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Aleks04 [339]
3 years ago
10

What mass of Cu(s) would be produced if 0.40 mol of Cu2O(s) was reduced completely with excess H2(g) ?

Chemistry
1 answer:
yan [13]3 years ago
4 0

Answer:

50.8g

Explanation:

Given parameters:

Nubmer of mole of Cu₂O = 0.4mol

Unknown:

Mass of Cu produced = ?

Solution:

The balanced reaction equation:

           Cu₂O  +  H₂   ⇄ 2Cu + H₂O

From the reaction, we know that:

            1 mole of Cu₂O produced 2 moles of Cu

             0.4 mol of Cu₂O would give 0.8mol of Cu

Now we know the number of moles of Cu produced because the limiting reactant here is Cu₂O. It was used up in the reaction in which the hydrogen gas was in excess.

To find the mass of Cu produced, we use the equation below :

              Mass of Cu = Number of moles of Cu x molar mass of Cu

Molar mass of Cu = 63.5gmol⁻¹

              Mass of Cu = 0.8 x 63.5 = 50.8g

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Potassium and chlorine
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A pharmacist has a 12% solution of boric acid and a 20% solution of boric acid. How much of each must he use to make 80 grams of
Iteru [2.4K]

Answer:

  • <u><em>He must use 50g of the 12% solution and 30 g of the 20% solution.</em></u>

<u><em></em></u>

Explanation:

Call x the amount of <em>12% boric acid </em>solution to be used.

  • The content of acid of that is: 0.12x

Since he wants to make <em>80 grams</em> of solution, the amount of <em>20% boric acid</em> solution to be used is 80 - x.

  • The content of acid of that is{ 0.20(80 - x).

The final solution is <em>15% </em>concentrated.

  • The content of boric acid of that is 0.15 × 80 g.

Now you have can write your equation:

         0.12x+0.20(80-x)=0.15\times 80

Solve:

           0.12x + 16 - 0.20x=12\\\\-0.08x=12-16\\\\0.08x=4\\\\x=4/0.08\\\\x=50

That is 50 grams of the 12% solution of boric acid.

Calculate the amount of the 20% solution of boric acid:

         80-x=80-50=30

That is 30 grams.

Then, he must use 50g of the 12% solution and 30 g of the 20% solution.

5 0
3 years ago
The basic structure of an amino acid contains the amine group, NH2, and a carboxylic acid group separated by one or more carbon
OLEGan [10]

The answer would be True

3 0
3 years ago
The radius of the smaller circle is 8 feet. The distance from the rim of the inner circle to the rim of the outer circle is 3 fe
postnew [5]

Answer:

178.98 sq. feet

Explanation:

The path and the garden has been shown in the figure below. The green area is the garden and the area in brown is the path.

It has been given that,

Radius of garden = 8 feet

So, the area of garden = 3.14 × 8 × 8 = 200.96 sq. feet

The total radius of the land including garden and path = 8 + 3 = 11 feet

So, the total are of land including garden and path = 3.14 × 11 × 11 = 379.94 sq. feet

So, the area of path = Total area of the land - area of garden

Area of path = 379.94 - 200.96 = 178.98 sq. feet

5 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
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