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Zepler [3.9K]
3 years ago
9

What is a product for 4 reaction type?

Chemistry
1 answer:
Crank3 years ago
6 0

Answer:

1= 2H₂ + O₂ → 2H₂O

2=CaCo₃ + heat → CaO +CO₂

3=CH₄ + 2O₂   → CO₂ +2H₂O

4=HCl + NaOH   → NaCl + H₂O

Explanation:

1 = Simple composition

The formation of water molecule is simple composition reaction. In this reaction two hydrogen atoms react with one oxygen atom and form one water molecules.

2H₂ + O₂ → 2H₂O

The amount of energy released is -285.83 KJ/mol. It is exothermic reaction.

2 = Simple decomposition reaction:

The break down of sodium hydrogen carbonate into sodium carbonate, carbondioxide and water is decomposition reaction. The decomposition reactions re mostly endothermic, because compound required energy to break.

2NaHCO₃ + heat → Na₂CO₃ + H₂O + CO₂

It is endothermic reaction.

Another example is:

CaCo₃ + heat → CaO +CO₂

3 = Combustion reaction

Consider the combustion of methane:

CH₄ + 2O₂   → CO₂ +2H₂O

The burning of methane is exothermic. The combustion reactions are exothermic because when fuel are burns they gives energy.

4 = Neutralization reaction

The neutralization reactions are those in which acid and base react to form the salt and the water. Some neutralization reactions are exothermic because they release heat. e.g

Consider the neutralization reaction of HCl and NaOH.

HCl + NaOH   → NaCl + H₂O

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The molar mass of a certain gas is 49 g. What is the density of the gas in g/L at STP?
snow_tiger [21]

Answer:

\boxed{\text{2.2 g/L}}

Explanation:

We can use the Ideal Gas Law to calculate the density of the gas.

   pV = nRT

      n = m/M           Substitute for n

   pV = (m/M)RT     Multiply both sides by M

pVM = mRT            Divide both sides by V

  pM = (m/V) RT

     ρ = m/V             Substitute for m/V

 pM = ρRT              Divide each side by RT

\rho = \frac{pM }{RT}

Data:

p = 1.00 bar

M = 49 g/mol

R = 0.083 14 bar·L·K⁻¹mol⁻¹

T = 0 °C = 273.15 K

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ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L

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8 0
4 years ago
12.5 g of copper are reacted with an excess of chlorine gas, and 25.4 g of copper(II) chloride are
Alika [10]

Answer:

Percent yield = 94.5%

Theoretical yield =  26.89 g

Explanation:

Given data:

Mass of copper = 12.5 g

Mass of copper chloride produced = 25.4 g

Theoretical yield = ?

Percent yield = ?

Solution:

Cu + Cl₂  →  CuCl₂

Number of moles of Copper:

Number of moles = mass/ molar mass

Number of moles = 12.5 g/ 63.55 g/mol

Number of moles = 0.2 mol

Now we will compare the moles of copper with copper chloride.

          Cu          :           CuCl₂

           1             :              1

          0.2          :            0.2

Theoretical yield:

Mass of copper chloride:

Mass = Number of moles × molar mass

Mass = 0.2 mol × 134.45 g/mol

Mass = 26.89 g

Percent yield:

Percent yield = Actual yield / theoretical yield  × 100

Percent yield = 25.4 g/26.89 g × 100

Percent yield = 94.5%

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