Any element can always be identified by its:

The combustion of carbohydrates and the combustion of fats are both exothermic processes, yet the combustion of carbohydrates is

Julli [10]

Explanation:

There are two aspects of a reaction:

Thermodynamics:

Combustion of the compound is reaction with oxygen.

The carbohydrates contain more content of oxygen as compared to fats.

Hence, carbohydrates have lot of oxygen content which are already partially oxidized as compared to fats. Hence, combustion of the carbohydrates is a faster process.

Kinetics:

The molecular reacts also shows that the combustion of the carbohydrates is a faster process.

7 0

3 years ago

Relate the strength of a weak acid to the strength of a conjugate base.

Elenna [48]

Answer:

There is a relationship between the strength of an acid (or base) and the strength of its conjugate base (or conjugate acid): The stronger the acid, the weaker its conjugate base. The weaker the acid, the stronger its conjugate base. The stronger the base, the weaker its conjugate acid.

explanation

The strength of an acid and a base is determined by how completely they dissociate in water. Strong acids (like stomach acid) break down or dissociate in water. Weak acids maintains their protons in water.

4 0

3 years ago

It takes 338. kJ/mom to break a carbon-chlorine single bond. Calculate the maximum wavelength of light for which a carbon-chlori

Vesnalui [34]

Answer: The maximum wavelength of light for which a carbon-chlorine Single bond could be broken by absorbing a single photon is 354 nm

Explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

$E=\frac{Nhc}{\lambda}$

where,

E = energy of the light = $338kJ=338000J$ (1kJ=1000J)

N= avogadro's number

h = Planck's constant

c = speed of light

$\lambda$ = wavelength of light

$338\times 1000J=\frac{6.023\times 10^{23}\times 6.6\times 10^{-34}\times 3\times 10^{8}}{\lambda}$

${\lambda}=0.354\times 10^{-6}m=354nm$ $(1nm=10^{-9}m$

Thus the maximum wavelength is 354 nm

6 0

3 years ago

I NEED HELP ASAP!!!!!!!!!!!!!!!!!!!

Hoochie [10]

Answer:

a) 0

Explanation:

Each of the small dots surrounding the C1 represents one electron. These are where electrivity comes from. Since there is the same number of electrons in both atoms, the difference is 0 (because 6 electrons-6electrons= 0).

4 0

1 year ago

Read 2 more answers

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

5.

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago