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pantera1 [17]
4 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]4 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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The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh
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When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ

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150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

150.0 mL \times \frac{1.02g}{mL}  = 153 g

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Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J  }{\frac{4.184J}{g.\° C }  \times 153g} = 5.87 \° C

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

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