Explanation:
im sorry but it depends on what the diagram looks like for the problem
Answer:
Explanation:
From the statement of the problem,
B₂S₃
+ H₂O
→ H₃BO₃
+ H₂S
B₂S₃ + H₂O → H₃BO₃ + H₂S
We that the above expression does not conform with the law of conservation of mass:
To obey the law, we need to derive a balanced reaction equation:
Let us use the mathematical method to obtain a balanced equation.
let the balanced equation be:
aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S
where a, b, c and d will make the equation balanced.
Conservating B: 2a = c
S: 3a = d
H: 2b = 3c + 2d
O: b = 3c
if a = 1,
c = 2,
b = 6,
2d = 2(6) - 3(2) = 6, d = 3
Now we can input this into our equation:
B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S
B₂S₃
+ 6H₂O
→ 2H₃BO₃
+ 3H₂S
Answer:
avogadro's constant
Explanation:
this is the fixed number of the atoms in the molecule of an element
avogadro's law states that equal volumes of gases<em> </em><em>at</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em> </em><em>contain</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>
<em>that</em><em> </em><em>is</em><em> </em><em>all</em><em> </em><em>gases</em><em> </em><em>with</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em>will</em><em> </em><em>always</em><em> </em><em>have</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>
Answer:
The answer is 2i on right hand side.
Explanation:
We should star by checking the equation from right.
First we check how many Zn r there in left hand side. Which is 1. Let us check how many Znr there in right hand side, there is 1.So Zn is balanced, and don't worry about Znplus2 on right hand side it is just the ions not how many zinc r there.
Now let us check how many I are there left hand side. Which is 2. Now how many I are there in right hand side? Only 1.
So we put 2 behind I.
Now there r 2 I on both sides.
Its simple actually.
1. 1 M , 2 M , 1 M
2. 10 mol , 0.1 mol , 0.5 mol
3. 0.5 L , 6.6 L , 5/21 L
M = mol/L