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pickupchik [31]
3 years ago
11

When 1.375 g of copper (II) oxide is reduced on heating in a current of hydrogen, the weight of copper remaining after the react

ion is complete is 1.098 g. What is the percent composition of oxygen?
Chemistry
1 answer:
nasty-shy [4]3 years ago
7 0

Answer:

20.11% of oxygen

Explanation:

The reaction is:

2CuO ⇒ 2Cu + O2

If you have an initial amoung of a sample, thi may be have impurities, so you have to start from products to know what you had at the begining.

to make a relation between reagents and products you need to change the mass to mol, then you can relate them with the molar ratio.

1,098gCu x \frac{1mol Cu}{63.546g Cu} x \frac{2mol CuO}{2molCu} x \frac{1mol O}{1molCuO} x \frac{16gO}{1molO} = 0,276gO

Now you have the initial amount of oxygen, to convert to percent you must divide the mass of oxygen between the mass of the sample and multiply by 100

percent composition of O = \frac{0.276gO}{1.375g sample} *100 = 20.11

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7. If the pOH of an RbOH solution is 6.32, what is the concentration (molarity) of the base?
liubo4ka [24]

4.79 x 10⁻⁷moldm⁻³

Explanation:

Given parameters:

    pOH of RbOH = 6.32

Unknown:

Molarity of the base = ?

Solution:        

    The pH or pOH scale is used for expressing the level of acidity alkalinity of aqueous solutions.

                      pOH = -log[OH⁻]

  we know the pOH to be 6.32

               6.32 = -log[OH⁻]

               [OH⁻]  = inverse log₁₀(6.32)

                [OH⁻] =  4.79 x 10⁻⁷moldm⁻³

Learn more:

pH brainly.com/question/12985875

#learnwithBrainly

6 0
3 years ago
How many significant figures are in $10,000,210<br> 7<br> 3
nevsk [136]

Answer:

7

Explanation:

7 0
3 years ago
Read 2 more answers
Name four products of incomplete combustion<br><br> One :<br> Two :<br> Three :<br> Four :
melamori03 [73]

Answer:

Carbon Monoxide / Carbon Dioxide / Sulfur and Nitrogen Dioxide

Explanation:

8 0
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What type of pressure would you expect to find at the equator?
vesna_86 [32]
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Explanation: Air at the equator is warmer and the air above expands, becomes less dense and rises
6 0
3 years ago
How do you work out question 1a?
Sliva [168]

Answer:

-125 kJ

Explanation:

You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.

                     H₂C=CH₂   +    H₂ ⟶    H₃C-CH₃

Bonds:       4C-H + 1C=C     1H-H     6C-H + 1C-C

D/kJ·mol⁻¹:  413       612        436       413      347

The formula relating ΔHrxn and bond dissociation energies (D) is

ΔHrxn = Σ(Dreactants) – Σ(Dproducts)

(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

<em>For the reactant</em>s:

Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ

<em>For the products:</em>

Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ

<em>For the system</em> :

ΔHrxn = 2700 - 2825 = -125 kJ

4 0
3 years ago
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