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aleksandr82 [10.1K]
3 years ago
5

10. Identify the products in the combustion of methane.

Chemistry
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

a. CO2 and H20

Explanation:

Chemically, this combustion process consists of a reaction between methane and oxygen in the air. When this reaction takes place, the result is carbon dioxide (CO2), water (H2O), and a great deal of energy. The following reaction represents the combustion of methane:

CH4[g] + 2 O2[g] -> CO2[g] + 2 H2O[g] + energy

One molecule of methane, (the [g] referred to above means it is gaseous form), combined with two oxygen molecules, react to form a carbon dioxide molecule, and two water molecules usually given off as steam or water vapor during the reaction and energy.

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3241004551 [841]

Answer:

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5 0
3 years ago
A gas occupies a volume of 85.0 liters at a pressure of 2.24 atm and a temperature of 22.5 degrees celsius. How many moles of ga
White raven [17]

Answer:

n = 7.86 mol

Explanation:

This question can be solved using the ideal gas law of PV = nRT.

Temperature must be in K, so we will convert 22.5C to 295 K ( Kelvin = C + 273).

R is the ideal gas constant of 0.0821.

(2.24atm)(85.0L) = n(0.0821)(295K)

Isolate n to get:

n = (2.24atm)(85.0L)/(0.0821)(295K)

n = 7.86 mol

8 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
How many atoms is in 1 molacule of uranium?
Minchanka [31]

Answer:

One mole of U238 ( 6.022 X 10^23 atoms) is 238 grams; one mole of U235 is 235 grams. The difference in molecular weight between a mole of U235 and U238 is 3 grams, the secret to enriching uranium.

Explanation:

4 0
3 years ago
What is the melting point of a solution in which 2.5 grams of sodium chloride is added to 230 mL of water?
Lemur [1.5K]
To get the melting point of a solution so, we will use this formula:

ΔT = - mKf

when:

m is molality of the solution

Kf is cryoscopic constant  of water  = 1.86 C/m

and ΔT is the change in melting point (T2 - 0 °C)

so, now we need to calculate the molality to substitute:

when the molality = moles NaCl / Kg of water

 and when moles NaCl  = mass / molar mass

                                       = 2.5 g / 58 g/mol

                                       = 0.043 mol
∴ Kg water = volume *density /1000

                   = 230 mL * 1 g/mL / 1000

                  = 0.23 Kg
∴ molality = 0.043 / 0.23 =0.187 M

by substitution:

T2-0°C = - 0.187 * 1.86

∴T2 = - 0.348 °C

5 0
3 years ago
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