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DerKrebs [107]
3 years ago
12

What do the objects in motion have in common ​

Chemistry
1 answer:
Alona [7]3 years ago
6 0

Answer:

All objects in motion do posses kinetic energy.

Explanation:

You might be interested in
1. what is volume?
gavmur [86]
1) D) The amount of space a substance takes up 
2) A) The amount of matter in a substance 
3 0
4 years ago
Read 2 more answers
Starting with 195 g Li2O and 106 g H2O, decide which reactant is present in limiting quantities. Given: Li2O+H2O→2LiOH
kkurt [141]

Taking into account the stoicionetry reaction and the definition of limiting reagent, Li₂O is present in limiting quantities.

The balanced reaction is:

Li₂O + H₂O → 2 LiOH

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Li₂O: 1 mole
  • H₂O: 1 mole
  • LiOH: 2 moles

The molar mass of each compound is:

  • Li₂O: 29.88 g/mole
  • H₂O: 18 g/mole
  • LiOH: 23.95 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Li₂O: 1 mole× 29.88 g/mole= 29.88 g
  • H₂O: 1 mole× 18 g/mole= 18 g
  • LiOH: 2 moles× 23.95 g/mole= 47.9 g

On the other side, the limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 18 grams of H₂O reacts with 29.88 grams of Li₂O, 106 grams of H₂O react with how much mass of Li₂O?

mass of Li_{2} O=\frac{29.88 grams of Li_{2} Ox106 grams of H_{2} O}{18 grams of H_{2} O}

mass of Li₂O= 175.96 grams

But 175.96 grams of Li₂O are not available, 195 grams are available. Since you have less moles than you need to react with 106 grams of H₂O, Li₂O will be the limiting reagent.

Finally, Li₂O is present in limiting quantities.

Learn more:

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  • <u>brainly.com/question/11848702?referrer=searchResults</u>

4 0
3 years ago
Convert 0.30 m to mm.
scoundrel [369]

Answer:

0.3 Meters = 300 Millimeters

Explanation:

Multiply the length value by 1000

Hopefully, this helps! :D

7 0
3 years ago
How are sublimation in evaporation similar
diamong [38]

Evaporation is when a liquid turns into a gas, while sublimation is when a solid turns into a gas. Both involve absorbing energy to break apart the bonds of the substance. On the other hand, an example of sublimation is dry ice.

3 0
3 years ago
Read 2 more answers
The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most rem
son4ous [18]

Answer:

10 kg Al(OH)₃

Explanation:

There is some info missing. I think this is the original question.

<em>The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th-century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today. </em>

<em>In the first step, aluminum hydroxide reacts to form alumina (Al₂O₃) and water: 2 Al(OH)₃(s) → Al₂O₃(s) + 3H₂O(g). In the second step, alumina (Al₂O₃ and carbon react to form aluminum and carbon dioxide: 2Al₂O₃(s)+3C(s)→4Al(s)+3CO₂(g). Suppose the yield of the first step is 63% and the yield of the second step is 89%. </em>

<em>Calculate the mass of aluminum hydroxide required to make 2.0 kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.</em>

<em />

Let's consider the 2 steps in the synthesis of Al.

Step 1: 2 Al(OH)₃(s) → Al₂O₃(s) + 3 H₂O(g)

Step 2: 2 Al₂O₃(s) + 3 C(s) → 4 Al(s) + 3 CO₂(g)

In Step 2, the percent yield of Al is 89% and the real yield is 2.0 kg. The theoretical yield is:

2.0 kg (R) × (100 kg (T) / 89 kg (R)) = 2.2 kg = 2.2 × 10³ g

In Step 2, the mass of Al is 4 × 26.98 g = 107.9 g and the mass of Al₂O₃ is 2 × 101.96 g = 203.92g. The mass of Al₂O₃ that produced 2.2 × 10³ g of Al is:

2.2 × 10³ g Al × (203.92g Al₂O₃ / 107.9 g Al) = 4.2 × 10³ g Al₂O₃

In Step 1, the percent yield of Al₂O₃ is 63% and the real yield is 4.2 × 10³ g. The theoretical yield is:

4.2 × 10³ g (R) × (100 g (T)/ 63 g (R)) = 6.7 × 10³ g

In Step 1, the mass of Al₂O₃ is 101.96 g and the mass of Al(OH)₃ is 2 × 78.00 g = 156.0 g. The mass of Al(OH)₃ that produced 6.7 × 10³ g of Al₂O₃ is:

6.7 × 10³ g Al₂O₃ × (156.0 g Al(OH)₃ / 101.96 g Al₂O₃) = 1.0 × 10⁴ g Al(OH)₃ = 10 kg Al(OH)₃

7 0
4 years ago
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