Which character of Life distinguishes a fire and a flowering plant

Which of the following is the correct balanced equation for the reaction in which methane (CH4) burns in atmospheric oxygen (0₂)

Talja [164]

Answer: $\text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}$

Explanation:

The unbalanced equation is

$\text{CH}_{4}+\text{O}_{2} \longrightarrow \text{CO}_{2}+\text{H}_{2}\text{O}$

Balancing this equation, we get:

$\boxed{\text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}}$

8 0

1 year ago

Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h

andrew11 [14]

Answer : The change in entropy is $6.05\times 10^3J/K$

Explanation :

Formula used :

$\Delta S=\frac{m\times L_v}{T}$

where,

$\Delta S$ = change in entropy = ?

m = mass of water = 1.00 kg

$L_v$ = heat of vaporization of water = $2256\times 10^3J/kg$

T = temperature = $100^oC=273+100=373K$

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

$\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}$

$\Delta S=6048.25J/K=6.05\times 10^3J/K$

Therefore, the change in entropy is $6.05\times 10^3J/K$

5 0

2 years ago

Which types of electron orbitals will have higher energy than a 4d orbital?

alexdok [17]

Answer:

D) 4f

Explanation:

To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.

1s

2s 2p

3s 3p 3d 3f

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d 6f

7s 7p 7d 7f

In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, <u>4d</u><em>,</em> 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

Therefore, 4f is the correct answer.

8 0

2 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the 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.

Hence, 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. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

= -314 kJ + (-80) kJ

= -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.

4 0

3 years ago

Suppose the mass of a jelly bean is less than the mass of a gum drop. If you counted out 10 of each kind of candy and measured t

MAVERICK [17]

Answer: Option (a) is the correct answer.

Explanation:

Since it is given that the mass of a jelly bean is less than the mass of a gum drop.

So, when we counted out 10 of each kind of candy and measured the mass of each kind of candy, the mass of the jellybeans would be less than the mass of the gumdrops.

This is because mass of jelly bean is less and even if we take take or more jelly beans then also their total mass will remain less than the total mass of same number of gum drops.

7 0

3 years ago

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Which character of Life distinguishes a fire and a flowering plant​

Which character of Life distinguishes a fire and a flowering plant