It indicates the number of moles of reactants and products
Explanation:
The coefficients in front of the reactants and products in a chemical reaction represents the number of moles of reactants and products.
Every reaction is made up of equal number of moles of reactants and products. Thus, chemical equations are written in such a way to obey the law of conservation of matter.
The numbers used are usually whole numbers and the are very important in stoichiometry.
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Answer:
(C) through the atmosphere
Explanation:
Butyl Alcohol on reaction with Oxygen produces Butyric Acid and Water.
The Balance Chemical equation is as follow,
C₄H₉OH + O₂ → C₄H₇O₂H + H₂O
Result: The equation given in statement is already balanced.
Answer: B. It’s a dilute strong base.
Explanation:
1) Definition of acids and bases: as per Bronsted-Lowry model, an acid is a substance that donates hydrogen ions and a base is a substance that accepts hydrogen ions.
Ca(OH)₂ does not have hydrogen ions to donate, but it can accept hydrogen ions to form H₂O according to this equation: H⁺ + OH⁻ → H₂O.
Hence, Ca(OH)₂ is a base.
2) Definition of strong base: a strong base is a base that dissociates completely into metallic and hydroxide ions in aqueous solutions, while a weak base dissociates partially.
Hence, Ca(OH)₂ is a strong base.
3) Definition of dilute: it refers to a solution meaning that the substance is not pure and the concentration is low. Since, the solution the Ca(OH)₂ is 0.02 M means that it is dilute.
Therefore, we have found that the description of 0.02 M Ca(OH)₂ is that is is a dilute strong base (option B).
Answer:
36.8 L
Explanation:
We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 80 °C
T(K) = 80 + 273
T(K) = 353 K
Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:
Number of mole (n) = 1.27 moles
Temperature (T) = 353 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 1.27 × 0.0821 × 353
V = 36.8 L
Thus, the volume occupied by the helium gas is 36.8 L
