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Kay [80]
3 years ago
8

The oxidation of copper(I) oxide, Cu 2 O ( s ) , to copper(II) oxide, CuO ( s ) , is an exothermic process. 2 Cu 2 O ( s ) + O 2

( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol Calculate the energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure.
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

The energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure is -10.142 kJ

Explanation:

Here we have

2Cu₂O ( s ) + O₂ ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol

In the above reaction, 2 Moles of Cu₂O (copper (I) oxide) react with one mole of O₂ to produce 4 moles of CuO, with the release of − 292.0 kJ/mol of energy

Therefore,

1 Moles of Cu₂O (copper (I) oxide) react with 0.5 mole of O₂ to produce 2 moles of CuO, with the release of − 146.0 kJ  of energy

We have 9.94 g of Cu₂O with molar mass given as 143.09 g/mol

Hence the number of moles in 9.94 g of Cu₂O is given as

9.94/143.09 = 6.95 × 10⁻² moles of Cu₂O

6.95 × 10⁻² moles of Cu₂O will therefore produce 6.95 × 10⁻² ×  − 146.0 kJ mol  or -10.142 kJ.

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leva [86]
<h3>Answer:</h3>

0.819 mol Ag

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})
  2. Divide:                              \displaystyle 0.818665 \ mol \ Ag

<u>Step 4: Check</u>

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

0.818665 mol Ag ≈ 0.819 mol Ag

8 0
2 years ago
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3 years ago
would a cloud or a rock be more dense? PLSSSSSS HELP PLSSS I WILL MARK BRAINLIEST AND ANSWER SOME OF YOUR QUESTIONS
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A rock is definitely more dense. If you were to put a cloud in water it would float/stay above it and a rock would sink to the bottom
6 0
3 years ago
determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass
Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin  is  calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

K_{f}or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

             ΔT_{f} = i ×K_{f} ×m

T⁰-T_{s}  = i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water     in     kg}

in which,

ΔT_{f} = change in freezing point

ΔT_{s} = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T_{s} = 1 ×1.86°C/m ×\frac{30.7}{92.09g/mol * 0.376kg}

T_{s} = -1.65°C

Thus, the freezing point depression of a solution is  -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

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8 0
10 months ago
Predict what will be observed in each experiment below. experiment predicted observation.
irga5000 [103]

Answer:

The correct answer is -

1.  a) The bubbles will shrink, some may vanish.

2. a) Can A will make a louder and stronger fizz than can B.

Explanation:

In the first question, it is given that the bottle is not opened and therefore, squeezing the bottle filled with a carbonated drink will increase the pressure on the carbonated liquid which forces the bubbles to dissolve or displace or vanish as it moves to empty space.

Thus, the correct answer would be - The bubbles will shrink, some may vanish

In the second question, there are two different conditions for two different unopened cans of carbonated water that are different temperatures one at the garage with higher temperature and one in the fridge at low temperature. As it is known that higher the temperature less will be solubility of gas in liquid so gas in can A will be less soluble which means it has more gas and it will make louder and stronger fizz than B which was stored at low temperature.

thus, the correct answer would be - Can A will make a louder and stronger fizz than can B.

8 0
2 years ago
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