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mars1129 [50]
2 years ago
15

The minerals hematite (Fe₂O₃) and magnetite (Fe₃O₄) exist in equilibrium with atmospheric oxygen:4Fe₃O₄(s) + O₂(g) ⇄ 6Fe₂O₃(s) K

p = 2.5×10⁸⁷ at 298K (c) Calculate Kc at 298 K.
Chemistry
1 answer:
iren2701 [21]2 years ago
3 0

Kc at 298K is Kc = 1 × 10^84.

Calculation:

The relation between Kp and Kc is

      Kp = KcRT

The value of Kp is known

We can calculate the value of Kc with this

R =  8.3145 J mol^−1 K^−1 .

T = 298K

      Kp = KcRT

      2.5×10⁸⁷ = Kc × 8.3145 × 298

      2.5×10⁸⁷ = Kc × 2477.572

      Kc = 1 × 10^84

To learn more click the given link

brainly.com/question/25651917

#SPJ4

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A piece of aluminum weighing 10.0 grams is placed into a graduated cylinder that has an initial volume (before immersion) of 35.
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Answer:

The density of the metal sample is 2.70 \frac{grams}{mL}

Explanation:

Density is a magnitude that allows you to measure the amount of mass in a given volume of a substance. Then, the expression for the calculation of density is the ratio between the mass of a body and the volume it occupies:

d=\frac{m}{V}

Density, according to the International System of Units, is usually expressed in kilograms per cubic meter (kg / m3) or gram per cubic centimeter (g / cm3). Although it can also be expressed in any other unit of mass by volume.

So, for the calculation of density you know that:

  • mass of the sample= 10 grams
  • volume of the sample = volume on the cylinder after immersion - volume on the cylinder before immersion= 38.7 mL - 35.0 ml = 3.7 mL

Then:

d=\frac{10 grams}{3.7 mL}

d=2.70 \frac{grams}{mL}

<u><em>The density of the metal sample is 2.70 </em></u>\frac{grams}{mL}<u><em></em></u>

5 0
3 years ago
A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass
valkas [14]

Answer:

\large \boxed{\text{0.0035 mol/L}}

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

                   Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

    V/mL:   3.00      7.00

c/mol·L⁻¹:  0.050   0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant  

\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}

(ii) From NH₃

\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}

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Answer:

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Explanation:

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