A combustion reaction is a reaction that reacts in the presence of oxygen molecules. Methane will release -3115 kJ/mol of heat.
<h3>What is a combustion reaction?</h3>
A combustion reaction includes the reaction between the chemical reactant and oxygen molecule to produce the product. The combustion reaction between methane and oxygen is given as:
CH₄(g) + 2O₂ (g) → CO₂(g) + 2H₂O (l), ΔH = -890 kJ/mol
The stoichiometry coefficient from the reaction gives 1 mole of methane releases -890 kJ/mol enthalpy.
So, 3.5 moles methane will release = 3.5 × -890 = -3115 kJ/mol
Therefore, -3115 kJ/mol of heat is released.
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D) 1 and 4
Chromosomes are made up of DNA. DNA strands contain short segments called genes.
The bottom of a cricket wing is covered with teeth-like ridges that make<span> it rough. The upper surface of the wing is like a scraper. When </span>crickets<span> rub the upper and lower parts of their wings together, they create a chirping</span>sound<span> called “stridulating."</span>
Therefore option c , i.e. The substances in both test tubes are reactive only at high temperatures. is the only statement which is NOT supported by the student's observations.
<h3>What is the reaction between Magnesium and Hydrogen ?</h3>
Magnesium reacts with hydrochloric acid to produce hydrogen gas
Mg (s) + 2 HCl (aq) → MgCl₂ (aq) + H₂ (g)
In this reaction, the magnesium and acid are gradually used up , which can be seen in the test tube 2 .
A chemical reaction is taking place in Test tube 2 ,
Hydrogen gas is released in test tube 2 ,
Energy is released in the reaction involving hydrochloric acid and we can see in test tube 2 the reaction is going on
therefore option C i.e. The substances in both test tubes are reactive only at high temperatures. is the only statement which is NOT supported by the student's observation.
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The molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.
<h3>How to calculate molecular formula?</h3>
The molecular formula can be calculated from the empirical formula. The empirical formula of the compound is calculated as follows:
- C = 38.7% = 38.7g
- H = 9.76% = 9.76g
- O = 51.5% = 51.5g
Next, we convert the mass to moles by dividing by their atomic mass:
- C = 38.7 ÷ 12 = 3.23mol
- H = 9.76 ÷ 1 = 9.76mol
- O = 51.5÷ 16 = 3.22mol
Next, we divide by the smallest (3.22)
Hence, the empirical formula of the compound is CH3O
If the molar mass of the compound is 62g/mol;
(CH3O)n = 62
31n = 62
n = 2
(CH3O)2 = C2H6O2
Therefore, the molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.
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