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3 years ago

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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zubka84 [21]

Multiple or divide them

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3 years ago

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How would Lithium gain full valence levels

Lubov Fominskaja [6]

Answer:

In order to have a full outer shell atoms can GAIN or LOSE electrons. Atoms with very few VALENCE electrons, such as Lithium, would need to gain many electrons for a full shell so it is easier for them to LOSE electrons and become POSITIVE.

Explanation:

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2 years ago

The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of

attashe74 [19]

Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

6 0

2 years ago

The concentration of the stock solution she needs is 100 milli molar (mm) and she needs to make 1.2 milli liters (ml) solution o

Lera25 [3.4K]

Molarity is defined as the number of moles of solute in 1 L of solution
molarity of stock solution to be prepared - 100 x 10⁻³ mol/L
volume of stock solution to be prepared - 1.2 mL
Therefore number of moles in 1.2 mL - 100 x 10⁻³ mol/L x 1.2 x 10⁻³ L
number of moles of drug - 1.2 x 10⁻⁴ mol
mass of drug required - 1.2 x 10⁻⁴ mol x 181.6 g/mol = 21. 8 mg
21.8 g of drug is required to make the stock solution

6 0

2 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

Anastasy [175]

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

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3 years ago