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2 years ago

Please help. 1.600x10^8 / 8.000x10^11

Chemistry

2 answers:

sweet-ann [11.9K]2 years ago

3 0

Answer:

2x10^-4 = 2E-4

Explanation:

patriot [66]2 years ago

3 0

The answer would be 2e18.

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A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104 $[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0

2 years ago

The rate constant for the decomposition of nitrogen dioxide NO2(g) LaTeX: \longrightarrow⟶ NO (g) + 1/2 O2(g) with a laser beam

aleksley [76]

Answer :

The time taken by the reaction is 19.2 seconds.

The order of reaction is, second order reaction.

Explanation :

The general formula to determine the unit of rate constant is:

$(Concentration)^{(1-n)}(Time)^{-1}$

Unit of rate constant Order of reaction

$Concentration/Time$ 0

$Time^{-1}$ 1

$(Concentration)^{-1}Time^{-1}$ 2

As the unit of rate constant is $M^{-1}min^{-1}$ . So, the order of reaction is second order.

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $1.95M^{-1}s^{-1}$

t = time = ?

$[A_t]$ = final concentration = 0.97 M

$[A_o]$ = initial concentration = 2.48 M

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

$1.95\times t=\frac{1}{0.97}-\frac{1}{2.48}$

$t=0.32min=0.32\times 60s=19.2s$

Therefore, the time taken by the reaction is 19.2 seconds.

8 0

2 years ago

What mass of HgO is required to produce 0.692 mol of O2? 2HgO(s) -> 2Hg(l) + O2(g)

Vika [28.1K]

The answer for the following problem is mentioned below.

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO → 2 Hg + $O_{2}$

where,

HgO represents mercuric oxide

Hg represents mercury

$O_{2}$ represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

2×216 grams of HgO → 1 mole of oxygen molecule

? → 0.692 moles of oxygen molecule

= $\frac{2*216*0.692}{1}$

= 298.944 grams of HgO

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

7 0

3 years ago

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Which of the following is true about a compound and its elements?

BaLLatris [955]

They can change properties completely
They can be separated
They form a new set of elements and compounds
The elements become part of the original compounds

6 0

3 years ago

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A gecko used its sticky feet to crawl from the bottom of a window to the top of

sp2606 [1]

Answer:

Decreased

Explanation:

You're faster when you go downhill than you are uphill.

8 0

3 years ago