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pantera1 [17]
3 years ago
6

What is the oxidation state of an individual nitrogen atom in nh4oh ? express the oxidation state numerically (e.g., +1)?

Chemistry
1 answer:
Nonamiya [84]3 years ago
6 0
In NH4OH, the compounds that make it up are NH4+ and OH-
Therefore N exists in the ammonium form.
In the ammonium ion 4H atoms are connected to N. 
 N is more electronegative than H, therefore when H bonds to N, H is the more positive atom therefore each H has a charge of +1, since there are 4 H atoms the charge contributed by the 4H atoms are +1 * 4 = +4
the overall charge of NH4 is +1
Charge of N (+) +4 =  +1
Charge of N = +1 - 4
Therefore oxidation state of N in NH4 is  = -3
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Explanation:

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where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

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Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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3 years ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
3 years ago
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