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3 years ago

In the following reaction, which species is reduced? Au(s) + 3NO3-(aq) + 6H+(aq) → Au3+(aq) + NO(g) + 3H2O (l) H+ N+5 O2- H2O Au

Chemistry

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

NO3-

Explanation:

Given the reaction equation;

Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).

We can consider the oxidation states of species on the left and right hand sides of the reaction equation;

Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.

NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.

H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.

Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.

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Give the quantum number set for one electron in the 3p sub level of a sulfur (S)

vesna_86 [32]

Answer:- Atomic number for sulfur is 16 and it's electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^4$ . Here, there are total for electrons in 3p and the set of quantum numbers for these 4 electrons would be as..

For the first electron of 3p-

n = 3, l = 1, ml = -1 and ms = +(1/2)

for the second electron of 3p-

n = 3, l = 1, ml = 0 and ms = +(1/2)

for the third electron of 3p-

n = 3, l = 1, ml = +1 and ms = +(1/2)

and for the fourth electron of 3p-

n = 3, l = 1, ml = -1 and ms = -(1/2)

3 0

3 years ago

Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

5 0

3 years ago

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of copper (II) chloride.

$M_2\text{ and }V_2$ are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

$M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?$

Putting values in above equation, we get:

$0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL$

Hence, the volume of stock solution needed are, 12.5 mL

8 0

3 years ago

How many liters of CO2 are formed by burning 7 x 1025 molecules of C2H4 with excess oxygen?

S_A_V [24]

Answer:

ESTA!!!!1

Explanation:

5 0

3 years ago

05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises

OLga [1]

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

6 0

3 years ago