What is the oxidation state of nitrogen in NaNO2?
(1) +1 (3) +3
(2) +2 (4) +4

Answers

Answer 1

Answer: The answer is (3) +3. The Na ion usually has oxidation state of +1. The O ion usually has oxidation state of -2. And the total of oxidation state needs to be zero. Then we can calculate the oxidation state of nitrogen.

Answer 2

Answer:

Final answer:

The oxidation state of nitrogen in NaNO2 is +3. This figure is calculated by balancing the charges in the molecule to zero, with sodium assigned +1 and oxygen assigned -2 as their oxidation states.

Explanation:

The oxidation state of an element in a compound refers to the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. For NaNO2, the sodium (Na) has an oxidation state of +1, and the oxygen (O) has an oxidation state of -2 each. The molecule is neutral, with charge of zero. Thus, if we let x represent the oxidation state of nitrogen (N), we set up the equation:

+1 + x + 2(-2) = 0

Solving for x gives us:

x = +1 - 2*(-2) = +3.

So, the oxidation state of nitrogen in NaNO2 is +3.

Learn more about Oxidation State here:

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At -18.6 °C, a common temperature for household freezers, what is the maximum mass of sucralose (C₁₂H₁₉Cl₃O₈) in grams you can add to 2.00 kg of pure water and still have the solution freeze? Assume that sucralose is a molecular solid and does not ionize when it dissolves in water.
$K_f$ = 1.86 °C/m.

Answers

Answer : The maximum mass of sucralose is, 7952.8 grams.

Explanation : Given,

Molal-freezing-point-depression constant $(K_f)$ for water = $1.86^oC/m$

Mass of water (solvent) = 2.00 kg

Molar mass of sucralose = 397.64 g/mole

Formula used :

$\Delta T_f=i* K_f* m\n\nT^o-T_s=i* K_f*\frac{\text{Mass of sucralose}}{\text{Molar mass of sucralose}* \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-18.6^oC$

$\Delta T^o$ = freezing point of water = $0^oC$

i = Van't Hoff factor = 1 (for sucralose non-electrolyte)

$K_f$ = freezing point constant for water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$(0-(-18.6)^oC)=1* (1.86^oC/m)* \frac{\text{Mass of sucralose}}{397.64g/mol* 2.00kg}$

$\text{Mass of sucralose}=7952.8g$

Therefore, the maximum mass of sucralose is, 7952.8 grams.

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Answers

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