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faltersainse [42]
3 years ago
14

What important points do you need to remember about combustion?

Chemistry
1 answer:
goldenfox [79]3 years ago
7 0

Explanation:

A substance burning in the presence of oxygen and leads to the formation of heat and light is called combustion.

Some important points about combustion are as follows:

  • Fuel should be present for burning.
  • Air should be present for the supply of oxygen.
  • There should be heat (or ignition temperature) to initiate the chemical reaction.
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Could someone plz help me? :(
kvv77 [185]

Answer:

I think it's B " Constraints are more important than criteria ".

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3 years ago
What is the answer to “An element is:”
Jobisdone [24]
A. A pure chemical substance consisting of one type of atom
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2 years ago
Read 2 more answers
In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0
3 years ago
A cylinder of argon contains 50 L of Ar at 12.4 atm and 127°C . How many moles of argon are in the cylinder
joja [24]

Answer:

18.9 moles

Explanation:

We have the following data:

V = 50 L

P = 12.4 atm

T= 127°C + 273 = 400 K

R = 0.082 L.atm/K.mol (it is the gas constant)

We use the ideal gas equation to calculate the number of moles n of the gas:

PV = nRT

⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol

6 0
3 years ago
Calculate the energy of an electron in the n = 2 level of a hydrogen atom.
ANEK [815]

Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.

Explanation:

Given: n = 2

The relation between energy and n^{th} orbit of an atom is as follows.

E = - \frac{13.6}{n^{2}} eV

Substitute the values into above formula as follows.

E = - \frac{13.6}{n^{2}} eV\\= - \frac{13.6}{(2)^{2}}\\= - 3.40 eV

The negative sign indicates that energy is being released.

Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.

5 0
3 years ago
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