta n}

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

N_2_{(g)}+3H_2_{(g)}\rightleftharpoons2NH_3_{(g)}

Given: Kc = 0.50

Temperature = 400^oC=[400+273]K=673K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(3+1) = -2

Thus, Kp is:

K_p= 0.50\times (0.082057\times 673)^{-2}

K_p= 0.00016

6 0
3 years ago
Please Help Me!!!
sergejj [24]

Answer:

The products are carbon dioxide and water

Explanation:

Step 1: Data given

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Step 2: The complete combustion of C3H7OH:

For the combustion of 1-propanol, we need O2.

The products of this combustion are CO2 and H2O.

C3H7OH + O2→ CO2 + H2O

On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3

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On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.

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On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).

To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.

2C3H7OH + 9O2→ 6CO2 + 8H2O

For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O

The products are carbon dioxide and water

7 0
3 years ago
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