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2 years ago

Is Cl2 an electrophile? Why/why not?

Chemistry

1 answer:

Triss [41]2 years ago

6 0

Answer:

electrophile is pe do file

Explanation:

elecricity = needing a professor= creepy profesor= pe edo

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Which particle is most likely present in the nucleus of the atom

k0ka [10]

Oh protons and neutrons

8 0

3 years 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 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

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

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

Don't know how to solve

faltersainse [42]

Happy Valentines Day! <3

8 0

3 years ago

What elements are found in the compound represented in the diagram

DedPeter [7]

hydrogen and carbon, hope that helped

3 0

4 years ago

An ideal gas in a 1.25-gallon container is at a temperature of 125 degrees Celsius and pressure of 2.5 atmospheres. If the gas i

Tresset [83]

You will need the equation PV = nRT

P = Pressure in kPa
V = Volume in L
n = moles
R = 8.314 (constant)
T = Temperature in Kelvin

First convert 2.5 atm into kPa:

2.5 X 101.3 = 253.25 kPa

Convert 125 Celsius into Kelvin:

125 + 273 = 398 K

Convert Gallons to Litres:

1.25 X 3.79 = 4.74 L

Plug your values into the equation to solve for n:

(253.25)(4.74) = n(8.314)(398)

n = (253.25)(4.74)/(8.314)(398)

n = 0.362 moles

Now use M = m/n to solve for the mass of O2

M = Molar Mass
M = mass
n= moles

32 = m/(0.362)

m = (32)(0.362)

m = 11.58g

8 0

3 years ago