Which method would be the best to use in order to separate a mixture of water and rubbing alcohol?

Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrog

vitfil [10]

Answer:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

Explanation:

To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.

Nitrite ion (NO₂⁻)

1 × N + 2 × O = -1

1 × N + 2 × (-2) = -1

N = +3

Nitrous oxide (NO)

1 × N + 1 × O = 0

1 × N + 1 × (-2) = 0

N = +2

Nitrate ion (NO₃⁻)

1 × N + 3 × O = -1

1 × N + 3 × (-2) = -1

N = +5

Ammonia (NH₃)

1 × N + 3 × H = 0

1 × N + 3 × (+1) = 0

N = -3

Nitrogen gas (N₂)

2 × N = 0

N = 0

The order of increasing nitrogen oxidation state is:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

7 0

4 years ago

During the water cycle, liquid water from bodies, such as oceans, lakes, and streams, evaporates and moves upward in the atmosph

Rufina [12.5K]

Property of water?? I have no idea

4 0

4 years ago

Read 2 more answers

Suppose you increase your walking speed from 7 m/s to 13 m/s in a period of 1 s. what is your acceleration?

GREYUIT [131]

A = change in velocity / time
so change in velocity = 13-7 = 6
so 6/1 =6:
acceleration = 6m/s^2

7 0

3 years ago

The early ancestors of the horse (shown in the picture) were adapted to life in tropical forests. Gradually the forests disappea

Gre4nikov [31]

So here you go are taller with long legs and are capable of great speed to escape predators

8 0

3 years ago

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A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i

Sergeeva-Olga [200]

Answer:

$m_{I^-}=1.18gI^-$

Explanation:

Hello,

In this case, the reaction is given as:

$I^-+AgNO_3\rightarrow AgI+NO_3^-$

Thus, starting by the yielded grams of silver iodide, we obtain:

$n_{I^-}=2.185gAgI*\frac{1molAgI}{234.77gAgI}*\frac{1molI}{1molAgI}=9.31x10^{-3}molI^-$

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

$m_{I^-}=9.31x10^{-3}molI^-\frac{127gI^-}{1molI^-} =1.18gI^-$

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.

Best regards.

7 0

4 years ago