4. What is a scientific law, and how does it differ from a theory?

The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

Answer: The moles of $CO_2$ added to the system is 7.13 moles

Explanation:

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago

When the pH of a solution is changed from 4 to 3,

Ivenika [448]

The correct answer is option 2. When the pH of a solution is changed from 4 to 3, the hydronium ions increases by a factor of 10. pH is a scale used to measure the acidity or alkalinity of an aqueous solution. It can be used to calculate the amount of hydronium ions in the solution. It is expressed as: pH = -log [H+].

7 0

3 years ago

Please help help help help

Anni [7]

Answer:

50000 dollars or 5 e5

Explanation:

Mr. Garibay has $5.0 * 10^4$ dollars in his bank.

Scientific notation represents the data in the format of expanded number or with e character.

For instance, 5.0 * 10^4 dollars = 50000 dollars or 5 e5

3 0

3 years ago

In the periodic table how many elements could be classified as metals in group IV?

lord [1]

4 elements are classified in group 4 like titanium, zirconium, hafnium and ruthorfordium

8 0

4 years ago

A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the ph

Debora [2.8K]

A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the pH is 5.81.

Given,

Volume of Triethylamine (C₂H₅)₃N = 29.4 ml = 0.0294 L

Molarity of (C₂H₅)₃N = 0.275 M

Molarity of HBr = 0.375M

Solution:

(C₂H₅)₃N + HBr ⇔ (C₂H₅)₃ NH⁺ + Br⁻

⇒ Volume of HBr = mole of (C₂H₅)₃N / Molarity of HBr

⇒ Volume of HBr = 0.00572 mol / 0.375 mol/l

⇒ Volume of HBr = 0.0152 L

∴ Total Volume = 0.0294 + 0.0152

⇒ Total Volume = 0.0446 L

Concentration of (C₂H₅)₃NH⁺ = $\frac{0.00572}{0.0446}$ ⇒ 0.128 M,

(C₂H₅)₃NH⁺ + H₃O⁺ ⇄ (C₂H₅)₃N + H₃O⁺

let's assume, (C₂H₅)₃N = $x$ ,

⇒ H₃O⁺ = $x$ , and

⇒ (C₂H₅)₃NH⁺ = 0.128 - $x$

∴ Now, k = kw / kb

⇒ k = 1.9 × 10⁻¹¹

and, k = [(C₂H₅)₃N][H₃O⁺] / (C₂H₅)₃NH⁺

⇒ 1.9 × 10⁻¹¹ = $x^{2}$ / (0.128 - $x$ )

Thus, $x$ = 1.55 × 10⁻⁶ M,

Hence, pH = - log[x]

⇒ - log [1.55 × 10⁻⁶ ]

⇒ - log (1.55) - log (10⁻⁶)

⇒ - log (1.55) + 6log10

⇒ - 0.19 + 6

⇒ 5.81 = pH

Therefore, pH = 5.81.

To learn more about equivalence point here

brainly.com/question/14782315

#SPJ4

7 0

2 years ago