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2 years ago

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How many moles of nitrogen gas would be produced if 4.92 moles of copper(II) oxide were reacted with excess ammonia in the follo

wing chemical reaction? 2 NH₃(g) + 3 CuO (s) → 3 Cu(s) + N₂(g) + 3 H₂O(g)\

1 answer:

WITCHER [35]2 years ago

3 0

Answer:

4.92/3

Explanation:

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A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)

miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s) E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻ E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0

3 years ago

Cuantos gramos necesito para preparar una solución de 500 ml al 35% de NaCl

Iteru [2.4K]

A patient who is prescribed a dose inhaler will find that it must be filled with a) medicine in powder form only. Works with lower (not upper) respiratory diseases only. Full of medicine used to give a fixed amount of medicine per oral inhalation. d) Medication in the form of a spray only.

6 0

3 years ago

If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?

Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 412.072 \ g \ Cl_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

412.072 g Cl₂ ≈ 412 g Cl₂

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3 years ago

The atomic number is the number of _____ an atom

ololo11 [35]

The Rutherford–Bohr model of the hydrogen atom (Z = 1) or a hydrogen-like ion (Z > 1). In this model it is an essential feature that the photon energy (or frequency) of the electromagnetic radiation emitted (shown) when an electron jumps from one orbital to another, be proportional to the mathematical square of atomic charge (Z2). Experimental measurement by Henry Moseley of this radiation for many elements (from Z = 13 to 92) showed the results as predicted by Bohr. Both the concept of atomic number and the Bohr model were thereby given scientific credence. The atomic number is the number of _z_ an atom.

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3 years ago

Read 2 more answers

The gasses in a hair spray can are at temperature 300k and a pressure of 30 atm, it

Sergeeva-Olga [200]

Answer:

900 K

Explanation:

Recall the ideal gas law:

$\displaystyle PV = nRT$

Because only pressure and temperature is changing, we can rearrange the equation as follows:
$\displaystyle \frac{P}{T} = \frac{nR}{V}$

The right-hand side stays constant. Therefore:

$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$

The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.

Substitute and solve for <em>T</em>₂:

$\displaystyle \begin{aligned} \frac{(30\text{ atm})}{(300\text{ K})} & = \frac{(90\text{ atm})}{T_2} \\ \\ T_2 & = 900\text{ K}\end{aligned}$

Hence, the temperature must be reach 900 K.

7 0

2 years ago