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svetoff [14.1K]
3 years ago
11

What is the amount of aluminum chloride produced from 40 moles of aluminum and excess chlorine?

Chemistry
1 answer:
Marrrta [24]3 years ago
6 0

Answer:

is it 26.98

Explanation:

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2. What is the mass of one mole of Na?
borishaifa [10]
Sodium, Atomic mass: 22.989769 g
You can see in a periodic table
5 0
3 years ago
Wuy Piolcululugu<br> Give the names of two organs in the chest.<br> 1. ...............<br> 2 m
Arada [10]

Answer:

The organs present inside the chest are :

1. The lungs

2. The heart

Explanation:

The chest cavity is also called as the thoracic cavity. It is the second largest hollow space of the body.In the bottom , it is enclosed by the diaphragm.

This cavity actually contain three space each round with mesothelium , pleural cavity and precardial cavity.

This contain the lungs , the tracheobronchial tree , the heart , the blood vessels which transport the blood between the heart and the lungs.

It also contain the esophagus .

Esophagus is the path through which the food passes from the mouth to the stomach.

8 0
3 years ago
..can someone answer these questions ive already posted this question 16 times and I dont have much points left so please anyone
noname [10]

Answer

  1. A
  2. D
  3. D
  4. A
  5. B
  6. A

This is what I got, but i'm mot sure if I'm right

7 0
3 years ago
Read 2 more answers
Which two energy sources were used the most in
Naily [24]

Answer:

The 1st and 2nd ones on the top

Explanation:

Hope this helps:)

4 0
3 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
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