Mechanism of 1-iodobutane reacts with pyridine

Answer: The mass of Si in kilograms is,

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si =

Density of Fe =

Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms =

Si concentration in kilograms =

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms =

Si concentration in kilograms =

Hence, the mass of Si in kilograms is,

Answer:

0.31%

Explanation:

For the chemical reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0.043 L multiplied by 0.117 M of sodium thiosulfate gives 5.031x10⁻³ moles of S₂O₃²⁻

5.031x10⁻³ moles of S₂O₃²⁻ produces:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2.5156x10⁻³ moles of I₂ equates to moles of NaClO

2.5156x10⁻³ moles of NaClO times yields 0.187 g of NaClO

Thus, the mass percentage composition is:

= 0.31%

I hope this helps!

Step 1: Convert density from g/mL to g/L; 0.807 g/mL is equivalent to 807 g/L. Step 2: Calculate Moles of N₂; Density = Mass / Volume, or Mass = Density × Volume. Plugging in values, Mass = 807 g/L × 1 L gives us Mass = 807 g. Similarly, Moles = Mass / M.mass, which leads to Moles = 807 g / 28 g.mol⁻¹, giving us Moles = 28.82 moles. Step 3: Apply the Ideal Gas Law to determine Volume of gas occupied; P V = n R T, thus V = n R T / P. Remember to convert temperature to Kelvin (25 °C + 273 = 298 K). Hence, V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm, resulting in V = 704.76 L.

The correct answer is: c. N2O, N2O4, N2O5.

According to the law of multiple proportions, also referred to as Dalton's Law, when two elements form compounds, the mass ratios of the second element that combine with a specific mass of the first yield small whole number ratios.

1) For NO, the mass ratio m(N): m(O) is 14: 16, simplified to 7: 8.

2) In N₂O, the ratio m(N): m(O) equates to 2·14: 16, which simplifies to 7: 4.

3) For NO₂, the masses yield m(N): m(O) = 14: 2·16, simplifying to 7: 16.

4) In N₂O₅, the ratio is (2·14): (5·16), which simplifies to 7: 20.

5) For NO₄, the mass ratio is m(N): m(O) = 14: (4·16), which simplifies to 7: 32.

6) N₂O₄ gives a ratio of m(N): m(O) as (2·14): (4·16), simplifying to 7: 16.

A) This means m(N): m(O) = 5.6 g: 3.2 g, simplifying results in 1.75: 1, which further translates to m(N): m(O) = 7: 4.

B) Here, m(N): m(O) is 3.5 g: 8.0 g. Dividing reveals a ratio of 1: 2.285, which alters to m(N): m(O) = 7: 16.

C) Lastly, for m(N): m(O) = 1.4 g: 4.0 g, then adjustments yield a ratio of m(N): m(O) = 7: 20.