What is the difference between a strong electrolyte and a weak electrolyte

What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, c2h6? in this reaction,

Vitek1552 [10]

Balance equation for combustion of ethane will be:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

To balance the equation:

1. Balance the number of carbon atom on both side:

C₂H₆(g) + O₂(g)--------> CO₂(g) + H₂O(g)

1. balance the number of carbon on both side, as in reactant there are 2 but in product one,

so , multiply the CO₂, by 2 in the product.

2. Balance the number of hydrogen on both side as in reactant the number of hydrogen is 3 but in product it is 6 so, multiplythe number of H₂O by 3,

so multiply the number of H₂O by 3 in product.

3. Balance the number of oxygen on both side , as 1 and 2 step increases the number of oxygen and it becomes 7 , so to balance the number of oxygen on both side by mutiplying the number of O₂ by 7/2 in reactant .

4. Now, doubling the equation will give balance equation that is:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

8 0

3 years ago

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m

(b) $n_1$ =4 to $n_2$ = 20

$\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )$ = 6580.62

Therefore, $\lambda$ = 1 / 6580.62 = 0.000152 m

(c) $n_1$ =3 to $n_2$ = 10

$\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )$ = 11089.56

Therefore, $\lambda$ = 1 / 11089.56 = 0.00009 m

(d) $n_1$ =2 to $n_2$ = 1

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )$ = - 82257.75

Therefore, $\lambda$ = 1 /82257.75 = - 0.0000121 m

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

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8 0

1 year ago

Which correctly describes the relationship between genes and chromosomes?

Vlada [557]

Answer:

Chromosomes are structures within cells that contain a person's genes.

Explanation:

Genes are segments of deoxyribonucleic acid (DNA) that contain the code for a specific protein that functions in one or more types of cells in the body. Chromosomes are structures within cells that contain a person's genes. Genes are contained in chromosomes, which are in the cell nucleus.

6 0

2 years ago

How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the wor

tangare [24]

Answer is: 4.45 grams of methane gas need to be combusted.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V = n·R·T.
p = 1.1 atm.
T = 301 K.
V(H₂O) = 12.5 L.
R = 0,08206 L·atm/mol·K.
n(H₂O) = 1.1 atm · 12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.

4 0

3 years ago

Volume occupied 3.52x10^32 moluchles of Mathane (CH4) 1) At STP

Naddika [18.5K]

Answer:

volume = 13097674418.528dm³

Explanation:

n = (3.52)*10^32/(6.02)*10^23)

n = (584717607.97)

n = volume /molar volume

molar volume at stp = 22.4dm³

volume= 584717607.97 x 22.4

volume = 13097674418.528dm³

6 0

2 years ago