Following the key in the diagram (see the attached image), the only particle diagram that represents a mixture of three substances is diagram 2.
To simplify it, let us replace the key in the diagram as follows;
- atom of one element = A
- atom of different element = B
Diagram 1 consists of only AA and AB
Diagram 2 consists of AA, BB, and AB.
Diagram 3 consists of AA and ABA
Diagram 4 consists of AA and BAB
Thus, only diagram 2 has a mixture of 3 substances.
More on mixtures can be found here: brainly.com/question/6594631
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper
Answer:
products are copper and aluminium chloride
reactants are aluminium and copper chloride
Explanation:
products are formed at the end of a reaction
reactants are what's reacting to form the products ie the things at the start of the equation
Answer:
<h3>A</h3>
Explanation:
<h2>sorry try lang baka mali</h2><h2 />
