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3 years ago

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

1 answer:

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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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

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3 years ago

A cylinder of compressed gas has a volume of 350 ml and a pressure of 931 torr. What volume in Liters would the gas occupy if al

hodyreva [135]

Answer:

$0.384\ \text{L}$

Explanation:

$P_1$ = Initial pressure = 931 torr = $931\times \dfrac{101.325}{760}=124.12\ \text{kPa}$

$P_2$ = Final pressure = 113 kPa

$V_1$ = Initial volume = 350 mL

$V_2$ = Final volume

From the Boyle's law we have

$P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{124.12\times 350}{113}\\\Rightarrow V_2=384.44\ \text{mL}=0.384\ \text{L}$

The volume the gas would occupy is $0.384\ \text{L}$ .

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