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VLD [36.1K]
3 years ago
7

Acetylene gas (ethyne; HC≡CH) burns with oxygen in an oxyacetylene torch to produce carbon dioxide, water vapor, and the heat ne

eded to weld metals. The heat of combustion for acetylene is −1259 kJ/mol. Calculate the C≡C bond energy._________ kJ/mol.
Chemistry
1 answer:
makkiz [27]3 years ago
4 0

Answer : The bond energy of C\equiv C is 800 kJ/mol.

Explanation :

The given chemical reaction is:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

For 1 mole, the reaction will be:

C_2H_2+2.5O_2\rightarrow 2CO_2+H_2O

As we know that:

The enthalpy change of reaction = E(bonds broken) - E(bonds formed)

\Delta H=[(B.E_{C\equiv C})+(2\times B.E_{C-H})+(2.5\times B.E_{O=O})]-[(2\times B.E_{C=O})+(2\times B.E_{H-O})]

Given:

\Delta H = heat of combustion = -1259 kJ/mol

B.E_{C-H} = 413 kJ/mol

B.E_{O=O} = 498 kJ/mol

B.E_{O-H} = 467 kJ/mol

B.E_{C=O} = 799 kJ/mol

Now put all the given values in the above expression, we get:

-1259kJ/mol=[(B.E_{C\equiv C})+(2\times 413kJ/mol)+(2.5\times 498kJ/mol)]-[(4\times 799kJ/mol)+(2\times 467kJ/mol)]

B.E_{C\equiv C}=800kJ/mol

Therefore, the bond energy of C\equiv C is 800 kJ/mol.

You might be interested in
Please balance the equation, putting the correct coefficient in each box.
Mariulka [41]

Answer: Bi(OH)_3+3HNO_3\rightarrow 3H_2O+Bi(NO_3)_3

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced equation will be:

Bi(OH)_3+3HNO_3\rightarrow 3H_2O+Bi(NO_3)_3

5 0
3 years ago
A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

7 0
3 years ago
The number of protons plus the number of neutrons in an atom is called its 7 number. for carbon-12, this number is
emmasim [6.3K]
The mass of Carbon-12 is 12.
8 0
3 years ago
Helppp VVV URGENTT GENIUSES HELP MEEE
marta [7]

Answer: Hydrogen

Explanation: Im pretty sure its Hydrogen since P is the cathode and it has a - charge meaning positively charged ions will be attracted to it and Hydrogen is the only gas with a positive charge in the answers.

6 0
2 years ago
How many moles of potassium hydroxide are needed to completely react with 2.94 moles of aluminum sulfate
ArbitrLikvidat [17]

Answer:- Third choice is correct, 17.6 moles


Solution:- The given balanced equation is:


Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4


We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.


From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.


It is a simple mole to mole conversion problem. We solve it using dimensional set up as:


2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})


= 17.6 mol KOH


So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.



6 0
3 years ago
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