Answer:
Mass of water produced is 22.86 g.
Explanation:
Given data:
Mass of hydrogen = 2.56 g
Mass of oxygen = 20.32 g
Mass of water = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 20.32 g/ 32 g/mol
Number of moles = 0.635 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 2.56 g/ 2 g/mol
Number of moles = 1.28 mol
Now we will compare the moles of water with oxygen and hydrogen.
O₂ : H₂O
1 : 2
0.635 ; 2×0.635 = 1.27
H₂ : H₂O
2 : 2
1.28 : 1.28
The number of moles of water produced by oxygen are less thus it will be limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 1.27 × 18 g/mol
Mass = 22.86 g
It is a scientific hypothesis. A scientific hypothesis must be testable, however there is a significantly more grounded necessity that a testable speculation must meet before it can truly be viewed as logical. This foundation comes essentially from crafted by the rationalist of science Karl Popper, and is called "falsifiability".
Answer: silicon Si, Germanium GE
Explanation:
The statement above is true. He conducted the oil-drop experiment which lead him to determine the charge of the electron. He suspended charged droplets into an oil which is in between two electrodes and balancing the upward force with the downward forces.
The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
