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kifflom [539]
3 years ago
9

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

)]2+ that would be formed from 6.00 mL of 0.00180 M KSCN 5.00 mL 0.240 M Fe(NO3)3 and 14.00 mL of 0.050 M HNO3.
Question 3 options:

0.240 M

4.32 x 10^-4

0.0480

0.0460 M
Chemistry
1 answer:
viktelen [127]3 years ago
7 0

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}

\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol

Moles of KSCN = Moles of K^+ = Moles of SCN^- = 1.08\times 10^{-5}mol

Now we have to calculate the concentration of [Fe(SCN)]^{2+}

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M

Thus, the concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

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Explanation:

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Observando la siguiente reacción química: 3Fe + 4H2 O Fe3 O4 + H2(g) 4. Contesta ¿Por qué si se lleva a cabo en el recipiente ab
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Answer:

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Explanation:

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<em>3 moles de hierro con 4 moles de agua producen 1 mol de óxido de hierro y 4 moles de hidrógeno (gas)</em>

Cuando la reacción se está produciendo con un recipiente abierto, todo el gas de hidrógeno está escapando a la atmósfera y no habrá forma de que reaccione con el óxido de hierro.

Pero, si el recipiente está cerrado, el hidrógeno no podrá escapar y podrá reaccionar con el óxido de hierro, así:

Fe₃O₄ + 4H₂(g) → 3Fe + 4H₂O

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3 years ago
What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4
Lorico [155]

Answer:

3.2 L

Explanation:

Given data:

Mass of oxygen = 3.760 g

Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)

Temperature = 19°C (19+273.15 = 292.15 K)

R = 8.314 Nm K⁻¹ mol⁻¹

Volume occupied = ?

Solution:

Number of moles of oxygen:

Number of moles = mass/ molar mass

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The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant

T = temperature in kelvin

V = nRT/P

V = 0.12 mol ×  8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²

V = 291.472 Nm /88400 Nm⁻²

V = 0.0032 m³

m³ to L:

V = 0.0032×1000 = 3.2 L

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