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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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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0

4 years ago

10. Which term is same for one mole of O2 and one mole of ethane C2H6

melamori03 [73]

Answer:

b) Number of molecules

Explanation:

3 0

3 years ago

What changes happens when you run ?

oksano4ka [1.4K]

Answer:

It increases endurance, builds muscle to maintain an optimal body fat composition, promotes cardiovascular health, strengthens the heart, and even improves your overall mood. Combined with weight training, running can help you maintain the perfect balance of mental stimulation and physical fitness.

Explanation:

4 0

3 years ago

A sample of pure oxalic acid (H2C2O4.2H2O) weighs 0.2000 g and requires 30.12 ml of KOH solution for

ValentinkaMS [17]

The molarity of KOH is 0.1055 M

<u><em> calculation</em></u>

Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH

H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O

step 2: find the moles of H₂C₂O₄.2H₂O

moles = mass÷ molar mass

from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol

= 0.2000 g ÷ 126 g/mol =0.00159 moles

step 3: use the mole ratio to calculate the moles of KOH

H₂C₂O₄.2H₂O : KOH is 1:2

therefore the moles of KOH =0.00159 x 2 = 0.00318 moles

step 4: find molarity of KOH

molarity = moles/volume in liters

volume in liters = 30.12/1000=0.03012 L

molarity is therefore = 0.00318/0.03012 =0.1055 M

5 0

3 years ago

Calculate the number of atoms

ra1l [238]

D.....................

5 0

2 years ago