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4 years ago

How is carbon monoxide formed in a car engine?

1 answer:

3 0

Carbon monoxide is produced in an engine when incomplete combustion happened. In a combustion engine intake valve, it combines the fuel and air (nitrogen and oxygen) and when spark plug ignites it pushes the piston downwards. After that piston will push itself upward and pushes the air that was generated by the combination of those components in the exhaust valve. This is a repeated cycle which keeps the engine running. In a perfect combustion process the air being released are carbon dioxide, water and nitrogen. But when an incorrect ratio of the combination of fuel and air in the engine is processed, carbon monoxide is being generated.

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In a single replacement reaction between 2.57 moles Aluminum and 3.59 moles of Hydrochloric acid, how many moles of Hydrogen can

Alenkasestr [34]

Answer:

1.795 mole of H2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Al + 6HCl —> 2AlCl3 + 3H2

Step 2:

Determination of the limiting reactant.

From the balanced equation above,

2 moles of Al reacted with 6 moles

Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.

From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.

Step 3:

Determination of the number of mole H2 produced from the reaction.

Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.

The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:

From the balanced equation above,

6 moles of HCl reacted to produce 3 moles of H2.

Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.

From the calculations made above, 1.795 mole of H2 is produced from the reaction.

5 0

3 years ago

When the oxide of generic metal m is heated at 25.0 °c, only a negligible amount of m is produced. o2m(s ----> m(s o2(g delta

Len [333]

(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K = $e^{ \frac{-dG0}{RT} }$
= $e^{ \frac{105500}{8.314 x 298.15} }$
= 3.05 x 10¹⁸

6 0

3 years ago

What is the weight in grams of 3.36 × 10Æ³ molecules of copper sulfate (CuSOÁ)?

spin [16.1K]

Molar mass ( CuSO₄) = 159.609 g/mol

159.609 g ----------------- 6.02 x 10²³ molecules
? g ------------------ 3.36 x 10²³ molecules

mass = ( 3.36 x10²³) x 159.609 / 6.02 x 10²³

mass = 5.36 x 10²⁴ / 6.02 x 10²³

mass = 8.90 g

hope this helps!

6 0

3 years ago

the diagram below shows DNA molecules. What do the broken lines connected to hydrogen atoms represent in this diagram?

uranmaximum [27]

Answer is: hydrogen bonds.

Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.

According to the principle of base pairing hydrogen bonds could form between adenine and thymine (two hydrogen bonds between this nucleobases) and guanine and cytosine (three hydrogen bonds between this nucleobases).

Adenine and guanine are purine derivatives and thymine and cytosine are pyrimidine derivates.

5 0

3 years ago

Read 2 more answers

The essential oil found in cloves, eugenol, can be isolated by steam distillation because it is insoluble in water and has a mea

Damm [24]

We know that:
Molar Mass H2O: 18 g/mol
Molar Mass of Eugenol: 164 g/mol 
Boiling point of H2O: 100 degrees C 
Boiling point of Eugenol: 254 degrees C 
Density of water: 1.0 g/mL 
Density of Eugenol: 1.05 g/mL 

Using formula:
V= [mole fraction x molar mass] / density 

mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 
morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 

V% = Vorg/(Vorg + VH2O) * 100 
(0.8278/18.7324) * 100 = 4.419% 

Yotal volume = 30 mL; therefore,
0.0442 = (volume eugenol/30) 

(m eug/mH2O) = (peug*164/pH2O*18) 
(m eug/30) = (4*164/760*18) 
m eug = about 1.44g and 

volume = mass/density
= 1.44/1.05
= about 1.37 mL

6 0

3 years ago