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2 years ago

Could someone help me with this?

Chemistry

1 answer:

Varvara68 [4.7K]2 years ago

7 0

Solving part-1 only

KMnO_4

Transition metal is Manganese (Mn)

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

$\\ \tt\Rrightarrow x+1-8=0$

$\\ \tt\Rrightarrow x-7=0$

$\\ \tt\Rrightarrow x=+7$

x is the oxidation number

Purple as per the color of potassium permanganate

$\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}$

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An unknown compound contains 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by mass. A mass spectrometry analysis reveals that

tiny-mole [99]

Answer:

Molecular formula is C₂₆H₃₆O₄

Explanation:

The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.

In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H

In 412 g of compound we would have:

(412 . 75.69) / 100 = 311.8 of C

(412 . 15.51) / 100 = 63.9 g of O

(412 . 8.80) / 100 = 36.2 g of H

Now, we can determine the moles of each, that are contained in 1 mol of compound.

312 g / 12 g/mol 26 C

64 g / 16 g/mol = 4 O

36 g / 1 g/mol = 36 H

Molecular formula is C₂₆H₃₆O₄

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Consider 2 NOCl(g) \Longleftrightarrow⟺ 2 NO(g) + Cl2 (g) At 25oC under conditions other than equilibrium, there are 1.20 moles

nekit [7.7K]

Answer:

a) Q = 6.1875x10⁻³

b) The direction of the reaction is to form the products.

c) [Cl₂]e = 0.094 M

[NO]e = 0.190 M

[NOCl]e = 0.140 M

Explanation:

a) Q is the reaction quotient, and for a generic reaction aA + bB ⇄ cC + dD it is

$Q = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}$

Which is the same equation for Kc, but in Kc expressions, the concentrations are in the equilibrium. Q is calculated at any time. So, for the reaction given

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

$Q = \frac{[Cl2]x[[NO]^2}{[NOCl]^2}$

[Cl₂] = 0.220/5.00 = 0.044 M

[NO] = 0.450/5.00 = 0.090 M

[NOCl] = 1.20/5.00 = 0.240 M

Q = (0.044)x(0.090)²/(0.240)²

Q = 6.1875x10⁻³

b) Q < Kc, which means that there are fewer products to what are needed to the equilibrium. So the direction of the reaction is to form the products.

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

0.240 0.090 0.044 Initial

-2x +2x +x Reacts (stoichiometry is 2:2:1)

0.240-2x 0.090+2x 0.044+x Equilibrium

$Kc = \frac{(0.044+x)x(0.090+2x)^2}{(0.240 - 2x)^2}$

$1.86x10^{-1} = \frac{(0.044+x)*(8.1x10^{-3} +0.36x + 4x^2)}{(0.0576 - 0.96x +4x^2)}$

3.564x10⁻⁴+0.01584x+0.176x²+8.1x10⁻³x+0.36x²+4x³ = 0.010714-0.17856x+0.744x²

4x³ - 0.208x² + 0.2025x - 0.01036 = 0

Solving this third grade equation in a computer program:

x = 0.05 M

So:

[Cl₂]e = 0.044 + 0.05 = 0.094 M

[NO]e = 0.090 + 2x0.05 = 0.190 M

[NOCl]e = 0.240 - 2x0.05 = 0.140 M

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