Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
Explanation:
An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.
A Reductant thus exactly the opposite.
Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:
I2 --> 2I-.
The oxidation number reduced from 0 to -1.
In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.
Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
Actually when we burn metal salts, we cannot actually
really see distinct lines to appear because in reality, they are not really
visible to the human eye. There is only a certain range of wavelength of light that
our eyes can see.
To state in other way, some metal salts will give off light which has wavelengths that are outside of the visible region of the electromagnetic spectrum.<span> </span>
Answer:
<span>Formula for sodium hydroxide is represented by <u>NaOH</u>.
Explanation:
The reaction of Sodium oxide with water is as follow,
Na</span>₂O + H₂O → 2 NaOH
<span>
Na</span>₂O is considered as a strong basic oxide as it contains O²⁻ which has high tendency to bind with hydrogen atoms. This reaction is an exothermic reaction and is conducted in cold water to produce NaOH.<span>
</span>
Answer:
34,6g of (NH₄)₂SO₄
Explanation:
The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:
ΔT = kb×m
Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.
For the problem:
ΔT = 109,7°C-108,3°C = 1,4°C
kb = 1.07 °C kg/mol
Solving:
m = 1,31 mol/kg
As mass of X = 600g = 0,600kg:
1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:
0,785 moles of ions×
= 0,262 moles of (NH₄)₂SO₄
As molar mass of (NH₄)₂SO₄ is 132,14g/mol:
0,262 moles of (NH₄)₂SO₄×
= <em>34,6g of (NH₄)₂SO₄</em>
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I hope it helps!
Answer:
Explanation:
A neutral titanium atom will have 22 electrons. Therefore, its electron configuration will be 1s2 2s2 2p6 3s2 3p6 4s2 3d2.
