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3 years ago

The oxidation numbers of nitrogen in nh3, hno3, and no2 are, respectively:

Chemistry

2 answers:

vovikov84 [41]3 years ago

3 0

The oxidation numbers for Nitrogen are respectively -3, +5, +4

julia-pushkina [17]3 years ago

3 0

Answer:

The oxidation number are

NH₃: -3

HNO₃ : +5

NO₂ : +4

Explanation:

The oxidation number is calculated considering that

a) oxidation number of hydrogen is +1 in all compounds except hydrides

b) oxidation number of oxygen is -2 in all compound except peroxides, superoxides and compound of fluorine.

a) NH₃ : let the oxidation number of nitrogen is "x"

x + 3 (+1) = 0

Therefore x = -3

b) HNO₃

Let the oxidation number of nitrogen is "x"

+1 + x +3(-2) = 0

x = -5.

c) NO₂

Let oxidation number of nitrogen ix "x"

x + 2(-2)= 0

x = +4

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How many moles of CO gas are in 34.6 L?

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moles of CO gas : 1.545

<h3>Further explanation</h3>

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

There are 2 conditions that are usually used as a reference : STP and RTP

Assuming the STP state :

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then for 34.6 L of CO gas :

$\tt moles=\dfrac{34.6}{22.4}=1.545$

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2 years ago

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2 years ago

Nicotine is a toxic substance present in tobacco leaves. There are two lone pairs in the structure of nicotine. In general, loca

Harlamova29_29 [7]

Answer:

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8 0

3 years ago

A sample of a vapor occupies a volume of 500 mL at 65°C. If pressure remains constant, what is the volume of the gas at standard

DedPeter [7]

Answer:

403 mL

Explanation:

First, I will assume that the mole is 1, because you are not specifing this.

Now, with the innitial data, we need to get the pressure:

T = 65+273 = 338 K

V = 500 / 1000 = 0.5 L

Now if:

PV = nRT

Then:

P = nRT/V and V = nRT/P

Let's calculate the P:

P = 1 * 0.082 * 338 / 0.5 = 55.432 atm

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8 0

3 years ago

Consider the balanced equation for the following reaction:

Bad White [126]

<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

$9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g$

To calculate the theoretical yield of lithium chlorate, we use the equation:

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

$81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g$

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0

3 years ago