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fenix001
2 months ago
14

Consider the general reversible reaction. Lower A upper A plus lower B upper B double-headed arrow lower C upper C plus Lower d

upper D. What is the equilibrium constant expression for the given system? K e q equals StartFraction lowercase C StartBracket upper C EndBracket lowercase d StartBracket upper D EndBracket over lowercase A StartBracket upper A EndBracket lowercase B StartBracket upper B EndBracket EndFraction. K e q equals StartFraction StartBracket upper C EndBracket StartBracket upper D EndBracket over StartBracket upper A EndBracket StartBracket upper B EndBracket EndFraction. K e q equals StartFraction StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b over StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D EndFraction. K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.
Chemistry
2 answers:
Alekssandra [3K]2 months ago
6 0
K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction. To clarify the query, let's express the components in the format of a standard chemical reaction equation: aA + bB ⇄cC + dD. We should bear in mind that the equilibrium constant, commonly represented as Keq, is defined as the ratio of the product concentrations of products raised to the power of their respective molar coefficients to that of reactants raised to their respective molar coefficients. Hence, based on the previous description, we can assert that Keq= [C]^c [D]^d / [A]^a [B]^b. This is expressed in words in the answer provided above.
lions [2.9K]2 months ago
4 0
Keq= [C]^c [D]^d / [A]^a [B]^b.
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A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista
eduard [2782]

Details:

The equation to calculate work done is defined as follows.

W = -k \frac{q_{1}q_{2}}{d}

where, k = proportionality constant = 8.99 \times 10^{9} Jm/C^{2}

q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C

q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C

d = separation distance = 0.45 nm = 0.45 \times 10^{-9} m

Now we will insert the given values into the formula above to compute the work done as follows.

W = -k \frac{q_{1}q_{2}}{d}

= \frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}

= 3.68 \times 10^{-18} J

Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is 3.68 \times 10^{-18} J.

7 0
2 months ago
A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2
castortr0y [3046]

Answer:

11.55 moles of CO2 gas result from 3.85 moles of propane.

15.4 moles of H2O result from 3.85 moles of propane.

b) 0.5176 moles of water result from 0.647 moles of oxygen gas.

0.1294 moles of propane are reacted.

Explanation:

C_3H_7+5O_2\rightarrow 3CO_2+4H_2O

a) Amount of propane = 3.85 moles

The reaction indicates that for every mole of propane, 3 moles of carbon dioxide are produced.

Thus, 3.85 moles of propane yield:

\frac{3}{1}\times 3.85 mol=11.55 mol of carbon dioxide.

11.55 moles of CO2 gas result from 3.85 moles of propane.

The reaction also specifies that 1 mole of propane produces 4 moles of water.

Therefore, 3.85 moles of propane will generate:

\frac{4}{1}\times 3.85 mol=15.4 mol of water.

15.4 moles of H2O result from 3.85 moles of propane.

b) Amount of oxygen gas = 0.647 moles

The reaction states that 5 moles of oxygen yield 4 moles of water.

As a result, 0.647 moles of oxygen will produce:

\frac{4}{5}\times 0.647 mol=0.5176 mol of water.

0.5176 moles of water result from 0.647 moles of oxygen gas.

The reaction indicates that 5 moles of oxygen gas interact with 1 mole of propane.

Then, 0.647 moles of oxygen will cause:

\frac{1}{5}\times 0.647 mol=0.1294 mol of propane to react.

0.1294 moles of propane are reacted.

5 0
2 months ago
The [H3O+] in a solution is increased to twice the original concentration. Which change could occur in the pH? 2.0 to 4.0 1.7 to
KiRa [2933]
Answer: second option: 1.70 to 1.40

Explanation:

1) pH is defined using the formula pH = - log [H₃O⁺]

2) Given that the initial concentration is x and after doubling it becomes 2x, we calculate:

pHi = - logx
pHf = - log 2x = - log 2 - logx

Thus, pHf - pHi = - log2 - logx - (- logx) = - log2 ≈ - 0.30

⇒ pHi - pHf = 0.30, indicating that the final pH (with twice the hydronium ions) is 0.30 lower than the starting pH.

3) The only option that indicates a 0.30 decline in pH is the second one: from 1.70 to 1.40. Therefore, that is the correct choice.


8 0
3 months ago
Read 2 more answers
A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes combustion, the e
castortr0y [3046]

Response:

9.606 g

Clarification:

Step 1: Write the balanced combustion equation

C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the moles for 11.27 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

11.27 g × (1 mol/18.02 g) = 0.6254 mol

Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O

The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol

Step 4: Calculate the mass for 0.2085 moles of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

0.2085 mol × 46.07 g/mol = 9.606 g

7 0
3 months ago
The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its
KiRa [2933]

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.

7 0
2 months ago
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