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Phantasy [73]
3 years ago
9

A 1.00 L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium was reached, it was f

ound that 1.30 mol gaseous NO was present. Assume that the reaction occurs under these conditions. Calculate the value of the equilibrium constant, K, for the following reaction. SO2(g) NO2(g) equilibrium reaction arrow SO3(g) NO(g)
Chemistry
1 answer:
Nataly_w [17]3 years ago
6 0

Answer:

Explanation:

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

Initial (M) 2.00 2.00 0 0

Change (M) −x −x +x +x

Equil (M) 2.00 − x 2.00 − x x x

2 2 c 3

2

c 2

[SO ][NO ]

[SO ][NO]

(2.00 )

=

= −

K

x K

x

Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of

NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.

2.00 − x = 1.30

x = 0.70

Substituting back into the equilibrium constant expression:

2c 2 2c 2

(2.00 )

(0.70)

(2.00 0.70)

= − = −

x KxK

Kc = 0.290

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A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
3 years ago
How many years old are you if you have lived 1 billion seconds 31 years round this answer to 3 sig figs
stich3 [128]

Answer:

age = 63.2 years

Explanation:

Given data

Number of years = 31

number of seconds = 1 billion

Age = ?

Solution

First we convert seconds to minutes

minutes = 1000000000 / 60

minutes = 16666666.667

Now convert minutes to hours

hours = 16666666.667 / 60

hours = 277777.778

Now convert hours to days

days = 277777.778 / 24

days = 11574.074

Now convert days to months

months = 11574.074 / 30

months = 385.802

Now convert months to years

years = 385.803 / 12

years = 32.15

Now add these years with given years

age = 31 + 32.15

age = 63.2 years

5 0
3 years ago
Equivalent massof sodium​
andre [41]

Answer:

Sodium

Formula : Na

Equivalent mass:23.0

Mark me as Branliest

8 0
3 years ago
I need help! how many moles are in 3.4*10^-7 grams of silicon dioxide, SiO2. I got 2.0^17 but I think it's wrong :(
blondinia [14]
Molar mass SiO2 = 28 + 32 = 60 

<span>so moles sand = 3.4 x 10-7 / 60</span>
8 0
3 years ago
The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3 * xH2O.
lesantik [10]

Answer:

44.7 kWh

Explanation:

Let's consider the reduction of Al₂O₃ to Al in the Bayer process.

6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻

We can establish the following relations:

  • The molar mass of Al is 26.98 g/mol.
  • 2 moles of Al are produced when 6 moles of e⁻ circulate.
  • 1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
  • 1 V = 1 J/c
  • 1 kWh = 3.6 × 10⁶ J

When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh

6 0
3 years ago
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