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12345 [234]
3 years ago
9

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

superoxide.
(a)H3PO4
(b)Al(OH)3
(c)SeO2
(d)KNO2
(e)In2S3
(f)P4O6

Chemistry
1 answer:
natka813 [3]3 years ago
5 0

Answer:

Explanation:

You need to remember that the oxidation number of H is +1, except when it is in a metal hydrites like NaH, where its oxidation number is -1. Then, the oxidation number of O is -2, but in peroxides is -1. So with these rules you just have to multiply the ox. number with the name of atoms and all the elements in the reaction must sum 0.

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When 0.100 mol of carbon is burned in a closed vessel with8.00
antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of O_2 = 8.00 g

Molar mass of O_2 = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of O_2.

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C+O_2\rightarrow 2CO

From the balanced reaction we conclude that

As, 2 mole of C react with 1 mole of O_2

So, 0.1 moles of C react with \frac{0.1}{2}=0.05 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and C is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO

From the reaction, we conclude that

As, 2 mole of C react to give 2 mole of CO

So, 0.1 moles of C react to give 0.1 moles of CO

Now we have to calculate the mass of CO

\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO

\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g

Therefore, the mass of carbon monoxide form can be 2.8 grams.

5 0
3 years ago
Hey please answer this for me thanks.
Aleks [24]

Explanation:

Moles of N2 = 84/28 = 3.0mol

Moles of H2 = 29/2 = 14.5mol

Hence amount of ammonia produced = 6.0 * 17 = 102g.

7 0
3 years ago
What is the molar mass of carbon (C)?
stira [4]

The molar mass of carbon (C) is 12.0107 u.

7 0
3 years ago
Read 2 more answers
The number of protons in an atom is known as its atomic
Fynjy0 [20]
Atomic number should be the answer
4 0
3 years ago
A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
  • brainly.com/question/12625048
6 0
2 years ago
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